Solutions and Colloids

Colligative Properties

Calculating Concentrations

Concentration reported as molality (moles of solute divided by the mass in kilograms of the solvent) and mole fraction (moles of solvent divided by moles of solution) is independent of the volume of the solvent.

Concentrations can be calculated in several different sets of units. Molarity (M) is the number of moles of a solute dissolved in 1 liter of solution: 1M=1mol/L1\;{\rm M=1}\;{\rm{mol}}/{\rm L}. Molality (m) is concentration expressed as the number of moles of a solute divided by the mass of solvent in kilograms: 1m=1mol/kg1\;{\rm m=1}\;{\rm{mol}}/{\rm{kg}}. Molality is linked with the mass of the solvent, and molarity is linked with the volume of the solvent. Because volume is a function of temperature, it changes with the rise or fall in temperature. But mass is not a function of temperature. Neither the number of moles of the solute nor the mass of the solvent is affected by changes in temperature. Mass, and therefore molality, does not depend on temperature. Liquids expand as they warm, so if a 1 M solution is prepared at 20°C and the temperature is raised to 25°C, the volume of the solvent will increase slightly, and the molar concentration (in M) therefore will decrease slightly—the same mass of solute in a larger volume of solvent. The molal concentration (in m) of that same solution would remain unchanged—the same mass of solute in the same mass of solvent.

Mole fraction (χ\chi) is concentration expressed as the moles of solvent divided by the total number of all moles in solution. For example, in a solution that contains 0.50 mol H2O, 0.70 mol CH3OH, and 0.30 mol CH3CH2OH, their respective mole fractions are:
χH2O=0.50mol(0.50+0.70+0.30)mol=0.33χCH3OH=0.70mol(0.50+0.70+0.30)mol=0.47χCH3CH2OH=0.30mol(0.50+0.70+0.30)mol=0.20\begin{aligned}\chi_{\rm {H}_2{O}}&=\frac{0.50\;{\rm{mol}}}{(0.50+0.70+0.30)\;{\rm{mol}}}=0.33\\\\\chi_{\rm{CH}_3{OH}}&=\frac{0.70\;{\rm{mol}}}{(0.50+0.70+0.30)\;{\rm{mol}}}=0.47\\\\\chi_{\rm{CH}_3{CH}_2{OH}}&=\frac{0.30\;{\rm{mol}}}{(0.50+0.70+0.30)\;{\rm{mol}}}=0.20\end{aligned}
The mole fraction is a proportion. It therefore has no unit, and the sum of all mole fractions of a solution is always 1.
Step-By-Step Example
Calculation of Molality
A 1.00 M solution of sodium hydroxide (NaOH) at 25.0°C has a density of 1.040 g/mL. What is the molality of the solution?
Step 1
Calculate the molar mass of NaOH using the atomic masses of each element from the periodic table.
molar mass of NaOH=(1)(22.99g/mol)+(1)(16.00g/mol)+(1)(1.01g/mol)=40.00g/mol\begin{aligned}{\text{molar mass of }\rm{NaOH}}& = (1)(22.99\;{\rm{g/mol}})+(1)(16.00\;{\rm{g/mol}})+(1)(1.01\;{\rm{g/mol}})\\&= 40.00\;{\rm{g/mol}}\end{aligned}
Step 2
Rewrite the concentration using the molar mass of NaOH to convert the given molarity to mass of NaOH per liter of solution.
1.00MNaOH=(1.00molNaOH1L)(40.00gNaOH1molNaOH)=40.00gNaOH1L\begin{aligned}1.00\;{\rm{M}}\;\rm{NaOH}&=\left(\frac{1.00\;{\rm{mol}}\;\rm{NaOH}}{1\;\rm{L}}\right)\!\left(\frac{40.00\;{\rm{g}}\;\rm{NaOH}}{1\;{\rm{mol}}\;\rm{NaOH}}\right)\\&=\frac{40.00\;{\rm{g}}\;\rm{NaOH}}{1\;{\rm{L}}}\end{aligned}
Step 3
Use the density of the solution to find the mass of 1 L of solution:
(1Lsoln)(1.040gmL)(1,000mLL)=1040g\left(1\;\rm{L}\;\rm{soln}\right)\!\left(\frac{1.040\;\rm{g}}{\rm{mL}}\right)\!\left(\frac{1{,}000\;\rm{mL}}{\rm{L}}\right)=1040\;\rm{g}
Step 4
Because 1 L of solution contains 40.00 g NaOH, the remaining mass must therefore be water.
mass of water=mass of solutionmass of NaOH=1,040g40.00gNaOH=1,000gwater=1.000kgwater\begin{aligned}\text{mass of water}&=\text{mass of solution}-\text{mass of }\rm{NaOH}\\&=1{,}040\;{\rm{g}}-40.00\;{\rm{g}}\;\rm{NaOH}\\&=1{,}000\;{\rm{g}}\;\text{water}\\&=1.000\;{\rm{kg}}\;\text{water}\end{aligned}
Solution
Use the mass of water to rewrite the concentration as molality.
molality=1.00molNaOH1.000kgH2O=1.00m{\text{molality}}=\frac{1.00\;{\rm{mol}}\;{\rm{NaOH}}}{1.000\;{\rm{kg}}\;{{\rm H}_2}{\rm{O}}}=1.00\;{\rm{m}}

The solution component can be described in terms of the mole fraction. Remember that the mole fraction is unitless. It is a ratio of the number of moles of each component to the total number of moles. NaOH, however, is a strong electrolyte, so 1 mol of NaOH completely dissociates and yields 2 mol of ions:
NaOHNa++OH{\rm{NaOH}}\rightarrow{\rm{Na}^+}+{\rm{OH}^-}
The mole fractions must therefore be calculated for the ions, not for NaOH. The first step is to use the molar mass of water, 18.02 g/mol, to determine the moles of water present:
(1,000gH2O)(1mol18.02g)=55.49molH2O(1{,}000\;{\rm{g}}\;{\rm{H}}_2{\rm{O})}\left(\frac{1\,{\rm{mol}}}{18.02\;{\rm{g}}}\right)=55.49\;{\rm{mol}}\;{\rm{H}}_2{\rm{O}}
The total number of moles in solution is the moles of water plus 1.00 mol Na+ and 1.00 mol OH:
total moles of solution=55.49mol+1.00mol+1.00mol=57.49mol\begin{aligned}\text{total moles of solution}&=55.49\;{\rm{mol}}+1.00\;{\rm{mol}}+1.00\;{\rm{mol}}\\&=57.49\;{\rm{mol}}\end{aligned}
The mole fractions are therefore:
χH2O=55.49mol57.49mol=0.965χNa=χOH=1.00mol57.49mol=0.0174\begin{gathered}{\chi_{\rm{H}_2{O}}=\frac{55.49\;{\rm{mol}}}{57.49\;{\rm{mol}}}=0.965}\\\\{\chi_{\rm{Na}}=\chi_{\rm{OH}}=\frac{1.00\;{\rm{mol}}}{57.49\;{\rm{mol}}}=0.0174}\end{gathered}
Molality and mole fraction are ratios of the number of solute particles to the number of solvent particles in a given solution. This ratio is important when working with a solution's colligative properties, which are physical properties that depend only on the number of solute particles present—the total concentration of solute particles—and not the identity of those particles. Vapor pressure lowering, freezing point depression, boiling point elevation, and osmotic pressure are all colligative properties of substances.

Vapor Pressures, Melting Points, and Boiling Points

The vapor pressures and melting points of solutions are lower than those of the pure solvent. The boiling points are elevated.
Vapor pressure is the pressure exerted by the particles that have evaporated off the surface of a solution. In 1887 François-Marie Raoult proposed that in a liquid solution, all the solution components, the solvent, and any solutes, contribute to the overall vapor pressure of the solution (Ptot) in proportion to their mole fractions (χ\chi) in the solution. Today this is known as Raoult's law and can be mathematically stated as
Ptot=(χA)(PA)+(χB)(PB)+{P_{\rm{tot}}}=\left({\chi_A}\right)\!\left({P_A^\circ}\right)+\left({\chi_B}\right)\!\left({P_B^\circ}\right)+\mathellipsis
where each PAP_A^\circ term is the vapor pressure of a pure solute at the same temperature and pressure of the solution. The value of PA=(χA)(PA)P_A=\left(\chi_A\right)\!\left(P_A^\circ\right) is called the partial pressure of component A. Because the χi<1 \chi_i<1, the partial pressure of a solution component is always smaller than P°, so mixing a solution lowers the vapor pressure of the solution's components. Note that nowhere in these equations is the identity of A indicated. It is not needed. The lowering of a substance's vapor pressure is a colligative property that depends only on the amount of that substance present in solution. It depends only on the mole fraction.

Because the vapor pressure of a solute A is lowered when it is part of a solution, the phase diagram (graph relating a substance's phases of matter, temperature, and pressure) of that solute is changed. Remember that where the vapor pressure curve intersects the sublimation curve defines the beginning of the solid-liquid transition curve—that is, the freezing curve—of a substance. If the vapor pressure is lowered, the curve intersects the sublimation (phase transition from a solid to a gas at a constant temperature and pressure) curve at a lower temperature, and the freezing point of the substance is also lowered. This is called freezing point depression.

In addition, because the vapor pressure is lowered, the vapor-liquid transition curve intersects the P=1atm P=1\;{\rm{atm}} at a higher temperature than the curve for the pure substance. Because the boiling point of a substance is defined as the temperature at which Pvap=1atm P_{\rm{vap}}=1\;{\rm{atm}}, this means the boiling point for a component in a solution has been raised. This is called boiling point elevation.

Recall that freezing point depression and boiling point elevation are colligative properties. Therefore, the amount by which the freezing point is lowered and the boiling point is elevated is proportional only to the mole fraction of the solute present, as with vapor pressure. This is represented mathematically by
ΔTf=(Kf)(χ)=KfmΔTb=(Kb)(χ)=Kbm\begin{gathered}{\Delta{T}_f=\left(K_{f}\right)\!\left(\chi\right)=K_{f}\rm{m}}\\\\{\Delta{T}_b=\left(K_{b}\right)\!\left(\chi\right)=K_{b}\rm{m}}\end{gathered}
where Kf and Kb are proportionality constants that depend on the solvent and m is the molality (moles of solute per mass, in kilograms, of solvent) of the solute. In dilute solutions, the mole fraction of a solute is proportional to its molality, so m can be substituted into these equations for χ\chi. The constants Kf and Kb are properties of the solvent that are determined experimentally.

Because ΔTf{\Delta T}_f and ΔTb{\Delta T}_b do not depend on the identity of the species (any specific type of chemical particle, such as a water molecule or a sodium ion) in solution, if Kf for the solvent is known, the molar mass of a solute can be calculated with precise measurement of ΔTf{\Delta T}_f.

Step-By-Step Example
Using Freezing Point Depression to Calculate the Molar Mass of a Solute
Consider a solution prepared by adding 4.988 g acetone to 997.5 g water. This solution freezes at –0.16°C, and the Kf for water is 1.86 (°C kg water)/(mol acetone). Use these data to calculate an experimental value for the molar mass of acetone.
Step 1
Use the change in temperature and Kf to calculate the molality of acetone.
ΔTf=Kfmacetonemacetone=ΔTfKf=0.16°C1.86°Ckgwatermolacetone=0.08602molacetonekgwater\begin{aligned}\Delta{T}_f&=K_{f}\rm{m_{\rm{acetone}}}\\\rm{m_{\rm{acetone}}}&=\frac{\Delta{T}_f}{K_{f}}\\&=\frac{-0.16\degree\rm{C}}{1.86\,\displaystyle\frac{\degree\rm{C}\;{\rm{kg}}\;\text{water}}{{\rm{mol}}\;{\text{acetone}}}}\\&=0.08602\,\frac{{\rm{mol}}\;\text{acetone}}{\rm{kg}\;\text{water}}\end{aligned}
Step 2
To calculate the molar mass of acetone, the amount of acetone used in the experiment expressed in both grams and in moles must be determined. The amount in grams is given. Use the molarity of acetone and the given mass of water in the experiment to determine the number of moles of acetone.
moles of acetone=macetone(mass of water)=(0.08602molacetonekgwater)(997.5gwater)(1kg1,000g)=0.08580molacetone\begin{aligned}\text{moles of acetone}&=\rm{m}_{\rm{acetone}}(\text{mass of water})\\&=\left(0.08602\,\frac{{\rm{mol}}\;{\text{acetone}}}{{\rm{kg}}\;\text{water}}\right)\!(997.5\;{\rm{g}}\;\text{water})\!\left(\frac{1\;{\rm{kg}}}{1{,}000\;{\rm{g}}}\right)\\&=0.08580\;{\rm{mol}}\;\text{acetone}\end{aligned}
Solution
Use the mass of acetone and the moles of acetone to calculate the experimental value of the molar mass.
experimental molar mass=mass of acetonemoles of acetone=4.988g0.08580mol=58.1g/mol\begin{aligned}\text{experimental molar mass}&=\frac{\text{mass of acetone}}{\text{moles of acetone}}\\&=\frac{4.988\;{\rm{g}}}{0.08580\;{\rm{mol}}}\\&=58.1\;{\rm{g/mol}}\end{aligned}
The chemical formula for acetone is (CH3)2CO, and the true molar mass is 58.09 g/mol, so the experimental value is very close to the actual value.

The van 't Hoff Factor

The increased mole number introduced by dissociated ions affects the colligative properties of a solution.

The mole fraction of an electrolytic solution depends on the degree to which the electrolyte dissociates, and 1 mol of a strong electrolyte produces more than 1 mole of solute particles in solution. The number of moles depends on the stoichiometry (quantitative relationships among different elements and substances) of the electrolyte. For example, NaCl produces 2 moles, and MgCl2 produces 3 moles. Weak electrolytes, such as CH3COOH, only partially dissociate, so the number of particles resulting from 1 mol of a weak electrolyte is more than 1 but less than the stoichiometric ratio.

Because colligative properties depend only on the number of particles in solution, electrolytes dissolved in water have a greater effect on the ΔTf\Delta T_f and ΔTb\Delta T_b than nonelectrolytes. This can be accounted for by including a factor i, called the van 't Hoff factor, in the equations for ΔT\Delta T:
ΔTf=iKfmΔTb=iKbm\begin{gathered}{\Delta T_f={i}K_{f}{\rm{m}}}\\\\{\Delta T_b={i}K_{b}{\rm{m}}}\end{gathered}
For nonelectrolyte solutions, such as acetone, i=1 i=1. For electrolytes, i is approximately equal to the number of moles produced by the dissociation of the electrolyte, depending on the concentration of the solution.
That the van 't Hoff factor depends on solution concentration may seem surprising, but it is because the motions of ions in solution are affected by one another. Ionic particles in solution are surrounded by particles with the opposite charge, so their mobility is reduced. Colligative properties depend on freely moving particles, so the effect of colligative properties is reduced in an ionic solution. The more dilute the solution, the more the ions are able to move, and the closer the van 't Hoff factor gets to its stoichiometric value.
This chart plots the van 't Hoff factor as a function of concentration. The van 't Hoff factor for most ionic solutions does not approach the stoichiometric ratio except at very low concentrations, so colligative properties are diminished in electrolytic solutions.

Distillation and Osmosis

The reduced vapor pressure in a solution is responsible for distillation (separation of liquids by boiling) and osmotic pressure (pressure needed to stop the flow of water through a semipermeable membrane).

Because chemical purification is often the goal in chemistry, it is sometimes desirable to separate solutions of two liquids. This can be accomplished by a process called fractional distillation, which is the separation of two liquids by boiling in stages. Raoult's law states that the partial pressure of a solute is proportional to its mole fraction in solution. The vapor pressures and liquid composition of a solution exist in an equilibrium that can be plotted as a function of the mole fraction. This is called a graph of the liquid-vapor equilibrium.

Simple distillation is separation by boiling a solution, collecting the vapor, and recondensing it. Salt water can be purified through this process. When it boils, the vapor pressure of water is so much lower than the vapor pressure of salt that the water boils off before the salt, and the resulting condensate is pure water.

Fractional distillation is used when the condensate from simple distillation is not pure, as in the case of a benzene-toluene mixture. The vapor pressures of benzene and toluene are close enough to each other that the vapor that is collected is a mixture of benzene and toluene vapors. However, because the Pvap values are not the same, the mole fraction of benzene in the vapor solution is higher than the χB\chi_{\rm{B}} in the liquid solution. Thus, if this vapor solution is condensed and redistilled, the vapor will be even more concentrated in benzene. Pure benzene can be collected after multiple distillation stages, and pure toluene is left behind.

Liquid-Vapor Equilibrium

This chart plots the liquid and vapor composition for a toluene-benzene solution as a function of the mole fraction of benzene. It also includes the partial pressures of benzene and toluene. Note that the sum of Pvap at point A (partial pressure of toluene when the liquid is 50 percent benzene) plus the Pvap at point B (partial pressure of benzene when the liquid is 50 percent benzene) equals the Pvap at point C, the total Pvap. The red line from point C to D is called the tie line and connects the mole fraction of benzene in the liquid phase with the corresponding mole fraction in the vapor phase. When the liquid is 50 percent benzene, the vapor is 80 percent benzene.
Distillation is possible because the vapor pressure of a species is lowered when it is part of a solution. Osmosis also occurs because of the lowering of Pvap. Osmosis can be demonstrated using a U-shaped tube open to the atmosphere at both ends. The tube is divided into two sides by a semipermeable membrane, a barrier through which certain types of particles can pass but not others. In biological systems a semipermeable membrane may consist of a lipid (fat molecule) bilayer that restricts the flow of charged particles, but the membrane used in the osmosis demonstration is a simple film with microscopic holes restricting particles according to size. Sucrose (sugar) is poured into the water on one side of the tube, forming a sucrose solution. In osmosis, water (consisting of small molecules) can pass through the semipermeable membrane, but the sucrose (consisting of larger molecules) cannot. The vapor pressure over the pure water is higher than the vapor pressure over the sucrose solution because the Pvap of a solution is lower than the Pvap of a pure substance. This difference in Pvap causes a net force down on the pure water, which in turn causes the water molecules to pass through the semipermeable membrane into the sucrose solution until the mole fraction of water in the sucrose solution is enough to equalize the pressures. This flow of water can be stopped if an external pressure, perhaps in the form of a piston, presses on the sucrose solution. The pressure needed to stop the flow of water through a semipermeable membrane is the osmotic pressure (π\bold\pi), another colligative property. Osmotic pressure can be calculated from the molarity of the solution (M), which is calculated from the number of moles—not the identity—of the solute and the gas constant, where n is the number of moles of the solute.
π=MRT=nRTV{\rm\pi}=MRT=\frac{nRT}V
Because it is a colligative property, osmotic pressure can be used to determine molar mass. Changes in osmotic pressure are far more drastic than ΔT{\Delta T} in freezing point depression or boiling point elevation, so using osmotic pressure is a more accurate method.

Osmotic Pressure

If sucrose sugar (C12H22O11) is added to one side of a U-tube full of water divided by a semipermeable membrane, water will flow into the side that contains sugar in an attempt to equalize the concentrations. If pressure equal to the osmotic pressure (π\pi) is applied on the sugar side of the membrane, the liquid levels will return to equal heights.
Step-By-Step Example
Using Osmotic Pressure to Calculate the Molar Mass of a Solute
A sucrose solution with volume 500.0 mL is prepared using 0.168 g of sucrose at 25.0°C. The solution is experimentally determined to have an osmotic pressure of 18.2 mm Hg. Use this data to calculate the molar mass of sucrose, C12H22O11, and compare it to its known molecular weight of 342.34 g/mol.
Step 1
First, convert the osmotic pressure to atmospheres:
π=(18.2mmHg)(1atm760mmHg)=0.02395atm{\rm\pi=(18.2\;{mm\;Hg}})\!\left(\frac{1\;{\rm{atm}}}{760\;{\rm{mm\;Hg}}}\right)=0.02395\;\rm{atm}
Step 2
Then, use the equation for osmotic pressure to calculate the number of moles of sucrose.
π=nRTVn=πVRT=(0.02395atm)(0.5000L)(0.082057atmL/Kmol)(298.15K)=4.895×104molsucrose\begin{aligned}\pi&=\frac{nRT}V\\n&=\frac{\pi{V}}{RT}\\&=\frac{(0.02395\;{\rm{atm}})(0.5000\;\rm{L})}{(0.082057\;\rm{atm}{\cdot}\rm{L/K}{\cdot}\rm{mol})(298.15\;{\rm{K}})}\\&=4.895\times10^{-4}\;{\rm{mol}}\;{\text{sucrose}}\end{aligned}
Step 3
Finally, use the given mass of sucrose to calculate an experimental value for the molar mass of sucrose.
msucrose=mass of sucrosemoles of sucrose=0.168g4.895×104mol=343g/mol\begin{aligned}\rm{m}_{\rm{sucrose}}&=\frac{\text{mass of sucrose}}{\text{moles of sucrose}}\\&=\frac{0.168\;\rm{g}}{4.895\times10^{-4}\;\rm{mol}}\\&=343\;\rm{ g/mol}\end{aligned}
Solution
The experimental value, 343 g/mol, is close to the known molar mass of sucrose, 342.34 g/mol.

Osmotic pressure is of particular importance in biology because cell membranes are semipermeable. Therefore, any fluids introduced into the body must be isotonic, which means they must have the same osmotic pressure as bodily fluids. A solution is hypertonic if it is at a pressure higher than the osmotic pressure of bodily fluids. Water flows out of cells in a hypertonic solution, and the cells shrivel. A solution is hypotonic if it is at a pressure lower than the osmotic pressure of bodily fluids. The cells in a hypotonic solution swell. Finally, if a pressure greater than the osmotic pressure is applied to the more concentrated solution, water flows from the more concentrated side to the less concentrated side in order to equalize the pressure. This is called reverse osmosis and can be used to remove contaminants from water.