Solutions and Colloids

Degree of Solubility

Solubility and Temperature

Temperature changes affect the solubility of solids and gases differently. The solubility of solids increases with increasing temperature. Gas solubility decreases with increasing temperature.

The concentration at which no more solute dissolves in a solution for a given temperature and pressure is the saturation point. The molarity of a compound in a saturated solution is the molar solubility. An unsaturated solution is one in which more solute can be dissolved into the solution. If more solute is added to a saturated solution, the solute stays in its undissolved form.

A solution's solubility is a function of temperature. For a solid dissolved in a liquid, solubility usually increases with increasing temperature. More of the solid will dissolve in a warm solvent than in a cold one. This property can be used to produce a solution that is supersaturated, which means the solution contains more solute than can be dissolved in the solution under normal circumstances. First, a saturated solution of concentration Chot is made at a high temperature. If the resulting solution is allowed to cool slowly, the solvent can stay in solution even though the solute concentration Chot is greater than the solubility at the lower temperature, Ccool. Such a solution is unstable because the solvent particles still surround the solute particles, even though at the lower temperature the attractive forces between solute particles are stronger than the attractive forces between solvent and solute particles. Crystallization of the solute forms quickly if the solution is disturbed, for example, with the introduction of a solute crystal or even by a scratch on the wall of the container, because the solute particles are close enough to overcome the attraction of the solvent particles.

For gas solutions in a liquid solvent, solubility usually increases with decreasing temperature. The gas molecules are already dissociated and spread apart from one another, so they must be brought closer together in solution. Thus, the enthalpy change in each step is:

1. Gas molecules move together: ΔH1<0 \Delta H_1<0.

2. Solvent molecules spread apart: ΔH2>0\Delta H_2>0.

3. Solvent-gas molecules move together: ΔH3<0 \Delta H_3<0.

For this kind of solution, ΔHsoln\Delta H_{\rm{soln}} is the sum of two exothermic processes and one endothermic process, and ΔHsoln<0 \Delta H_{\rm{soln}}<0.

If the temperature of the solution is raised, the gas molecules become more energetic and tend to spread apart. This is why adding heat to a solution causes the gas molecules to bubble up and leave the solution (as in a pot of boiling water). The gas solubility decreases with increasing temperature.
The water solubility of sodium chloride (NaCl), a solid at room temperature, increases with increasing temperature. The solubility of carbon dioxide (CO2), a gas at room temperature, decreases with increasing temperature.

Solubility and Pressure

The total vapor pressure of a solution is the sum of the partial pressures of the individual gas components.
Pressure also affects solubility, but it is most noticeable for gases dissolved in liquids. Henry's law (formulated in 1803 by the British chemist William Henry) states that the amount of a gas that dissolves in a certain type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid at a specific temperature. This law can be written as an equation using C as the concentration of the gas, k as the constant of proportionality known as Henry's constant, and Pgas as the partial pressure of the gas.
C=kPgasC=kP_{\rm{gas}}
This principle can be understood by imagining a piston pressing down on a cylinder of gas over a liquid solution. If the piston pushes down with a greater force, the pressure on the gas increases, forcing the gas molecules closer together. Some of the molecules are dissolved into the solution because the gas molecules have a tendency to spread out as much as possible. Increased pressure increases the solubility. If the piston is lifted, the gas can expand and will come out of solution.
Step-By-Step Example
Application of Henry's Law
Henry's law can be used to calculate the change in pressure needed to decrease the concentration of carbon dioxide (CO2) in water at 20°C and 1.00 atm pressure from 958 mL/L CO2 to 535 mL/L CO2.
Step 1
To solve this problem, it is not necessary to know the numerical value of k. It is necessary to know only that the CO2 concentration and pressure are directly proportional. Start by rewriting Henry's law to isolate k.
CCO2=kPCO2k=CCO2PCO2\begin{aligned}C_{\rm{CO}_2}&=kP_{{\rm{CO}}_2}\\k&=\frac{C_{{\rm{CO}}_2}}{P_{{\rm{CO}}_2}}\end{aligned}
Step 2
Because this relationship is always true, the initial concentration and pressure of the gas can be related to the final concentration and pressure of the gas.
k=CCO2,iPCO2,i=CCO2,fPCO2,fk=\frac{C_{{\rm{CO}}_2,i}}{P_{{\rm{CO}}_2,i}}=\frac{C_{{\rm{CO}}_2,f}}{P_{{\rm{CO}}_2,f}}
Step 3
Raise both sides of the equation to the –1 power to invert the fractions. Then multiply both sides by CCO2,fC_{{\rm{CO}}_2,f} to solve for PCO2,fP_{{\rm{CO}}_2,f}.
PCO2,iCCO2,i=PCO2,fCCO2,fPCO2,iCCO2,fCCO2,i=PCO2,f\begin{aligned} \frac{P_{{\rm{CO}}_2,i}}{C_{{\rm{CO}}_2,i}}&=\frac{P_{{\rm{CO}}_2,f}}{C_{{\rm{CO}}_2,f}}\\\frac{P_{{\rm{CO}}_2,i} \,C_{{\rm{CO}}_2,f}}{C_{{\rm{CO}}_2,i}}&=P_{{\rm{CO}}_2,f}\end{aligned}
Solution
Substitute the known values, and solve for PCO2,fP_{{\rm{CO}}_2,f}.
PCO2,f=(1.00atm)(535mL/L)958mL/L=0.558atm\begin{aligned}P_{{\rm{CO}}_{2},f}&=\frac{(1.00\;{\rm{atm}})(535\;{\rm{mL/L})}}{958\;{\rm{mL/L}}}\\&=0.558\;{\rm{atm}}\end{aligned}

Henry's constant can be calculated if the pressure and concentration of a gas are known. Again consider the CO2 dissolved in water at 20°C and 1.00 atm pressure at a concentration of 958 mL/L H2O. The value of kCO2k_{\rm{CO}_2} can be calculated in units of mol/LPa given that 1.00 atm is 101.325 kPa and 1 mol of gas at standard temperature and pressure is 22.4 L (molar volume). First, convert CCO2,iC_{{\rm{CO}}_2,i} into mol/L:
CCO2,i=(958mLL)(L1,000mL)(mol22.4L)=0.04277mol/LC_{{\rm{CO}}_2,i}=\left(\frac{958\;\rm{mL}}{\rm{L}}\right)\!\left(\frac{\rm{L}}{1{,}000\;\rm{mL}}\right)\!\left(\frac{\rm{mol}}{22.4\;\rm{L}}\right)=0.04277\;\rm{mol}/\rm L
Then calculate k:
kCO2=CCO2PCO2=(0.04277mol/L)(101.325kPa)(1,000Pa1kPa)=4.22×107mol/LPak_{{\rm{CO}}_2}=\frac{C_{\rm{CO_2}}}{P_{\rm{CO_2}}}=\frac{(0.04277\;\rm{mol}/\rm{L})}{({101.325\;\rm{kPa}})\!\left(\frac{1\rm{,}000\;\rm{Pa}}{1\;\rm{kPa}}\right)}=4.22\times10^{-7}\;\rm{mol}/\rm{L}{\cdot}\rm{Pa}