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Chapter 7 – Solutions

Chapter 7: Solutions And Solution Stoichiometry

7.1 Introduction:

Recall from Chapter 1 that solutions are defined as homogeneous mixtures that are mixed so thoroughly that neither component can be observed independently of the other. Solutions are all around us. Air, for example, is a solution. If you live near a lake, a river, or an ocean, that body of water is not pure H2O but most probably a solution. Much of what we drink—for example, soda, coffee, tea, and milk are solutions. Solutions are a large part of everyday life. A lot of the chemistry occurring around us happens in solution. In fact, much of the chemistry that occurs in our own bodies takes place in solution, and many solutions—such as the Ringer’s lactate IV solution—are important in healthcare. In our understanding of chemistry, we need to understand a little bit about solutions. In this chapter, you will learn about the special characteristics of solutions, how solutions are characterized, and some of their properties.

Skills to Develop

  • Define these terms: solution, solute, and solvent.
  • Distinguish solutions, mixtures, and colloids.
  • Describe various types of solutions.
  • Distinguish unsaturated, saturated, and supersaturated solutions.

The major component of the solution is called solvent, and the minor component(s) are called solute. If both components in a solution are 50%, the term solute can be assigned to either component. When a gaseous or solid material dissolves in a liquid, the gas or solid material is called the solute. When two liquids dissolve in each other, the major component is called the solvent and the minor component is called the solute.

Many chemical reactions are carried out in solutions, and solutions are also closely related to our everyday lives. The air we breathe, the liquids we drink, and the fluids in our body are all solutions. Furthermore, we are surrounded by solutions such as the air and waters (in rivers, lakes and oceans).

On the topic of solutions, we include the following sections.

  1. Types of Solutions: gaseous, liquid and solid solutions are based on the states of the solution.
  2. Solution Stoichiometry: expressing concentration in various units (mass per unit volume, moles per unit volume, percentage and fractions), reaction stoichiometry calculations involving solutions.
  3. Solutions of Electrolytes: solutions of acids, bases, and salts in which the solutes dissociate into positive and negative hydrated ions.
  4. Metathesis or Exchange Reactions: reaction of electrolytes leading to neutral molecules, gases, and solids.

Solving problems of solution stoichiometry requires the concepts introduced in stoichiometry in Chapter 6, which also provides the basis for the discussion on reactions.

7.2 Types of Solutions

In Chapter 1, you were introduced to the concept of a mixture, which is a substance that is composed of two or more substances. Recall that mixtures can be of two types: Homogeneous and Heterogeneous, where homogeneous mixtures combine so intimately that they are observed as a single substance, even though they are not. Heterogeneous mixtures, on the other hand, are non-uniform and have regions of the mixture that look different from other regions of the mixture. Homogeneous mixtures can be further broken down into two classifications: Colloids and Solutions. A colloid is a mixture that contains particles with diameters ranging from 2 to 500 nm. Colloids appear uniform in nature and have the same composition throughout but are cloudy or opaque. Milk is a good example of a colloid. True solutions have particle sizes of a typical ion or small molecule (~0.1 to 2 nm in diameter) and are transparent, although they may be colored. This chapter will focus on the characteristics of true solutions.

Material exists in three states: solid, liquid, and gas. Solutions also exist in all these states:

  1. Gaseous mixtures are usually homogeneous and are commonly gas-gas solutions. For quantitative treatment of this type of solutions, we will devote a unit to gases. The atmosphere is a gaseous solution that consists of nitrogen, oxygen, argon, carbon dioxide, water, methane, and some other minor components.  Some of these components, such as  water, oxygen, and carbon dioxide may vary in concentration in different locations on the Earth depending on factors such as temperature and  altitude.
  2. When molecules of gas, solid or liquid are dispersed and mixed with those of liquid, the homogeneous (uniform) states are called liquid solutions. Solids, liquids and gases dissolve in a liquid solvent to form liquid solutions.  In this chapter, most of the chemistry that we will discuss occurs in liquid solutions where water is the solvent.
  3. Many alloys, ceramics, and polymer blends are solid solutions. Within a certain range, copper and zinc dissolve in each other and harden to give solid solutions called brass. Silver, gold, and copper form many different alloys with unique colors and appearances. Alloys and other solid solutions are important in the world of materials chemistry.

7.2 Solubility

The maximum amount of a substance that can be dissolved in a given volume of solvent is called solubility. Often, the solubility in water is expressed in gram/100 mL. A solution that has not reached its maximum solubility is called an unsaturated solution. This means that more solute could still be added to the solvent and dissolving would still occur.

A solution that has reached the maximum solubility is called a saturated solution. If more solute is added at this point, it will not dissolve into the solution. Instead it will remain precipitated as a solid at the bottom of the solution.  Thus, one can often tell that a solution is saturated if extra solute is present (this can exist as another phase, such as gas, liquid, or solid). In a saturated solution there is no net change in the amount of solute dissolved, but the system is by no means static. In fact, the solute is constantly being dissolved and deposited at an equal rate. Such a phenomenon is called equilibrium. For example:

In special circumstances, a solution may be supersaturated. Supersaturated solutions are solutions that have dissolved solute beyond the normal saturation point. Usually a condition such as increased temperature or pressure is required to create a supersaturated solution. For example, sodium acetate has a very high solubility at 270 K.  When cooled, such a solution stays dissolved in what is called a meta-stable state. However, when a seeding crystal is added to the solution, the extra solute will rapidly solidify. During the crystallization process, heat is evolved, and the solution becomes warm. Common hand warmers use this chemical process to generate heat.

Video showing the crystallization of a supersaturated solution of sodium acetate. Video by : North Carolina School of Science and Mathematics


So how can we predict the solubility of a substance?

One useful classification of materials is polarity. As you read about covalent and ionic compounds in Chapters 3 and 4, you learned that ionic compounds have the highest polarity forming full cations and anions within each molecule as electrons are donated from one atom to another. You also learned that covalent bonds could be polar or nonpolar in nature depending on whether or not the atoms involved in the bond share the electrons unequally or equally, respectively. Recall that the electronegativity difference can be used to determine the polarity of a substance.  Typically an ionic bond has an electronegativity difference of 1.8 or above, whereas a polar covalent bond is between 0.4 to 1.8, and a nonpolar covalent bond is 0.4 or below.

Figure 7.1 Electronegativity Difference Diagram. The diagram above is a guide for discerning what type of bond forms between two different atoms. By taking the difference between the electronegativity values for each of the atoms involved in the bond, the bond type and polarity can be predicted. Note that full ionic character is rarely reached, however when metals and nonmetals form bonds, they are named using the rules for ionic bonding.


Substances with zero or low electronegativity difference such as H2, O2, N2, CH4, CCl4 are nonpolar compounds, whereas H2O, NH3, CH3OH, NO, CO, HCl, H2S, PH3 higher electronegativity difference are polar compounds. Typically compounds that have similar polarity are soluble in one another. This can be described by the rule:

Like Dissolves Like.

This means that substances must have similar intermolecular forces to form solutions. When a soluble solute is introduced into a solvent, the particles of solute can interact with the particles of solvent. In the case of a solid or liquid solute, the interactions between the solute particles and the solvent particles are so strong that the individual solute particles separate from each other and, surrounded by solvent molecules, enter the solution. (Gaseous solutes already have their constituent particles separated, but the concept of being surrounded by solvent particles still applies.) This process is called solvation and is illustrated in Figure 7.2. When the solvent is water, the word hydration, rather than solvation, is used.

In general polar solvents dissolve polar solutes whereas nonpolar solvents will dissolve nonpolar solutes. Overall, the solution process depends on the strength of the attraction between the solute particles and the solvent particles.  For example, water is a highly polar solvent that is capable of dissolving many ionic salts. Figure 7.2 shows the solution process, where water act as the solvent to dissolve the crystalline salt, sodium chloride (NaCl). Note that when ionic compounds dissolve in a solvent they break apart into free floating ions in solution. This enables the compound to interact with the solvent. In the case of water dissolving sodium chloride, the sodium ion is attracted to the partial negative charge of the oxygen atom in the water molecule, whereas the chloride ion is attracted to the partial positive hydrogen atoms.

Figure 7.2: The Process of Dissolving. When an ionic salt, such as sodium chloride, shown in (A), comes into contact with water, the water molecules dissociate the ion molecules of the sodium chloride into their ionic state, shown as a molecular model in (B) the solid crystalline lattice of sodium chloride, and (C) the sodium chloride dissolved in the water solvent. (Photo of sodium chloride provided by Chris 73 ).


Many ionic compounds are soluble in water, however, not all ionic compounds are soluble. Ionic compounds that are soluble in water exist in their ionic state within the solution. You will notice in Figure 7.2 that the sodium chloride breaks apart into the sodium ion and the chloride ion as it dissolves and interacts with the water molecules. For ionic compounds that are not soluble in water, the ions are so strongly attracted to one another that they cannot be broken apart by the partial charges of the water molecules. The following table can be used to help you predict which ionic compounds will be soluble in water.

Table 7.1 Solubility Rules

The dissociation of soluble ionic compounds gives solutions of these compounds an interesting property: they conduct electricity. Because of this property, soluble ionic compounds are referred to as electrolytes. Many ionic compounds dissociate completely and are therefore called strong electrolytes. Sodium chloride is an example of a strong electrolyte. Some compounds dissolve but dissociate only partially, and solutions of such solutes may conduct electricity only weakly. These solutes are called weak electrolytes. Acetic acid (CH3COOH), the compound in vinegar, is a weak electrolyte. Solutes that dissolve into individual neutral molecules without dissociation do not impart additional electrical conductivity to their solutions and are called nonelectrolytes. Polar covalent compounds, such as table sugar (C12H22O11), are good examples of nonelectrolytes.

The term electrolyte is used in medicine to mean any of the important ions that are dissolved in aqueous solution in the body. Important physiological electrolytes include Na+, K+, Ca2+, Mg2+, and Cl. Sports drinks such as Gatoraid have combinations of these key electrolytes, to help replenish electrolyte loss following a hard workout.

Similarly, solutions can also be made by mixing two compatible liquids together. The liquid in the lower concentration is termed the solute, and the one in higher concentration the solvent. For example, grain alcohol (CH3CH2OH) is a polar covalent molecule that can mix with water. When two similar solutions are placed together and are able to mix into a solution, they are said to be miscible. Liquids that do not share similar characteristics and cannot mix together, on the other hand, are termed immiscible. For example, the oils found in olive oil, such as oleic acid (C18H34O2) have mainly nonpolar covalent bonds which do not have intermolecular forces that are strong enough to break the hydrogen bonding between the water molecules. Thus, water and oil do not mix and are said to be immiscible.

Other factor such as temperature and pressure also affects the solubility of a solvent. Thus, in specifying solubility, one should also be aware of these other factors.

7.3 Temperature and Solubility

When considering the solubility solids, the relationship of temperature and solubility is not simple or predictable. Figure 7.3 shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and CH3CO2Na) exhibit a dramatic increase in solubility with increasing temperature. Others (such as NaCl and K2SO4) exhibit little variation, and still others (such as Li2SO4) become less soluble with increasing temperature.

Figure 7.3 Solubilities of Several Inorganic and Organic Solids in Water as a Function of Temperature. Solubility may increase or decrease with temperature; the magnitude of this temperature dependence varies widely among compounds.


The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by fractional crystallization, the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate (CH3CO2Na) and 50 g of KBr, we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure 7.3, both compounds dissolve in water at 80°C, and all 50 g of KBr remains in solution at 0°C. Only about 36 g of CH3CO2Na are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of CH3CO2Na crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original CH3CO2Na in essentially pure form in only one step.

Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure 7.3 and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was KBr in this example) and preferably present in relatively small amounts.

The solubility of gases in liquids is much more predictable.  The solubility of gases in liquids decreases with increasing temperature, as shown in Figure 7.4. Attractive intermolecular interactions in the gas phase are essentially zero for most substances, because the molecules are so far apart when in the gaseous form. When a gas dissolves, it does so because its molecules interact with solvent molecules. Heat is released when these new attractive forces form.  Thus, if external heat is added to the system, it overcomes the attractive forces between the gas and the solvent molecules and decreases the solubility of the gas.

Figure 7.4 Solubilities of Several Common Gases in Water as a Function of Temperature at Partial Pressure of 1 atm. The solubilities of gases decrease with increasing temperature.


The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (Figure 7.5). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex, but the driving force is the loss of dissolved carbon dioxide (CO2) from solution. Hard water contains dissolved Ca2+ and HCO3 (bicarbonate) ions. Calcium bicarbonate [Ca(HCO3)2] is rather soluble in water, but calcium carbonate (CaCO3) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water:

2HCO3(aq) → CO22−(aq) + H2O(l) + CO2(aq)

Heating the solution decreases the solubility of CO2, which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale.

Figure 7.5 Boiler Scale in a Water Pipe. Calcium carbonate (CaCO3) deposits in hot water pipes can significantly reduce pipe capacity. These deposits, called boiler scale, form when dissolved CO2 is driven into the gas phase at high temperatures.

In thermal pollution, lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of O2 at higher temperatures (Figure 7.4), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water. In the Pacific Northwest, salmonid populations are extremely susceptible to changes in water temperature. Within these population, optimal water temperatures are between 12.8 and 17.8 oC (55-65 oF). In addition to reduced oxygen levels, salmon populations are much more susceptible to disease, predation, and parasite infections at higher water temperatures. Thus, thermal pollution and global climate change are creating real challenges to the survival and maintenance of these species. For more information on the effects of rising temperatures on salmonid populations visit the State of Washington’s Focus Publication.

A similar effect is seen in the rising temperatures of bodies of water such as the Chesapeake Bay, the largest estuary in North America, where global warming has been implicated as the cause. For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality.

7.4 Effects of Pressure on the Solubility of Gases: Henry’s Law

External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure 7.6, which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures.

Figure 7.6 A Model Depicting Why the Solubility of a Gas Increases as the Partial Pressure Increases at Constant Temperature. (a) When a gas comes in contact with a pure liquid, some of the gas molecules (purple spheres) collide with the surface of the liquid and dissolve. When the concentration of dissolved gas molecules has increased so that the rate at which gas molecules escape into the gas phase is the same as the rate at which they dissolve, a dynamic equilibrium has been established, as depicted here.  (b) Increasing the pressure of the gas increases the number of molecules of gas per unit volume, which increases the rate at which gas molecules collide with the surface of the liquid and dissolve. (c) As additional gas molecules dissolve at the higher pressure, the concentration of dissolved gas increases until a new dynamic equilibrium is established.


The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836):

C = kP

where C is the concentration of dissolved gas at equilibrium, P is the partial pressure of the gas, and k is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature. Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table 7.2


Video Tutorial on Henry’s Law from Kahn Academy

All Khan Academy content is available for free at www.khanacademy.org


As the data in Table 7.2 demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the elements of group 18, the Henry’s law constants increase smoothly from He to Ne to Ar. The table also shows that O2 is almost twice as soluble as N2. Although London dispersion forces are too weak to explain such a large difference, O2 is paramagnetic and hence more polarizable than N2, which explains its high solubility. (Note: When a substance is paramagnetic it is very weakly attracted by the poles of a magnet, but does not retain any permanent magnetism).

Table 7.2 Henry’s Law Constants for Selected Gases in Water at 20°C

The partial pressure of a gas can be expressed as concentration by writing Henry’s Law as Pgas = C/k.  This is important in many aspects of life including medicine where blood gases, like oxygen and carbon dioxide are commonly measured. Since partial pressure and concentration are directly proportional, if the partial pressure of a gas changes while the temperature remains constant, the new concentration of the gas within the liquid can be easily calculated using the following equation:

Where C1 and P1 are the concentration and partial pressure, respectively, of the gas at the initial condition, and C2 and P2 are the concentration and partial pressure, respectively, of the gas at the final condition. For example:

Practice Problem:  The concentration of CO2 in a solution is 0.032 M at 3.0 atm.  What is the concentration of CO2 at 5.0 atm of pressure?

Solution: To address this problem, first we must identify what we want to find.  This is the concentration of CO2 at 5.0 atm of pressure.  These two values represent C2 = ?? and P2 = 5.0 atm. At this point it will be easiest to rearrange our equation above to solve for C2. Next we need to identify the starting conditions, C1 = 0.032 M and P1 = 3.0 atm.  Then we can plug that values into the equation and solve for C2 :

Gases that react chemically with water, such as HCl and the other hydrogen halides, H2S, and NH3, do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, HCl reacts with water to give H+(aq) and Cl(aq), not dissolved HCl molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule. Overall, gases that react with water do not obey Henry’s Law.

Note the Pattern

Henry’s law has important applications. For example, bubbles of CO2 form as soon as a carbonated beverage is opened because the drink was bottled under CO2 at a pressure greater than 1 atm. When the bottle is opened, the pressure of CO2 above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of N2 to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases.

Due to the low Henry’s law constant for O2 in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the O2 concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind O2 reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds O2 and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four O2 molecules. Although the concentration of dissolved O2 in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved O2 concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for O2 result in dissolved oxygen concentrations comparable to those in normal blood.

7.5 Solid Hydrates:

Some ionic solids will accept a small number of water molecules into their crystal lattice structure and remain in a solid state.  These solids are called solid hydrates. Solid hydrates contain water molecules combined in a definite ratio as an integral part of the crystal that are either bound to a metal center or that have crystallized with the metal complex. Such hydrates are also said to contain water of crystallization or water of hydration.

A colorful example is cobalt(II) chloride, which turns from blue to red upon hydration, and can therefore be used as a water indicator.

Figure 7.7: Cobalt chloride as an example of a solid hydrate. Anhydrous cobalt chloride (upper left) and it’s crystal lattice structure (lower left) compared with cobalt chloride hexahydrate (upper right) and it’s crystal lattice (lower right). Notice that the water molecules shown in red (oxygen) and white (hydrogen) are integrated into the crystal lattice of the cobalt (II) chloride, shown in blue (cobalt) and green (chloride),  based on polarity.  The partially negative oxygen atoms are attracted to the positively charged cobalt while the partially positive hydrogen atoms are attracted to the negatively charged chloride ions.  Images provided by Wikipedia Commons (upper left and lower left), Benjah-bmm27 (upper right), and Smokefoot (lower right)

The notation used to represent a solid hydrate is: “hydrated compoundnH2O“, where n is the number of water molecules per formula unit of the salt. The n is usually a low integer, though it is possible for fractional values to occur. For example, in a monohydrate n is one, and in a hexahydrate n is 6. For the example in Figure 7.7, the hydrated cobalt chloride would be designated: “cobalt (II) chloride⋅6H2O”. Numerical prefixes of Greek origin that are used to designate solid hydrates are:

  • Hemi – 1/2
  • Mono – 1
  • Sesqui – 1½
  • Di – 2
  • Tri – 3
  • Tetra – 4
  • Penta – 5
  • Hexa – 6
  • Hepta – 7
  • Octa – 8
  • Nona – 9
  • Deca – 10
  • Undeca – 11
  • Dodeca – 12

A hydrate which has lost water is referred to as an anhydride; the remaining water, if any exists, can only be removed with very strong heating. A substance that does not contain any water is referred to as anhydrous. Some anhydrous compounds are hydrated so easily that they will pull water out of the atmosphere and become hydrated. These substances are said to be hygroscopic and can be used as drying agents or desiccants.

7.6 Solution Concentration

In chemistry, concentration is defined as the abundance of a constituent divided by the total volume of a mixture. All of us have a qualitative idea of what is meant by concentration. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. Quantitatively, the concentration of a solution describes the quantity of a solute that is contained in a particular quantity of that solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution, and are critical for many aspects of our lives, from measuring the correct dose of medicine to detecting chemical pollutants like lead and arsenic. Chemists use many different ways to define concentrations. In this section, we will cover the most common ways of presenting solution concentration.  These include: Molarity, Percent Concentrations, and Parts Per Million (PPM)/Parts Per Billion (PPB) measurements. 

7.6.1 Molarity

The most common unit of concentration is molarity, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) of a solution is the number of moles of solute present in exactly 1 L of solution. 

The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as M. Note that the volume indicated is the total volume of the solution and includes both the solute and the solvent.  For example, an aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the concentration of a solute. So

[sucrose] = 1.00 M

is read as “the concentration of sucrose is 1.00 molar.” The equation above can be used to calculate how much solute is required to make any amount of a desired solution. 

Example Problem:

Calculate the number of moles of sodium hydroxide (NaOH) needed to make 2.50 L of 0.100 M NaOH.

Given: (1) identity of solute = NaOH,  (2) volume = 2.50 L,  and (3) molarity of solution = 0.100 mol/L (Note: when calculating problems always write out the units of molarity as mol/L, rather than M. This will allow you to cancel out your units when doing the calculation.)

Asked for: amount of solute in moles

Strategy:  (1) Rearrange the equation above to solve for the desired unit, in this case for moles. (2) Double check all the units in the equation and make sure they match. Perform any conversions that are needed so that the units match. (3) Fill in values appropriately and do the math.

Solution:

(1) Rearrange the equation above to solve for moles.

(2) Double check all the units in the equation and make sure they match.

The given values for this equation are the volume 2.50 L and the molarity 0.100 mol/L.  The volume units for both of these numbers are in Liters (L) and thus, match. Therefore, no conversions need to be made.

(3)  Fill in values appropriately and do the math.

Preparation of Solutions

Note that in the example above, we still don’t have enough information to actually make the solution in the laboratory. There is no piece of equipment that can measure out the moles of a substance.  For this, we need to convert the number of moles of the sample into the number of grams represented by that number. We can then easily use a balance to weigh the amount of substance needed for the solution. For the example above:
To actually make the solution, it is typical to dissolve the solute in a small amount of the solvent and then once the solute is dissolved, the final volume can be brought up to 2.50 L.  If you were to add 10 g of NaOH directly to 2.50 L, the final volume would be larger than 2.50 L and the solution concentration would be less than 0.100 M. Remember that the final volume must include both the solute and the solvent.

Figure 7.8 illustrates the procedure for making a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the solvent is not specified. Since the solute occupies space in the solution, the volume of the solvent needed is less than the desired total volume of solution.

Figure 7.8: Preparation of a Solution of Known Concentration Using a Solid Solute. To make a solution, start by addition a portion of the solvent to the flask.  Next, weigh out the appropriate amount of solute and slowly add it to the solvent.  Once it is dissolved in the solvent, the volume of the solution can be brought up to the final solution volume. For the volumetric flask shown, this is indicated by the black line in the neck of the flask. In this case, it indicates 500 mL of solution. Volumetric flasks exist in many different sizes to accommodate different solution volumes. Graduated cylinders can also be used to accurately bring a solution to its final volume. Other glassware, including beakers and Erlenmeyer flasks are not accurate enough to make most solutions.  


Example Molarity Calculation

The solution in Figure 7.8 contains 10.0 g of cobalt(II) chloride dihydrate, CoCl2·2H2O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl2·2H2O?

Given: mass of solute and volume of solution

Asked for: concentration (M)

Strategy:

1. We know that Molarity equals moles/Liter

2. To calculate Molarity, we need to express:

  • the mass in the form of moles
  • the volume in the form of Liters
  • Plug both into the equation above and calculate

Solution:

  1. Converting the mass into moles. We can use the molar mass to convert the grams of CoCl2·2H2O to moles.
  • The molar mass of CoCl2·2H2O is 165.87 g/mol (and includes the two water molecules as they are part of the crystal lattice structure of this solid hydrate!)

   2. Convert the volume into Liters

    3. Plug values into the Molarity equation:


7.6.2 Parts Per Solutions

In the consumer and industrial world, the most common method of expressing the concentration is based on the quantity of solute in a fixed quantity of solution. The “quantities” referred to here can be expressed in weight, in volume, or both (i.e., the weight of solute in a given volume of solution.) In order to distinguish among these possibilities, the abbreviations (w/w), (v/v) and (w/v) are used.

In most applied fields of Chemistry, (w/w) measure is often used, and is commonly expressed as weight-percent concentration, or simply “percent concentration”. For example, a solution made by dissolving 10 g of salt with 200 g of water contains “1 part of salt per 20 g of water”.  It is usually more convenient to express such concentrations as “parts per 100”, which we all know as “percent”.  “Cent” is the Latin-derived prefix relating to the number 100
(L. centum), as in century or centennial. It also denotes 1/100th (from L. centesimus) as in centimeter and the monetary unit cent. So the solution described above is a “5% (w/w) solution” of NaCl in water.

In clinical chemistry, (w/v) is commonly used, with weight expressed in grams and volume in mL (see Problem Example 1 below).

Percent means parts per 100; we can also use parts per thousand (ppt) for expressing concentrations in grams of solute per kilogram of solution. For more dilute solutions, parts per million (ppm) and parts per billion (109; ppb) are used. These terms are widely employed to express the amounts of trace pollutants in the environment.

7.6.X Preparing Dilutions

A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution, which is a prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred when making solutions of very weak concentrations,  because the alternative method, weighing out tiny amounts of solute, can be difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.

The procedure for preparing a solution of known concentration from a stock solution is shown in Figure 7.9. It requires calculating the amount of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does not change the amount of solute present, only the volume of the solution is changing. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution can therefore be expressed mathematically as:

 

Where Ms is the concentration of the stock solution, Vs is the volume of the stock solution, Md is the concentration of the diluted solution, and Vd is the volume of the diluted solution.

Figure 7.9 Preparation of a Solution of Known Concentration by Diluting a Stock Solution. (a) A volume (Vs) containing the desired amount of solute (Ms) is measured from a stock solution of known concentration. (b) The measured volume of stock solution is transferred to a second volumetric flask. (c) The measured volume in the second flask is then diluted with solvent up to the volumetric mark [(Vs)(Ms) = (Vd)(Md)].


Example of Dilution Calculations

What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of 0.400 M solution?

Given: volume and molarity of dilute solution, and molarity of stock solution

Asked for: volume of stock solution

Strategy and Solution:

For Dilution problems, as long as you know 3 of the variables, you can solve for the 4th variable.

  1. Start by rearranging the equation to solve for the variable that you want to find. In this case, you want to find the volume of the stock solution, Vs

2. Next, check to make sure that like terms have the same units.  For example, Md and Ms are both concentrations, thus, to be able to perform the calculations, they should be in the same unit (in this case they are both listed in Molarity). If the concentrations were different, say one was given in Molarity and the other in percent  or one was in Molarity and the other was in Millimolarity, one of the terms would need to be converted so that they match.  That way, the units will cancel out and leave you with units of volume, in this case.

   3. Finally, fill in the equation with known values and calculate the final answer.

Note that if 333mL of stock solution is needed, that you can also calculate the amount of solvent needed to make the final dilution. (Total volume – volume of stock solution = volume of solvent needed for the final dilution. In this case 2,500 mL – 333 mL = 2,167 mL of water needed to make the final dilution (this should be done in a graduated cylinder or volumetric flask.

Ion Concentrations in Solution

Thus far, we have been discussing the concentration of the overall solution in terms of total solute divided by the volume of the solution. Let’s consider in more detail exactly what that means when considering ionic and covalent compounds.  When ionic compounds dissolve in a solution, they break apart into their ionic state. Cations and anions associate with the polar water molecules. Recall that solutions that contain ions are called electrolyotes, due to their ability to conduct electricity.  For example, ammonium dichromate (NH4)2Cr2O7 is an ionic compound that contains two NH4+ ions and one Cr2O72− ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH4+ and Cr2O72− ions.  If we consider this this solution mathematically, we can see that for every ammonium dichromate molecule that dissolves, there will be three resulting ions that form (the two NH4+ ions and the one Cr2O72−  ion).   This can also be thought of on a larger molar scale.  When 1 mole of (NH4)2Cr2O7 is dissolved, it  results in 3 moles of ions (1 mol of Cr2O72− anions and 2 mol of NH4+ cations) within the solution.  To discuss the relationship between the concentration of a solution and the resulting number of ions, the term equivalents is used.

One equivalent is defined as the amount of an ion that provides 1 mole of electrical charge (+ or -).  It is calculated by dividing the molarity of the solution by the total charge

Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr2O72− anions and 2 mol of NH4+ cations (see Figure 4.9 “Dissolution of 1 mol of an Ionic Compound”).

Figure 4.9 Dissolution of 1 mol of an Ionic Compound

In this case, dissolving 1 mol of (NH4)2Cr2O7 produces a solution that contains 1 mol of Cr2O72− ions and 2 mol of NH4+ ions. (Water molecules are omitted from a molecular view of the solution for clarity.)

When we carry out a chemical reaction using a solution of a salt such as ammonium dichromate, we need to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH4)2Cr2O7, then the concentration of Cr2O72− must also be 1.43 M because there is one Cr2O72− ion per formula unit. However, there are two NH4+ ions per formula unit, so the concentration of NH4+ ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH4)2Cr2O7 produces three ions when dissolved in water (2NH4+ + 1Cr2O72−), the total concentration of ions in the solution is 3 × 1.43 M = 4.29 M.

Example 6

What are the concentrations of all species derived from the solutes in these aqueous solutions?

  1. 0.21 M NaOH
  2. 3.7 M (CH3)CHOH
  3. 0.032 M In(NO3)3

Given: molarity

Asked for: concentrations

Strategy:

A Classify each compound as either a strong electrolyte or a nonelectrolyte.

B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.

Solution:

  1. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution:

    NaOH(s)−→−−H2O(l)Na+(aq)+OH(aq)

 

  • B Because each formula unit of NaOH produces one Na+ ion and one OH ion, the concentration of each ion is the same as the concentration of NaOH: [Na+] = 0.21 M and [OH] = 0.21 M.

  • A The formula (CH3)2CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 “Aqueous Solutions” that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.

    B The only solute species in solution is therefore (CH3)2CHOH molecules, so [(CH3)2CHOH] = 3.7 M.

  • A Indium nitrate is an ionic compound that contains In3+ ions and NO3 ions, so we expect it to behave like a strong electrolyte in aqueous solution:

    In(NO3)3(s)−→−−H2O(l)In3+(aq)+3NO3(aq)

 

  1. B One formula unit of In(NO3)3 produces one In3+ ion and three NO3 ions, so a 0.032 M In(NO3)3 solution contains 0.032 M In3+ and 3 × 0.032 M = 0.096 M NO3—that is, [In3+] = 0.032 M and [NO3] = 0.096 M.

Exercise

What are the concentrations of all species derived from the solutes in these aqueous solutions?

  1. 0.0012 M Ba(OH)2
  2. 0.17 M Na2SO4
  3. 0.50 M (CH3)2CO, commonly known as acetone

Answer:

  1. [Ba2+] = 0.0012 M; [OH] = 0.0024 M
  2. [Na+] = 0.34 M; [SO42−] = 0.17 M
  3. [(CH3)2CO] = 0.50 M

Key Equations

definition of molarity

Equation 4.4: molarity =moles of soluteliters of solution=mmoles of solutemilliliters of solution

relationship among volume, molarity, and moles

Equation 4.5: VLMmol/L=L (molL )= moles

relationship between volume and concentration of stock and dilute solutions

Equation 4.7: (Vs)(Ms) = moles of solute = (Vd)(Md)

Summary

The concentration of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed as molarity, the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

Key Takeaway

  • Solution concentrations are typically expressed as molarity and can be prepared by dissolving a known mass of solute in a solvent or diluting a stock solution.

Conceptual Problems

  1. Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers.

    1. NH3
    2. HF
    3. CH3CH2CH2OH
    4. Na2SO4

  2. Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers.

    1. CH3CO2H
    2. NaCl
    3. Na2S
    4. Na3PO4
    5. acetaldehyde
  3. Would you expect a 1.0 M solution of CaCl2 to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not?

  4. An alternative way to define the concentration of a solution is molality, abbreviated m. Molality is defined as the number of moles of solute in 1 kg of solvent. How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 m solution of sucrose? Explain your answer.

  5. What are the advantages of using solutions for quantitative calculations?

Answer

  1. If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately.

Numerical Problems

  1. Calculate the number of grams of solute in 1.000 L of each solution.

    1. 0.2593 M NaBrO3
    2. 1.592 M KNO3
    3. 1.559 M acetic acid
    4. 0.943 M potassium iodate
  2. Calculate the number of grams of solute in 1.000 L of each solution.

    1. 0.1065 M BaI2
    2. 1.135 M Na2SO4
    3. 1.428 M NH4Br
    4. 0.889 M sodium acetate
  3. If all solutions contain the same solute, which solution contains the greater mass of solute?

    1. 1.40 L of a 0.334 M solution or 1.10 L of a 0.420 M solution
    2. 25.0 mL of a 0.134 M solution or 10.0 mL of a 0.295 M solution
    3. 250 mL of a 0.489 M solution or 150 mL of a 0.769 M solution
  4. Complete the following table for 500 mL of solution.

    Compound Mass (g) Moles Concentration (M)
    calcium sulfate 4.86
    acetic acid 3.62
    hydrogen iodide dihydrate 1.273
    barium bromide 3.92
    glucose 0.983
    sodium acetate 2.42
  5. What is the concentration of each species present in the following aqueous solutions?

    1. 0.489 mol of NiSO4 in 600 mL of solution
    2. 1.045 mol of magnesium bromide in 500 mL of solution
    3. 0.146 mol of glucose in 800 mL of solution
    4. 0.479 mol of CeCl3 in 700 mL of solution
  6. What is the concentration of each species present in the following aqueous solutions?

    1. 0.324 mol of K2MoO4 in 250 mL of solution
    2. 0.528 mol of potassium formate in 300 mL of solution
    3. 0.477 mol of KClO3 in 900 mL of solution
    4. 0.378 mol of potassium iodide in 750 mL of solution
  7. What is the molar concentration of each solution?

    1. 8.7 g of calcium bromide in 250 mL of solution
    2. 9.8 g of lithium sulfate in 300 mL of solution
    3. 12.4 g of sucrose (C12H22O11) in 750 mL of solution
    4. 14.2 g of iron(III) nitrate hexahydrate in 300 mL of solution
  8. What is the molar concentration of each solution?

    1. 12.8 g of sodium hydrogen sulfate in 400 mL of solution
    2. 7.5 g of potassium hydrogen phosphate in 250 mL of solution
    3. 11.4 g of barium chloride in 350 mL of solution
    4. 4.3 g of tartaric acid (C4H6O6) in 250 mL of solution
  9. Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant.

    1. BaCl2(aq) + Na2SO4(aq) →
    2. Ca(OH)2(aq) + H3PO4(aq) →
    3. Al(NO3)3(aq) + H2SO4(aq) →
    4. Pb(NO3)2(aq) + CuSO4(aq) →
    5. Al(CH3CO2)3(aq) + NaOH(aq) →
  10. An experiment required 200.0 mL of a 0.330 M solution of Na2CrO4. A stock solution of Na2CrO4 containing 20.0% solute by mass with a density of 1.19 g/cm3 was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na2CrO4 using the stock solution.

  11. Calcium hypochlorite [Ca(OCl)2] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl)2 concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite?

  12. Phenol (C6H5OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol?

  13. If a tablet containing 100 mg of caffeine (C8H10N4O2) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution?

  14. A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered?

Answers

  1. 0.48 M ClO

  2. 1.74 × 10−3 M caffeine

     

 

 

    Again, the units of the solute and the solution must be the same. A hybrid concentration unit, mass/volume percent (% m/v), is commonly used for intravenous (IV) fluids (Figure 9.2.1

    ). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100:

    %m/v=massofsolute(g)volumeofsolution(mL)×100%(9.2.3)

           800px-Baxter_sodium_chloride_irrigation.JPG

    Figure 9.2.1

    : Mass/Volume Percent.The 0.9% NaCl concentration on this IV bag is mass/volume percent (left). Such solution is used for other purposes and available in bottles (right). Figures used with permission from Wikipedia

    Each percent concentration can be used to produce a conversion factor between the amount of solute, the amount of solution, and the percent. Furthermore, given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates.

    Example 9.2.2

    A sample of 45.0% v/v solution of ethanol (C2H5OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain?

    SOLUTION

    A percentage concentration is simply the number of parts of solute per 100 parts of solution. Thus, the percent concentration of 45.0% v/v implies the following:

    45.0%v/v45mLC2H5OH100mLsolution

    That is, there are 45 mL of C2H5OH for every 100 mL of solution. We can use this fraction as a conversion factor to determine the amount of C2H5OH in 115 mL of solution:

    115mLsolution×45mLC2H5OH100mLsolution=51.8mLC2H5OH

    The highest concentration of ethanol that can be obtained normally is 95% ethanol, which is actually 95% v/v.

    Exercise 9.2.2

    What volume of a 12.75% m/v solution of glucose (C6H12O6) in water is needed to obtain 50.0 g of C6H12O6?

    Example 9.2.3

    A normal saline IV solution contains 9.0 g of NaCl in every liter of solution. What is the mass/volume percent of normal saline?

    SOLUTION

    We can use the definition of mass/volume percent, but first we have to express the volume in milliliter units:

    1 L = 1,000 mL

    Because this is an exact relationship, it does not affect the significant figures of our result.

    %m/v=9.0gNaCl1,000mLsolution×100%=0.90%m/v

    Exercise 9.2.3

    The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass/volume percent of the bleach?

    In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm) and parts per billion (ppb). Both of these units are mass based and are defined as follows:

    ppm=massofsolutemassofsolution×1,000,000(9.2.4)
    ppb=massofsolutemassofsolution×1,000,000,000(9.2.5)

    Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt).

    Concentrations of trace elements in the body—elements that are present in extremely low concentrations but are nonetheless necessary for life—are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon’s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm.

    In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb is equivalent to 1 µg/L.

    Example 9.2.4

    If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg?

    SOLUTION

    A concentration of 21 ppb means “21 g of solute per 1,000,000,000 g of solution.” Written as a conversion factor, this concentration of Co is as follows:

    21ppbCo21gCo1,000,000,000gsolution

    We can use this as a conversion factor, but first we must convert 70.0 kg to gram units:

    70.0kg×1,000g1kg=7.00×104g

    Now we determine the amount of Co:

    7.00×104gsolution×21gCo1,000,000,000gsolution=0.0015gCo

    This is only 1.5 mg.

    Exercise 9.2.4

    An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million?

    Molarity

    Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams.

    Molarity is defined as the number of moles of a solute dissolved per liter of solution:

    molarity=numberofmolesofsolutenumberoflitersofsolution(9.2.6)

    Molarity is abbreviated M (often referred to as “molar”), and the units are often abbreviated as mol/L. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore

    1.5molNaCl0.500Lsolution=3.0MNaCl(9.2.7)

    which is read as “three point oh molar sodium chloride.” Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl(aq).”

    Before a molarity concentration can be calculated, the amount of the solute must be expressed in moles, and the volume of the solution must be expressed in liters, as demonstrated in the following example.

    Example 9.2.5

    What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL?

    SOLUTION

    Before we substitute these quantities into the definition of molarity, we must convert them to the proper units. The mass of NaOH must be converted to moles of NaOH. The molar mass of NaOH is 40.00 g/mol:

    25.0gNaOH×1molNaOH40.00gNaOH=0.625molNaOH

    Next, we convert the volume units from milliliters to liters:

    750mL×1L1,000mL=0.750L

    Now that the quantities are expressed in the proper units, we can substitute them into the definition of molarity:

    M=0.625molNaOH0.750L=0.833MNaOH

    Exercise 9.2.5

    If a 350 mL cup of coffee contains 0.150 g of caffeine (C8H10N4O2), what is the molarity of this caffeine solution?

    The definition of molarity can also be used to calculate a needed volume of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and volume. The following example illustrates this.

    Example 9.2.6

    1. What volume of a 0.0753 M solution of dimethylamine [(CH3)2NH] is needed to obtain 0.450 mol of the compound?
    2. Ethylene glycol (C2H6O2) is mixed with water to make auto engine coolants. How many grams of C2H6O2 are in 5.00 L of a 6.00 M aqueous solution?

    SOLUTION

    In both parts, we will use the definition of molarity to solve for the desired quantity.

    1. 0.0753M=0.450mol(CH3)2NHvolumeofsolution

    To solve for the volume of solution, we multiply both sides by volume of solution and divide both sides by the molarity value to isolate the volume of solution on one side of the equation:

    volumeofsolution=0.450mol(CH3)2NH0.0753M=5.98L

    Note that because the definition of molarity is mol/L, the division of mol by M yields L, a unit of volume.

    1. The molar mass of C2H6O2 is 62.08 g/mol., so

    6.00M=molesofsolute5.00L

    To solve for the number of moles of solute, we multiply both sides by the volume:

    moles of solute = (6.00 M)(5.00 L) = 30.0 mol

    Note that because the definition of molarity is mol/L, the product M × L gives mol, a unit of amount. Now, using the molar mass of C3H8O3, we convert mol to g:

    30.0mol×62.08gmol=1,860g

    Thus, there are 1,860 g of C2H6O2 in the specified amount of engine coolant.

    Note: Dimethylamine has a “fishy” odor. In fact, organic compounds called amines cause the odor of decaying fish.

    Exercise 9.2.6

    1. What volume of a 0.0902 M solution of formic acid (HCOOH) is needed to obtain 0.888 mol of HCOOH?
    2. Acetic acid (HC2H3O2) is the acid in vinegar. How many grams of HC2H3O2 are in 0.565 L of a 0.955 M solution?

    Using Molarity in Stoichiometry Problems

    Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit. Consider the following chemical equation:

    HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)

    Suppose we want to know how many liters of aqueous HCl solution will react with a given mass of NaOH. A typical approach to answering this question is as follows:

    steps.jpg
    Figure 9.2.2
    : Typical approach to solving Molarity problems

    In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving.

    Example 9.2.7

    How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows:

    HCl(aq) + NaOH(s) → H2O(ℓ) + NaCl(aq)

    SOLUTION

    We will follow the flowchart to answer this question. First, we convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 g/mol:

    185gNaOH×1molNaOH40.00gNaOH=4.63molNaOH

    Using the balanced chemical equation, we see that there is a one-to-one ratio of moles of HCl to moles of NaOH. We use this to determine the number of moles of HCl needed to react with the given amount of NaOH:

    4.63molNaOH×1molHCl1molNaOH=4.63molHCl

    Finally, we use the definition of molarity to determine the volume of 2.75 M HCl needed:

    2.75MHCl=4.63molHClvolumeofHClsolution

    volumeofHCl=4.63molHCl2.75MHCl=1.68L×1,000mL1L=1,680mL

    We need 1,680 mL of 2.75 M HCl to react with the NaOH.

    Exercise 9.2.7

    How many milliliters of a 1.04 M H2SO4 solution are needed to react with 98.5 g of Ca(OH)2? The balanced chemical equation for the reaction is as follows:

    H2SO4(aq)+Ca(OH)2(s)2H2O()+CaSO4(aq)(9.2.8)

    The general steps for performing stoichiometry problems such as this are shown in Figure 9.2.3

    . You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure 9.2.3

    indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end.

    9.3.jpg

    Figure 9.2.3

    : Diagram of Steps for Using Molarity in Stoichiometry Calculations. When using molarity in stoichiometry calculations, a specific sequence of steps usually leads you to the correct answer.

    Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table 9.2.2

    lists the typical concentrations of some of these solutes.

    Table 9.2.2
    : Approximate Concentrations of Various Solutes in Some Solutions in the Body*
    Solution Solute Concentration (M)
    blood plasma Na+ 0.138
    K+ 0.005
    Ca2+ 0.004
    Mg2+ 0.003
    Cl 0.110
    HCO3 0.030
    stomach acid HCl 0.10
    urine NaCl 0.15
    PO43− 0.05
    NH2CONH2 (urea) 0.30
    *Note: Concentrations are approximate and can vary widely.

    Looking Closer: The Dose Makes the Poison

    Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: the dose makes the poison. This means that what may be dangerous in some amounts may not be dangerous in other amounts.

    Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison.

    Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous!

    Equivalents

    Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na+(aq) is also 1 Eq/L because sodium has a 1+ charge. A 1 mol/L solution of Ca2+(aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L. (For more information about the ions present in blood plasma, see Chapter 3 “Ionic Bonding and Simple Ionic Compounds”, Section 3.3 “Formulas for Ionic Compounds”.)

    Dilution

    When solvent is added to dilute a solution, the volume of the solution changes, but the amount of solute does not change. Before dilution, the amount of solute was equal to its original concentration times its original volume:

    amount in moles = (concentration × volume)initial

    After dilution, the same amount of solute is equal to the final concentration times the final volume:

    amount in moles = (concentration × volume)final

    To determine a concentration or amount after a dilution, we can use the following equation:

    (concentration × volume)initial = (concentration × volume)final

    Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit.

    Example 9.2.8

    A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the final concentration of the diluted solution?

    SOLUTION

    Because the volume units are the same, and we are looking for the molarity of the final solution, we can use (concentration × volume)initial = (concentration × volume)final:

    (0.900 M × 125 mL) = (concentration × 1,125 mL)

    We solve by isolating the unknown concentration by itself on one side of the equation. Dividing by 1,125 mL gives

    concentration=0.900M×125mL1,125mL=0.100M

    as the final concentration.

    Exercise 9.2.8

    A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin is an anticoagulant drug) into a 250 mL IV bag, for a final volume of 255 mL. What is the concentration of the resulting heparin solution?

    Concept Review Exercises

    1. What are some of the units used to express concentration?

    2. Distinguish between the terms solubility and concentration.

    Answers

    1. % m/m, % m/v, ppm, ppb, molarity, and Eq/L (answers will vary)

    2. Solubility is typically a limit to how much solute can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent.

    Key Takeaways

    • Various concentration units are used to express the amounts of solute in a solution.
    • Concentration units can be used as conversion factors in stoichiometry problems.
    • New concentrations can be easily calculated if a solution is diluted.

    Exercises

    1. Define solubility. Do all solutes have the same solubility?

    2. Explain why the terms dilute or concentrated are of limited usefulness in describing the concentration of solutions.

    3. If the solubility of sodium chloride (NaCl) is 30.6 g/100 mL of H2O at a given temperature, how many grams of NaCl can be dissolved in 250.0 mL of H2O?

    4. If the solubility of glucose (C6H12O6) is 120.3 g/100 mL of H2O at a given temperature, how many grams of C6H12O6 can be dissolved in 75.0 mL of H2O?

    5. How many grams of sodium bicarbonate (NaHCO3) can a 25.0°C saturated solution have if 150.0 mL of H2O is used as the solvent?

    6. If 75.0 g of potassium bromide (KBr) are dissolved in 125 mL of H2O, is the solution saturated, unsaturated, or supersaturated?

    7. Calculate the mass/mass percent of a saturated solution of NaCl. Use the data from Table 9.2.1

    • “Solubilities of Various Solutes in Water at 25°C (Except as Noted)”, assume that masses of the solute and the solvent are additive, and use the density of H2O (1.00 g/mL) as a conversion factor.

    • Calculate the mass/mass percent of a saturated solution of MgCO3 Use the data from Table 9.2.1

     

    1. “Solubilities of Various Solutes in Water at 25°C (Except as Noted)”, assume that masses of the solute and the solvent are additive, and use the density of H2O (1.00 g/mL) as a conversion factor.

    2. Only 0.203 mL of C6H6 will dissolve in 100.000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of benzene in water.

    3. Only 35 mL of aniline (C6H5NH2) will dissolve in 1,000 mL of H2O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of aniline in water.

    4. A solution of ethyl alcohol (C2H5OH) in water has a concentration of 20.56% v/v. What volume of C2H5OH is present in 255 mL of solution?

    5. What mass of KCl is present in 475 mL of a 1.09% m/v aqueous solution?

    6. The average human body contains 5,830 g of blood. What mass of arsenic is present in the body if the amount in blood is 0.55 ppm?

    7. The Occupational Safety and Health Administration has set a limit of 200 ppm as the maximum safe exposure level for carbon monoxide (CO). If an average breath has a mass of 1.286 g, what is the maximum mass of CO that can be inhaled at that maximum safe exposure level?

    8. Which concentration is greater—15 ppm or 1,500 ppb?

    9. Express the concentration 7,580 ppm in parts per billion.

    10. What is the molarity of 0.500 L of a potassium chromate solution containing 0.0650 mol of K2CrO4?

    11. What is the molarity of 4.50 L of a solution containing 0.206 mol of urea [(NH2)2CO]?

    12. What is the molarity of a 2.66 L aqueous solution containing 56.9 g of NaBr?

    13. If 3.08 g of Ca(OH)2 is dissolved in enough water to make 0.875 L of solution, what is the molarity of the Ca(OH)2?

    14. What mass of HCl is present in 825 mL of a 1.25 M solution?

    15. What mass of isopropyl alcohol (C3H8O) is dissolved in 2.050 L of a 4.45 M aqueous C3H8O solution?

    16. What volume of 0.345 M NaCl solution is needed to obtain 10.0 g of NaCl?

    17. How many milliliters of a 0.0015 M cocaine hydrochloride (C17H22ClNO4) solution is needed to obtain 0.010 g of the solute?

    18. Aqueous calcium chloride reacts with aqueous silver nitrate according to the following balanced chemical equation:

      CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ca(NO3)2(aq)

      How many moles of AgCl(s) are made if 0.557 L of 0.235 M CaCl2 react with excess AgNO3? How many grams of AgCl are made?

    19. Sodium bicarbonate (NaHCO3) is used to react with acid spills. The reaction with sulfuric acid (H2SO4) is as follows:

      2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2H2O(ℓ) + 2CO2(g)

      If 27.6 mL of a 6.25 M H2SO4 solution were spilled, how many moles of NaHCO3 would be needed to react with the acid? How many grams of NaHCO3 is this?

    20. The fermentation of glucose to make ethanol and carbon dioxide has the following overall chemical equation:

      C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)

      If 1.00 L of a 0.567 M solution of C6H12O6 were completely fermented, what would be the resulting concentration of the C2H5OH solution? How many moles of CO2 would be formed? How many grams is this? If each mole of CO2 had a volume of 24.5 L, what volume of CO2 is produced?

    21. Aqueous sodium bisulfite gives off sulfur dioxide gas when heated:

      2NaHSO3(aq) → Na2SO3(aq) + H2O(ℓ) + SO2(g)

      If 567 mL of a 1.005 M NaHSO3 solution were heated until all the NaHSO3 had reacted, what would be the resulting concentration of the Na2SO3 solution? How many moles of SO2 would be formed? How many grams of SO2 would be formed? If each mole of SO2 had a volume of 25.78 L, what volume of SO2 would be produced?

    22. What is the concentration of a 1.0 M solution of K+(aq) ions in equivalents/liter?

    23. What is the concentration of a 1.0 M solution of SO42−(aq) ions in equivalents/liter?

    24. A solution having initial concentration of 0.445 M and initial volume of 45.0 mL is diluted to 100.0 mL. What is its final concentration?

    25. A 50.0 mL sample of saltwater that is 3.0% m/v is diluted to 950 mL. What is its final mass/volume percent?

    Answers

    1. Solubility is the amount of a solute that can dissolve in a given amount of solute, typically 100 mL. The solubility of solutes varies widely.

    1. 76.5 g

    1. 12.6 g

    1. 26.5%

    1. 0.203%

    1. 52.4 mL

    1. 0.00321 g

    1. 15 ppm

    1. 0.130 M

    1. 0.208 M

    1. 37.6 g

    1. 0.496 L

    1. 0.262 mol; 37.5 g

    1. 1.13 M C2H5OH; 1.13 mol of CO2; 49.7 g of CO2; 27.7 L of CO2

    1. 1.0 Eq/L

    1. 0.200 M

    9.3: The Dissolution Process

    Skills to Develop

    • To describe the dissolution process at the molecular level

    The Dissolution Process

    What occurs at the molecular level to cause a solute to dissolve in a solvent? The answer depends in part on the solute, but there are some similarities common to all solutes.

    Recall the rule that like dissolves like. This means that substances must have similar intermolecular forces to form solutions. When a soluble solute is introduced into a solvent, the particles of solute can interact with the particles of solvent. In the case of a solid or liquid solute, the interactions between the solute particles and the solvent particles are so strong that the individual solute particles separate from each other and, surrounded by solvent molecules, enter the solution. (Gaseous solutes already have their constituent particles separated, but the concept of being surrounded by solvent particles still applies.) This process is called solvation and is illustrated in Figure 9.3.1

    . When the solvent is water, the word hydration, rather than solvation, is used.

    9.4.jpg

    Figure 9.3.1

    : Solvation. When a solute dissolves, the individual particles of solute become surrounded by solvent particles. Eventually the particle detaches from the remaining solute, surrounded by solvent molecules in solution. Source: Photo © Thinkstock

    Ionic Compounds and Covalent Compounds as Solutes

    In the case of molecular solutes like glucose, the solute particles are individual molecules. However, if the solute is ionic, the individual ions separate from each other and become surrounded by solvent particles. That is, the cations and anions of an ionic solute separate when the solute dissolves. This process is referred to as dissociation (Figure 9.3.1

    ).

    The dissociation of soluble ionic compounds gives solutions of these compounds an interesting property: they conduct electricity. Because of this property, soluble ionic compounds are referred to as electrolytes. Many ionic compounds dissociate completely and are therefore called strong electrolytes. Sodium chloride is an example of a strong electrolyte. Some compounds dissolve but dissociate only partially, and solutions of such solutes may conduct electricity only weakly. These solutes are called weak electrolytes. Acetic acid (CH3COOH), the compound in vinegar, is a weak electrolyte. Solutes that dissolve into individual neutral molecules without dissociation do not impart additional electrical conductivity to their solutions and are called nonelectrolytes. Table sugar (C12H22O11) is an example of a nonelectrolyte.

    The link below will connect you to an animation that illustrates the differences between ionic compounds (electroytes) and covalent compounds (non-electrolytes) dissolved in water.

    The term electrolyte is used in medicine to mean any of the important ions that are dissolved in aqueous solution in the body. Important physiological electrolytes include Na+, K+, Ca2+, Mg2+, and Cl.

    Example 9.3.1

    The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte.

    1. potassium chloride (KCl)
    2. fructose (C6H12O6)
    3. isopropyl alcohol [CH3CH(OH)CH3]
    4. magnesium hydroxide [Mg(OH)2]

    SOLUTION

    Each substance can be classified as an ionic solute or a nonionic solute. Ionic solutes are electrolytes, and nonionic solutes are nonelectrolytes.

    1. Potassium chloride is an ionic compound; therefore, when it dissolves, its ions separate, making it an electrolyte.
    2. Fructose is a sugar similar to glucose. (In fact, it has the same molecular formula as glucose.) Because it is a molecular compound, we expect it to be a nonelectrolyte.
    3. Isopropyl alcohol is an organic molecule containing the alcohol functional group. The bonding in the compound is all covalent, so when isopropyl alcohol dissolves, it separates into individual molecules but not ions. Thus, it is a nonelectrolyte
    4. Magnesium hydroxide is an ionic compound, so when it dissolves it dissociates. Thus, magnesium hydroxide is an electrolyte.

    Exercise 9.3.1

    The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte.

    1. acetone (CH3COCH3)
    2. iron(III) nitrate [Fe(NO3)3]
    3. elemental bromine (Br2)
    4. sodium hydroxide (NaOH)

    Concept Review Exercise

    1. Explain how the solvation process describes the dissolution of a solute in a solvent.

    Answer

    1. Each particle of the solute is surrounded by particles of the solvent, carrying the solute from its original phase.

    Key Takeaway

    • When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other.

    Exercises

    1. Describe what happens when an ionic solute like Na2SO4 dissolves in a polar solvent.

    2. Describe what happens when a molecular solute like sucrose (C12H22O11) dissolves in a polar solvent.

    3. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent.

      1. NH4NO3
      2. CO2
      3. NH2CONH2
      4. HCl
    4. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H2O to some extent.

      1. CH3CH2CH2OH
      2. Ca(CH3CO2)2
      3. I2
      4. KOH
    5. Will solutions of each solute conduct electricity when dissolved?

      1. AgNO3
      2. CHCl3
      3. BaCl2
      4. Li2O
    6. Will solutions of each solute conduct electricity when dissolved?

      1. CH3COCH3
      2. N(CH3)3
      3. CH3CO2C2H5
      4. FeCl2

    Answers

    1. Each ion of the ionic solute is surrounded by particles of solvent, carrying the ion from its associated crystal.

      1. electrolyte
      2. nonelectrolyte
      3. nonelectrolyte
      4. electrolyte
      1. yes
      2. no
      3. yes
      4. yes

    9.4: Properties of Solutions

    Skills to Develop

    • To describe how the properties of solutions differ from those of pure solvents.

    Solutions are likely to have properties similar to those of their major component—usually the solvent. However, some solution properties differ significantly from those of the solvent. Here, we will focus on liquid solutions that have a solid solute, but many of the effects we will discuss in this section are applicable to all solutions.

    Colligative Properties

    Solutes affect some properties of solutions that depend only on the concentration of the dissolved particles. These properties are called colligative properties. Four important colligative properties that we will examine here are vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure.

    Molecular compounds separate into individual molecules when they are dissolved, so for every 1 mol of molecules dissolved, we get 1 mol of particles. In contrast, ionic compounds separate into their constituent ions when they dissolve, so 1 mol of an ionic compound will produce more than 1 mol of dissolved particles. For example, every mole of NaCl that dissolves yields 1 mol of Na+ ions and 1 mol of Cl ions, for a total of 2 mol of particles in solution. Thus, the effect on a solution’s properties by dissolving NaCl may be twice as large as the effect of dissolving the same amount of moles of glucose (C6H12O6).

    Vapor Pressure Depression

    All liquids evaporate. In fact, given enough volume, a liquid will turn completely into a vapor. If enough volume is not present, a liquid will evaporate only to the point where the rate of evaporation equals the rate of vapor condensing back into a liquid. The pressure of the vapor at this point is called the vapor pressure of the liquid.

    The presence of a dissolved solid lowers the characteristic vapor pressure of a liquid so that it evaporates more slowly. (The exceptions to this statement are if the solute itself is a liquid or a gas, in which case the solute will also contribute something to the evaporation process. We will not discuss such solutions here.) This property is called vapor pressure depression and is depicted in Figure 9.4.1

    .

    9.6.jpg

    Figure 9.4.1

    : Vapor Pressure Depression. The presence of solute particles blocks some of the ability for liquid particles to evaporate. Thus, solutions of solid solutes typically have a lower vapor pressure than the pure solvent.

    Boiling Point and Freezing Point Effects

    A related property of solutions is that their boiling points are higher than the boiling point of the pure solvent. Because the presence of solute particles decreases the vapor pressure of the liquid solvent, a higher temperature is needed to reach the boiling point. This phenomenon is called boiling point elevation. For every mole of particles dissolved in a liter of water, the boiling point of water increases by about 0.5°C.

    Some people argue that putting a pinch or two of salt in water used to cook spaghetti or other pasta makes a solution that has a higher boiling point, so the pasta cooks faster. In actuality, the amount of solute is so small that the boiling point of the water is practically unchanged.

    The presence of solute particles has the opposite effect on the freezing point of a solution. When a solution freezes, only the solvent particles come together to form a solid phase, and the presence of solute particles interferes with that process. Therefore, for the liquid solvent to freeze, more energy must be removed from the solution, which lowers the temperature. Thus, solutions have lower freezing points than pure solvents do. This phenomenon is called freezing point depression. For every mole of particles in a liter of water, the freezing point decreases by about 1.9°C.

    Both boiling point elevation and freezing point depression have practical uses. For example, solutions of water and ethylene glycol (C2H6O2) are used as coolants in automobile engines because the boiling point of such a solution is greater than 100°C, the normal boiling point of water. In winter, salts like NaCl and CaCl2 are sprinkled on the ground to melt ice or keep ice from forming on roads and sidewalks (Figure 9.4.2

    ). This is because the solution made by dissolving sodium chloride or calcium chloride in water has a lower freezing point than pure water, so the formation of ice is inhibited.

    9.7 resized.jpg

    Figure 9.4.2

    : Effect of Freezing Point Depression. The salt sprinkled on this sidewalk makes the water on the sidewalk have a lower freezing point than pure water, so it does not freeze as easily. This makes walking on the sidewalk less hazardous in winter. © Thinkstock

    Example 9.4.1

    Which solution’s freezing point deviates more from that of pure water—a 1 M solution of NaCl or a 1 M solution of CaCl2?

    SOLUTION

    Colligative properties depend on the number of dissolved particles, so the solution with the greater number of particles in solution will show the greatest deviation. When NaCl dissolves, it separates into two ions, Na+ and Cl. But when CaCl2 dissolves, it separates into three ions—one Ca2+ ion and two Cl ions. Thus, mole for mole, CaCl2 will have 50% more impact on freezing point depression than NaCl.

    Exercise 9.4.1

    Which solution’s boiling point deviates more from that of pure water—a 1 M solution of CaCl2 or a 1 M solution of MgSO4?

    Osmotic Pressure

    The last colligative property of solutions we will consider is a very important one for biological systems. It involves osmosis, the process by which solvent molecules can pass through certain membranes but solute particles cannot. When two solutions of different concentration are present on either side of these membranes (called semipermeable membranes), there is a tendency for solvent molecules to move from the more dilute solution to the more concentrated solution until the concentrations of the two solutions are equal. This tendency is called osmotic pressure. External pressure can be exerted on a solution to counter the flow of solvent; the pressure required to halt the osmosis of a solvent is equal to the osmotic pressure of the solution.

    Osmolarity (osmol) is a way of reporting the total number of particles in a solution to determine osmotic pressure. It is defined as the molarity of a solute times the number of particles a formula unit of the solute makes when it dissolves (represented by i):

    osmol=M×i(9.4.1)

    If more than one solute is present in a solution, the individual osmolarities are additive to get the total osmolarity of the solution. Solutions that have the same osmolarity have the same osmotic pressure. If solutions of differing osmolarities are present on opposite sides of a semipermeable membrane, solvent will transfer from the lower-osmolarity solution to the higher-osmolarity solution. Counterpressure exerted on the high-osmolarity solution will reduce or halt the solvent transfer. An even higher pressure can be exerted to force solvent from the high-osmolarity solution to the low-osmolarity solution, a process called reverse osmosis. Reverse osmosis is used to make potable water from saltwater where sources of fresh water are scarce.

    Example 9.4.2

    A 0.50 M NaCl aqueous solution and a 0.30 M Ca(NO3)2 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow.

    SOLUTION

    The solvent will flow into the solution of higher osmolarity. The NaCl solute separates into two ions—Na+ and Cl—when it dissolves, so its osmolarity is as follows:

    osmol (NaCl) = 0.50 M × 2 = 1.0 osmol

    The Ca(NO3)2 solute separates into three ions—one Ca2+ and two NO3—when it dissolves, so its osmolarity is as follows:

    osmol [Ca(NO3)2] = 0.30 M × 3 = 0.90 osmol

    The osmolarity of the Ca(NO3)2 solution is lower than that of the NaCl solution, so water will transfer through the membrane from the Ca(NO3)2 solution to the NaCl solution.

    Exercise 9.4.2

    A 1.5 M C6H12O6 aqueous solution and a 0.40 M Al(NO3)3 aqueous solution are placed on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and predict the direction of solvent flow.

    To Your Health: Dialysis

    The main function of the kidneys is to filter the blood to remove wastes and extra water, which are then expelled from the body as urine. Some diseases rob the kidneys of their ability to perform this function, causing a buildup of waste materials in the bloodstream. If a kidney transplant is not available or desirable, a procedure called dialysis can be used to remove waste materials and excess water from the blood.

    In one form of dialysis, called hemodialysis, a patient’s blood is passed though a length of tubing that travels through an artificial kidney machine (also called a dialysis machine). A section of tubing composed of a semipermeable membrane is immersed in a solution of sterile water, glucose, amino acids, and certain electrolytes. The osmotic pressure of the blood forces waste molecules and excess water through the membrane into the sterile solution. Red and white blood cells are too large to pass through the membrane, so they remain in the blood. After being cleansed in this way, the blood is returned to the body.

    A patient undergoing hemodialysis depends on osmosis to cleanse the blood of waste products that the kidneys are incapable of removing due to disease. Image used with permission from Wikipedia.

    Dialysis is a continuous process, as the osmosis of waste materials and excess water takes time. Typically, 5–10 lb of waste-containing fluid is removed in each dialysis session, which can last 2–8 hours and must be performed several times a week. Although some patients have been on dialysis for 30 or more years, dialysis is always a temporary solution because waste materials are constantly building up in the bloodstream. A more permanent solution is a kidney transplant.

    Cell walls are semipermeable membranes, so the osmotic pressures of the body’s fluids have important biological consequences. If solutions of different osmolarity exist on either side of the cells, solvent (water) may pass into or out of the cells, sometimes with disastrous results. Consider what happens if red blood cells are placed in a hypotonic solution, meaning a solution of lower osmolarity than the liquid inside the cells. The cells swell up as water enters them, disrupting cellular activity and eventually causing the cells to burst. This process is called hemolysis. If red blood cells are placed in a hypertonic solution, meaning one having a higher osmolarity than exists inside the cells, water leaves the cells to dilute the external solution, and the red blood cells shrivel and die. This process is called crenation. Only if red blood cells are placed in isotonic solutions that have the same osmolarity as exists inside the cells are they unaffected by negative effects of osmotic pressure. Glucose solutions of about 0.31 M, or sodium chloride solutions of about 0.16 M, are isotonic with blood plasma.

    The concentration of an isotonic sodium chloride (NaCl) solution is only half that of an isotonic glucose (C6H12O6) solution because NaCl produces two ions when a formula unit dissolves, while molecular C6H12O6 produces only one particle when a formula unit dissolves. The osmolarities are therefore the same even though the concentrations of the two solutions are different.

    Osmotic pressure explains why you should not drink seawater if you are abandoned in a life raft in the middle of the ocean. Its osmolarity is about three times higher than most bodily fluids. You would actually become thirstier as water from your cells was drawn out to dilute the salty ocean water you ingested. Our bodies do a better job coping with hypotonic solutions than with hypertonic ones. The excess water is collected by our kidneys and excreted.

    Osmotic pressure effects are used in the food industry to make pickles from cucumbers and other vegetables and in brining meat to make corned beef. It is also a factor in the mechanism of getting water from the roots to the tops of trees!

    Career Focus: Perfusionist

    A perfusionist is a medical technician trained to assist during any medical procedure in which a patient’s circulatory or breathing functions require support. The use of perfusionists has grown rapidly since the advent of open-heart surgery in 1953.

    Most perfusionists work in operating rooms, where their main responsibility is to operate heart-lung machines. During many heart surgeries, the heart itself must be stopped. In these situations, a heart-lung machine keeps the patient alive by aerating the blood with oxygen and removing carbon dioxide. The perfusionist monitors both the machine and the status of the blood, notifying the surgeon and the anesthetist of any concerns and taking corrective action if the status of the blood becomes abnormal.

    Despite the narrow parameters of their specialty, perfusionists must be highly trained. Certified perfusion education programs require a student to learn anatomy, physiology, pathology, chemistry, pharmacology, math, and physics. A college degree is usually required. Some perfusionists work with other external artificial organs, such as hemodialysis machines and artificial livers.

    Concept Review Exercises

    1. What are the colligative properties of solutions?

    2. Explain how the following properties of solutions differ from those of the pure solvent: vapor pressure, boiling point, freezing point, and osmotic pressure.

    Answers

    1. Colligative properties are characteristics that a solution has that depend on the number, not the identity, of solute particles.

    2. In solutions, the vapor pressure is lower, the boiling point is higher, the freezing point is lower, and the osmotic pressure is higher.

    Key Takeaway

    • `Exercises
    1. In each pair of aqueous systems, which will have the lower vapor pressure?

      1. pure water or 1.0 M NaCl
      2. 1.0 M NaCl or 1.0 M C6H12O6
      3. 1.0 M CaCl2 or 1.0 M (NH4)3PO4
    2. In each pair of aqueous systems, which will have the lower vapor pressure?

      1. 0.50 M Ca(NO3)2 or 1.0 M KBr
      2. 1.5 M C12H22O11 or 0.75 M Ca(OH)2
      3. 0.10 M Cu(NO3)2 or pure water
    3. In each pair of aqueous systems, which will have the higher boiling point?

      1. pure water or a 1.0 M NaCl
      2. 1.0 M NaCl or 1.0 M C6H12O6
      3. 1.0 M CaCl2 or 1.0 M (NH4)3PO4
    4. In each pair of aqueous systems, which will have the higher boiling point?

      1. 1.0 M KBr
      2. 1.5 M C12H22O11 or 0.75 M Ca(OH)2
      3. 0.10 M Cu(NO3)2 or pure water
    5. Estimate the boiling point of each aqueous solution. The boiling point of pure water is 100.0°C.

      1. 0.50 M NaCl
      2. 1.5 M Na2SO4
      3. 2.0 M C6H12O6
    6. Estimate the freezing point of each aqueous solution. The freezing point of pure water is 0.0°C.

      1. 0.50 M NaCl
      2. 1.5 M Na2SO4
      3. 2.0 M C6H12O6
    7. Explain why salt (NaCl) is spread on roads and sidewalks to inhibit ice formation in cold weather.

    8. Salt (NaCl) and calcium chloride (CaCl2) are used widely in some areas to minimize the formation of ice on sidewalks and roads. One of these ionic compounds is better, mole for mole, at inhibiting ice formation. Which is that likely to be? Why?

    9. What is the osmolarity of each aqueous solution?

      1. 0.500 M NH2CONH2
      2. 0.500 M NaBr
      3. 0.500 M Ca(NO3)2
    10. What is the osmolarity of each aqueous solution?

      1. 0.150 M KCl
      2. 0.450 M (CH3)2CHOH
      3. 0.500 M Ca3(PO4)2
    11. A 1.0 M solution of an unknown soluble salt has an osmolarity of 3.0 osmol. What can you conclude about the salt?

    12. A 1.5 M NaCl solution and a 0.75 M Al(NO3)3 solution exist on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and the direction of solvent flow, if any, across the membrane.

    Answers

      1. 1.0 M NaCl
      2. 1.0 M NaCl
      3. 1.0 M (NH4)3PO4
      1. 1.0 M NaCl
      2. 1.0 M NaCl
      3. 1.0 M (NH4)3PO4
      1. 100.5°C
      2. 102.3°C
      3. 101°C
    1. NaCl lowers the freezing point of water, so it needs to be colder for the water to freeze.

      1. 0.500 osmol
      2. 1.000 osmol
      3. 1.500 osmol
    1. It must separate into three ions when it dissolves.

    9.5: Chapter Summary

    To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter.

    A solution is a homogeneous mixture. The major component is the solvent, while the minor component is the solute. Solutions can have any phase; for example, an alloy is a solid solution. Solutes are soluble or insoluble, meaning they dissolve or do not dissolve in a particular solvent. The terms miscible and immiscible, instead of soluble and insoluble, are used for liquid solutes and solvents. The statement like dissolves like is a useful guide to predicting whether a solute will dissolve in a given solvent.

    The amount of solute in a solution is represented by the concentration of the solution. The maximum amount of solute that will dissolve in a given amount of solvent is called the solubility of the solute. Such solutions are saturated. Solutions that have less than the maximum amount are unsaturated. Most solutions are unsaturated, and there are various ways of stating their concentrations. Mass/mass percent, volume/volume percent, and mass/volume percent indicate the percentage of the overall solution that is solute. Parts per million (ppm) and parts per billion (ppb) are used to describe very small concentrations of a solute. Molarity, defined as the number of moles of solute per liter of solution, is a common concentration unit in the chemistry laboratory. Equivalents express concentrations in terms of moles of charge on ions. When a solution is diluted, we use the fact that the amount of solute remains constant to be able to determine the volume or concentration of the final diluted solution.

    Dissolving occurs by solvation, the process in which particles of a solvent surround the individual particles of a solute, separating them to make a solution. For water solutions, the word hydration is used. If the solute is molecular, it dissolves into individual molecules. If the solute is ionic, the individual ions separate from each other, forming a solution that conducts electricity. Such solutions are called electrolytes. If the dissociation of ions is complete, the solution is a strong electrolyte. If the dissociation is only partial, the solution is a weak electrolyte. Solutions of molecules do not conduct electricity and are called nonelectrolytes.

    Solutions have properties that differ from those of the pure solvent. Some of these are colligative properties, which are due to the number of solute particles dissolved, not the chemical identity of the solute. Colligative properties include vapor pressure depression, boiling point elevation, freezing point depression, and osmotic pressure. Osmotic pressure is particularly important in biological systems. It is caused by osmosis, the passage of solvents through certain membranes like cell walls. The osmolarity of a solution is the product of a solution’s molarity and the number of particles a solute separates into when it dissolves. Osmosis can be reversed by the application of pressure; this reverse osmosis is used to make fresh water from saltwater in some parts of the world. Because of osmosis, red blood cells placed in hypotonic or hypertonic solutions lose function through either hemolysis or crenation. If they are placed in isotonic solutions, however, the cells are unaffected because osmotic pressure is equal on either side of the cell membrane.

    Additional Exercises

    1. Calcium nitrate reacts with sodium carbonate to precipitate solid calcium carbonate:

      Ca(NO3)2(aq)+Na2CO3(aq)CaCO3(s)+NaNO3(aq)

    References