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7.4: Colligative Properties

  • Page ID
    236764
  • Skills to Develop

    • Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)
    • Perform calculations using the mathematical equations that describe the various colligative effects
    • Describe ion pairing and use the van't Hoff factor in calculations 
     

    Colligative Properties

    The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.

    Vapor Pressure Lowering

    The equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:

    \[ \text{liquid} \rightleftharpoons \text{gas} \label{11.5.4}\]

    Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure \(\PageIndex{4}\)). While this kinetic interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the greater entropy of a solution in comparison to its separate solvent and solute serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module.

    This figure contains two images. Figure a is labeled “pure water.” It shows a beaker half-filled with liquid. In the liquid, eleven molecules are evenly dispersed in the liquid each consisting of one central red sphere and two slightly smaller white spheres are shown. Four molecules near the surface of the liquid have curved arrows drawn from them pointing to the space above the liquid in the beaker. Above the liquid, twelve molecules are shown, with arrows pointing from three of them into the liquid below. Figure b is labeled “Aqueous solution.” It is similar to figure a except that eleven blue spheres, slightly larger in size than the molecules, are dispersed evenly in the liquid. Only four curved arrows appear in this diagram with two from the molecules in the liquid pointing to the space above and two from molecules in the space above the liquid pointing into the liquid below.

    Figure \(\PageIndex{4}\): The presence of nonvolatile solutes lowers the vapor pressure of a solution by impeding the evaporation of solvent molecules.

    The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult’s law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

    \[P_\ce{A}=X_\ce{A}P^\circ_\ce{A} \label{11.5.5}\]

    where PA is the partial pressure exerted by component A in the solution, \(P^\circ_\ce{A}\) is the vapor pressure of pure A, and XA is the mole fraction of A in the solution. (Mole fraction is a concentration unit introduced in the chapter on gases.)

    Elevation of the Boiling Point of a Solvent

    As described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Compared to pure solvent, a solution, therefore, will require a higher temperature to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, \(ΔT_b\), is called boiling point elevation and is directly proportional to the molal concentration of solute species:

    \[ΔT_b=K_bm \label{11.5.8}\]

    where

    • \(K_\ce{b}\) is the boiling point elevation constant, or the ebullioscopic constant and
    • \(m\) is the molal concentration (molality) of all solute species.

    Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of Kb for several solvents are listed in Table \(\PageIndex{1}\).

    Table \(\PageIndex{1}\): Boiling Point Elevation and Freezing Point Depression Constants for Several Solvents
    Solvent Boiling Point (°C at 1 atm) Kb (Cm−1) Freezing Point (°C at 1 atm) Kf (Cm−1)
    water 100.0 0.512 0.0 1.86
    hydrogen acetate 118.1 3.07 16.6 3.9
    benzene 80.1 2.53 5.5 5.12
    chloroform 61.26 3.63 −63.5 4.68
    nitrobenzene 210.9 5.24 5.67 8.1

    The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose (342 g/mol) and a 1 m aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent.

    Example \(\PageIndex{7}\): Calculating the Boiling Point of a Solution

    What is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?

    Solution

    Use the equation relating boiling point elevation to solute molality to solve this problem in two steps.

    1. Calculate the change in boiling point.
    \(ΔT_\ce{b}=K_\ce{b}m=2.53\:°\ce C\:m^{−1}×0.33\:m=0.83\:°\ce C\)
    • Add the boiling point elevation to the pure solvent’s boiling point.
      \(\mathrm{Boiling\: temperature=80.1\:°C+0.83\:°C=80.9\:°C}\)

    Exercise \(\PageIndex{7}\)

    What is the boiling point of the antifreeze described in Example \(\PageIndex{4}\)?

    Answer

    109.2 °C

    Example \(\PageIndex{8}\): The Boiling Point of an Iodine Solution

    Find the boiling point of a solution of 92.1 g of iodine, I2, in 800.0 g of chloroform, CHCl3, assuming that the iodine is nonvolatile and that the solution is ideal.

    Solution

    We can solve this problem using four steps.

    This is a diagram with three boxes connected with two arrows pointing to the right. The first box is labeled, “Molality of solution,” followed by an arrow labeled, “1,” pointing to a second box labeled, “Change in boiling point,” followed by an arrow labeled, “2,” pointing to a third box labeled, “New boiling point.”

    1. Convert from grams to moles of I2 using the molar mass of I2 in the unit conversion factor.
      Result: 0.363 mol
    2. Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms.
      Result: 0.454 m
    3. Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes.
      Result: 1.65 °C
    4. Determine the new boiling point from the boiling point of the pure solvent and the change.
      Result: 62.91 °C

     

    Check each result as a self-assessment.

    Exercise \(\PageIndex{8}\)

    What is the boiling point of a solution of 1.0 g of glycerin, C3H5(OH)3, in 47.8 g of water? Assume an ideal solution.

    Answer

    100.12 °C

    Distillation of Solutions

    Distillation is a technique for separating the components of mixtures that is widely applied in both in the laboratory and in industrial settings. It is used to refine petroleum, to isolate fermentation products, and to purify water. This separation technique involves the controlled heating of a sample mixture to selectively vaporize, condense, and collect one or more components of interest. A typical apparatus for laboratory-scale distillations is shown in Figure \(\PageIndex{5}\).

    Figure a contains a photograph of a common laboratory distillation unit. Figure b provides a diagram labeling typical components of a laboratory distillation unit, including a stirrer/heat plate with heat and stirrer speed control, a heating bath of oil or sand, stirring means such as boiling chips, a still pot, a still head, a thermometer for boiling point temperature reading, a condenser with a cool water inlet and outlet, a still receiver with a vacuum or gas inlet, a receiving flask for holding distillate, and a cooling bath.

    Figure \(\PageIndex{5}\): A typical laboratory distillation unit is shown in (a) a photograph and (b) a schematic diagram of the components. (credit a: modification of work by “Rifleman82”/Wikimedia commons; credit b: modification of work by “Slashme”/Wikimedia Commons)

    Oil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure \(\PageIndex{6}\).

    This figure contains a photo of a refinery, showing large columnar structures. A diagram of a fractional distillation column used in separating crude oil is also shown. Near the bottom of the column, an arrow pointing into the column shows a point of entry for heated crude oil. The column contains several layers at which different components are removed. At the very bottom, residue materials are removed as indicated by an arrow out of the column. At each successive level, different materials are removed proceeding from the bottom to the top of the column. The materials are fuel oil, followed by diesel oil, kerosene, naptha, gasoline, and refinery gas at the very top. To the right of the column diagram, a double sided arrow is shown that is blue at the top and gradually changes color to red moving downward. The blue top of the arrow is labeled, “small molecules: low boiling point, very volatile, flows easily, ignites easily.” The red bottom of the arrow is labeled, “large molecules: high boiling point, not very volatile, does not flow easily, does not ignite easily.”

    Figure \(\PageIndex{6}\): Crude oil is a complex mixture that is separated by large-scale fractional distillation to isolate various simpler mixtures.

    Depression of the Freezing Point of a Solvent

    Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (Figure \(\PageIndex{7}\)), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).

    This is a photo of damp brick pavement on which a white crystalline material has been spread.

    Figure \(\PageIndex{7}\): Rock salt (NaCl), calcium chloride (CaCl2), or a mixture of the two are used to melt ice. (credit: modification of work by Eddie Welker)

    The decrease in freezing point of a dilute solution compared to that of the pure solvent, ΔTf, is called the freezing point depression and is directly proportional to the molal concentration of the solute

    \[ΔT_\ce{f}=K_\ce{f}m \label{11.5.9}\]

    where

    • \(m\) is the molal concentration of the solute in the solvent and
    • \(K_f\) is called the freezing point depression constant (or cryoscopic constant).

    Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of Kf for several solvents are listed in Table \(\PageIndex{1}\).

    Example \(\PageIndex{9}\): Calculation of the Freezing Point of a Solution

    What is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in Example \(\PageIndex{4}\)?

    Solution

    Use the equation relating freezing point depression to solute molality to solve this problem in two steps.

    This is a diagram with three boxes connected with two arrows pointing to the right. The first box is labeled, “Molality of solution,” followed by an arrow labeled, “1,” pointing to a second box labeled, “Change in freezing point,” followed by an arrow labeled, “2” pointing to a third box labeled, “New freezing point.”

    1. Calculate the change in freezing point. \[ΔT_\ce{f}=K_\ce{f}m=5.12\:°\ce C\:m^{−1}×0.33\:m=1.7\:°\ce C\]
    2. Subtract the freezing point change observed from the pure solvent’s freezing point. [\mathrm{Freezing\: Temperature=5.5\:°C−1.7\:°C=3.8\:°C}\]

    Exercise \(\PageIndex{9}\)

    What is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?

    Answer

    −9.3 °C

    Colligative Properties and De-Icing

    Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride.

    Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft (Video \(\PageIndex{1}\)).

    Video \(\PageIndex{1}\): Freezing point depression is exploited to remove ice from the control surfaces of aircraft.

    Phase Diagram for an Aqueous Solution of a Nonelectrolyte

    The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid. Phase diagrams for water and an aqueous solution are shown in Figure \(\PageIndex{8}\).

    This phase diagram indicates the pressure in atmospheres of water and a solution at various temperatures. The graph shows the freezing point of water and the freezing point of the solution, with the difference between these two values identified as delta T subscript f. The graph shows the boiling point of water and the boiling point of the solution, with the difference between these two values identified as delta T subscript b. Similarly, the difference in the pressure of water and the solution at the boiling point of water is shown and identified as delta P. This difference in pressure is labeled vapor pressure lowering. The lower level of the vapor pressure curve for the solution as opposed to that of pure water shows vapor pressure lowering in the solution. Background colors on the diagram indicate the presence of water and the solution in the solid state to the left, liquid state in the central upper region, and gas to the right.

    Figure \(\PageIndex{8}\): These phase diagrams show water (solid curves) and an aqueous solution of nonelectrolyte (dashed curves).

    The liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, ΔP, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, ΔTb, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, ΔTf, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed solvent only, and so transitions between these phases are not subject to colligative effects.

    Determination of Molar Masses

    Changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.

    Example \(\PageIndex{10}\): Determining Molar Mass from Freezing Point Depression

    A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?

    Solution

    We can solve this problem using the following steps.

    This is diagram with five boxes oriented horizontally and linked together with arrows numbered 1 to 4 pointing from each box in succession to the next one to the right. The first box is labeled, “Freezing point of solution.” Arrow 1 points from this box to a second box labeled, “delta T subscript f.” Arrow 2 points from this box to to a third box labeled “Molal concentration of compound.” Arrow labeled 3 points from this box to a fourth box labeled, “Moles of compound in sample.” Arrow 4 points to a fifth box labeled, “Molar mass of compound.”

    1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene (Table 11.5.1).
      \(ΔT_\ce{f}=\mathrm{5.5\:°C−2.32\:°C=3.2\:°C}\)
    2. Determine the molal concentration from Kf, the freezing point depression constant for benzene (Table 11.5.1), and ΔTf.
      \(ΔT_\ce{f}=K_\ce{f}m\)
      \(m=\dfrac{ΔT_\ce{f}}{K_\ce{f}}=\dfrac{3.2\:°\ce C}{5.12\:°\ce C m^{−1}}=0.63\:m\)
    3. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
      \(\mathrm{Moles\: of\: solute=\dfrac{0.62\:mol\: solute}{1.00\cancel{kg\: solvent}}×0.0550\cancel{kg\: solvent}=0.035\:mol}\)
    4. Determine the molar mass from the mass of the solute and the number of moles in that mass.
      \(\mathrm{Molar\: mass=\dfrac{4.00\:g}{0.034\:mol}=1.2×10^2\:g/mol}\)

    Exercise \(\PageIndex{10}\)

    A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?

    Answer

    1.8 × 102 g/mol

    Example \(\PageIndex{11}\): Determination of a Molar Mass from Osmotic Pressure

    A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?

    Solution

    Here is one set of steps that can be used to solve the problem:

    This is a diagram with four boxes oriented horizontally and linked together with arrows numbered 1 to 3 pointing from each box in succession to the next one to the right. The first box is labeled, “Osmotic pressure.” Arrow 1 points from this box to a second box labeled, “Molar concentration.” Arrow 2 points from this box to to a third box labeled, “Moles of hemoglobin in sample.” Arrow labeled 3 points from this box to a fourth box labeled, “Molar mass of hemoglobin.”

    1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure.
      \[\Pi=\mathrm{\dfrac{5.9\:torr×1\:atm}{760\:torr}=7.8×10^{−3}\:atm}\]
      \[\Pi=MRT\]
      \(M=\dfrac{Π}{RT}=\mathrm{\dfrac{7.8×10^{−3}\:atm}{(0.08206\:L\: atm/mol\: K)(295\:K)}=3.2×10^{−4}\:M}\)
    2. Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.
      \(\mathrm{moles\: of\: hemoglobin=\dfrac{3.2×10^{−4}\:mol}{1\cancel{L\: solution}}×0.500\cancel{L\: solution}=1.6×10^{−4}\:mol}\)
    3. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
      \(\mathrm{molar\: mass=\dfrac{10.0\:g}{1.6×10^{−4}\:mol}=6.2×10^4\:g/mol}\)

    Exercise \(\PageIndex{11}\)

    What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?

    Answer

    2.7 × 104 g/mol

    Colligative Properties of Electrolytes

    As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms 2 moles of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does.

    Example \(\PageIndex{13}\): The Freezing Point of a Solution of an Electrolyte

    The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Assume that each of the ions in the NaCl solution has the same effect on the freezing point of water as a nonelectrolyte molecule, and determine the freezing temperature the solution (which is approximately equal to the freezing temperature of seawater).

    Solution

    We can solve this problem using the following series of steps.

    This is a diagram with six boxes oriented horizontally and linked together with arrows numbered 1 to 5 pointing from each box in succession to the next one to the right. The first box is labeled, “Mass of N a C l.” Arrow 1 points from this box to a second box labeled, “Moles of N a C l.” Arrow 2 points from this box to to a third box labeled, Moles of ions.” Arrow labeled 3 points from this box to a fourth box labeled, “Molality of solution.” Arrow 4 points to a fifth box labeled, “Change in freezing point.” Arrow 5 points to a sixth box labeled, “New freezing point.”

    • Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor. Result: 0.072 mol NaCl
    • Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor (2 mol ions/1 mol NaCl). Result: 0.14 mol ions
    • Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms. Result: 1.1 m
    • Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes. Result: 2.0 °C
    • Determine the new freezing point from the freezing point of the pure solvent and the change. Result: −2.0 °C

    Check each result as a self-assessment.

    Exercise \(\PageIndex{13}\)

    Assume that each of the ions in calcium chloride, CaCl2, has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of CaCl2 in 175 g of water.

    Answer

    −0.208 °C

    Assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na+ and 1.0 mol Cl) per each kilogram of water, and its freezing point depression is expected to be

    \[ΔT_\ce{f}=\mathrm{2.0\:mol\: ions/kg\: water×1.86\:°C\: kg\: water/mol\: ion=3.7\:°C.} \label{11.5.11}\]

    When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.

    To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The van’t Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved:

    \[i=\dfrac{\textrm{moles of particles in solution}}{\textrm{moles of formula units dissolved}} \label{11.5.12}\]

    Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table \(\PageIndex{2}\).

    Table \(\PageIndex{2}\): Expected and Observed van’t Hoff Factors for Several 0.050 m Aqueous Electrolyte Solutions
    Electrolyte Particles in Solution i (Predicted) i (Measured)
    HCl H+, Cl 2 1.9
    NaCl Na+, Cl 2 1.9
    MgSO4 Mg2+, \(\ce{SO4^2-}\) 2 1.3
    MgCl2 Mg2+, 2Cl 3 2.7
    FeCl3 Fe3+, 3Cl 4 3.4
    glucose (a non-electrolyte) C12H22O11 1 1.0

    In 1923, the chemists Peter Debye and Erich Hückel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure \(\PageIndex{9}\)). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table \(\PageIndex{2}\) are for 0.05 m solutions, at which concentration the value of i for NaCl is 1.9, as opposed to an ideal value of 2.

    >The diagram shows four purple spheres labeled K superscript plus and four green spheres labeled C l superscript minus dispersed in H subscript 2 O as shown by clusters of single red spheres with two white spheres attached. Red spheres represent oxygen and white represent hydrogen. In two locations, the purple and green spheres are touching. In these two locations, the diagram is labeled ion pair. All red and green spheres are surrounded by the white and red H subscript 2 O clusters. The white spheres are attracted to the purple spheres and the red spheres are attracted to the green spheres.

    Figure \(\PageIndex{9}\): Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less.

    Summary

    Video \(\PageIndex{2}\): An overview of concentration units for solutions.

    In addition to molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components’ masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as molarity are not as commonly used.

    Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted.

    Key Equations

    • ΔTb = Kbm
    • ΔTf = Kfm
    • Π = MRT

    Footnotes

    1. A nonelectrolyte shown for comparison.

    Glossary

    mass percentage
    ratio of solute-to-solution mass expressed as a percentage
    mass-volume percent
    ratio of solute mass to solution volume, expressed as a percentage
    parts per billion (ppb)
    ratio of solute-to-solution mass multiplied by 109
    parts per million (ppm)
    ratio of solute-to-solution mass multiplied by 106
    volume percentage
    ratio of solute-to-solution volume expressed as a percentage
    boiling point elevation
    elevation of the boiling point of a liquid by addition of a solute
    boiling point elevation constant
    the proportionality constant in the equation relating boiling point elevation to solute molality; also known as the ebullioscopic constant
    colligative property
    property of a solution that depends only on the concentration of a solute species
    freezing point depression
    lowering of the freezing point of a liquid by addition of a solute
    freezing point depression constant
    (also, cryoscopic constant) proportionality constant in the equation relating freezing point depression to solute molality
    ion pair
    solvated anion/cation pair held together by moderate electrostatic attraction
    molality (m)
    a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms
    van’t Hoff factor (i)
    the ratio of the number of moles of particles in a solution to the number of moles of formula units dissolved in the solution

    Contributors

    • Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).

    • Adelaide Clark, Oregon Institute of Technology
    • Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes.