Chemistry
Exercise 5
5. Solve simple problems based on solution equilibria using solubility product constants (Ksp), acid and base ionization constants (Ka and Kb).
Solution equilibria are involved in many environmental phenomena such as the solubility of gases and ionic substances in water, acid rain and the carbonate cycle. It is a complex area but the fundamentals of equilibria such as LeChaleliers principle, pH and the calculation of equilibrium concentrations should be understood.
a) Henrys law may be stated as c = k x P where c is the concentration (in mol/L or M) of a gas in solution, P is the partial pressure of the gas above the solution and k is a constant for a particular gas in a particular solvent. For aqueous solutions of CO2 at 25 C, k = 3.38 x 10-2 mol/L.atm. Calculate how many moles of CO2 will dissolve in 500 ml of water when the partial pressure of CO2 is 1.00 atm. How many moles of CO2 will dissolve in 500 ml of water which is exposed to the atmosphere (1.00 atm total pressure) which contains 0.0350% CO2?
1.6g x 10-2 mol CO2; 5.9 x 10-6 mol CO2
Using the above equation, c= k x P and c= n/v, the number of moles of CO2 that dissolves is given by
This question is similar to the above question except the partial pressure CO2 needs to be calculated;
PCO2= 1.00 atm x 0.00035 = 0.00035 atm
b) At 25C the solubility product Ksp of calcium carbonate, CaCO3, is 8.7 x 10-9. Calculate the maximum amount (in moles and grams) of CaCO3that will dissolve in 10.0 L of water at 25C. How much will dissolve in 10.0 L of a 0.100 M aqueous solution of sodium carbonate, Na2CO3, at 25C?
The solubility product (Ksp) is a measure of how much a given solid is dissociated at equilibrium, a large Ksp value indicates that the solid is highly soluble, while a low Ksp indicates little dissolution of the solid.
The dissolution of CaCO3 can be expressed using the following equation:
| CaCO3(s) | Ca2+(aq) |
+ | CO32-(aq) |
||
| Initial | 0 | 0 | |||
| Change | -x |
+x |
+x |
||
| Equilibrium | x |
x |
One molecule of CaCO3 yields 1 equivalent of Ca2+and CO32- and Ksp = [Ca2+][ CO32-] . Assuming that an equal amount of each ion is present at equilibrium and that amount is equal to the total amount of CaCO3 dissolved, the Ksp equation can be manipulated to predict the amount of solid that dissolves. By using x to represent the change in ion concentration from an initial concentratin of 0 to the equilibrium concentration,
Ksp = [x][x] = 8.7x10-9, solving for x, x=9.3x10-5
The amount of CaCO3that dissolves is equal to x, 9.3x10-5 mol/L
The amount of CaCO3 in 10 L =
(9.3x10-5 mol CaCO3/L)x(10L)x(100 g/mol CaCO3 = 9.3x10-2 g
Using the same Ksp equation, this question also incorporates the effects of the common ion effect (see 4f). the presence of Na2CO3 in solution will affect the overall solubility of CaCO3.
| CaCO3(s) | Ca2+(aq) |
+ | CO32-(aq) |
||
| Initial | 0 | 0.1M |
|||
| Change | -x |
+x |
+x |
||
| Equilibrium | x |
0.1+x |
Using equilibrium values in the Ksp equation results in
Ksp = [x][0.1+x] = 8.7x10-9
Assuming less than 5% dissolution (always check at the end of your calculation), x << 0.1 so the equation can be simplified to read:
Ksp = [x][0.1] = 8.7x10-9
Or
[x] = 8.7x10-8
In this case the amount of CaCO3 that dissolves in solution is 8.7x10-8 mol/L
The amount of CaCO3 in 10 L =
(8.7x10-8mol CaCO3/L)x(10L)x(100 g/mol CaCO3) = 8.7x10-5 g
c) Normal household ammonia consists of an approximately 0.80 M solution of ammonia, NH3, in water. The Kb value for ammonia is 1.75 X 10-5 at 25C. Calculate the pH of this solution.
pH = 11.6
Kb is similar to Ka, except instead of measuring the dissociation of an acid into a conjugate base and proton, Kb measures the dissociation of a base into its conjugate acid and a hydroxide ion
This problem can be solved in a similar manner to question 4f, by considering the initial and equilibrium concentrations of the species present in solution. The reaction of ammonia and water is given below
NH3 |
+ | H2O |
NH4 | + | OH- |
||
| Initial | 0.8M |
0 | 0 | ||||
| Change | -x |
+x |
+x |
||||
| Equilibrium | 0.8M-x |
x |
x |
Substituting the above information into the Kb formula and simplifying the equation;
x = 3.74 x 10-3
To determine the pH of the solution, solve for the OH- concentration and calculate pH
d) A 0.25 M solution of a weak acid HX has a pH = 2.86. What is the ionization constant Ka of the acid?
Ka = 7.3 x 10-6
This problem can be solved using the Ka equation and the known concentration of protons
Assuming that 1 equivalent of HX produces one equivalent each of H+ and X-, Ka is calculated as
