Exercise practice: Equilibrium

6.1 How the exercises are grouped

The chapter prints exercises 6.1 to 6.73. The exercise practice page groups them by skill and works a representative question in each group.

6.2 Group A: equilibrium constants

Exercise 6.2 gives the equilibrium concentrations of sulphur dioxide, oxygen and sulphur trioxide and asks for Kc of the reaction 2SO2 + O2 giving 2SO3. Write the equilibrium expression, square the sulphur trioxide concentration, and divide by the square of sulphur dioxide times oxygen.

Exercise 6.3 gives a vapour containing 40 percent iodine atoms by volume at a total pressure of 10^5 Pa. The partial pressure of each species follows from the volume fractions: the iodine atoms have 0.40 of the total pressure and the molecules 0.60. Kp is then the square of the atom partial pressure divided by the molecule partial pressure, which gives about 2.67 x 10^4 Pa.

Exercise 6.5 converts Kp to Kc. For 2NOCl giving 2NO + Cl2 the change in gas moles is plus one, so Kc equals Kp divided by RT. For calcium carbonate decomposing to calcium oxide and carbon dioxide, the solid species are left out and the change in gas moles is one, so the same conversion applies with the carbon dioxide pressure alone.

Exercise 6.33 uses Kc for the formation of ozone from oxygen. The reaction is 3O2 giving 2O3 with a very small Kc, so at atmospheric oxygen concentration the ozone concentration is extremely small, which is why the equilibrium lies far on the oxygen side.

6.3 Group B: Le Chatelier's principle

Exercises 6.15 to 6.26 test the direction of shifts. Adding a reactant pushes the equilibrium forward, removing a product also pushes it forward, and increasing the pressure pushes it towards fewer gas moles. Raising the temperature pushes it in the endothermic direction and changes K. A catalyst changes only the rate.

Exercise 6.1 is a physical example. A liquid in equilibrium with its vapour is suddenly given a larger volume. The vapour pressure immediately falls because the same vapour occupies more space, the rate of evaporation becomes larger than the rate of condensation, and the system shifts towards more vapour. When equilibrium returns, the vapour pressure is back at the same value as before, because vapour pressure depends only on temperature and not on the volume of the container.

6.4 Group C: acids, bases and pH

Exercise 6.41 gives a hydrogen ion concentration of 3.8 x 10^-3 M and asks for the pH. The pH is minus log of that concentration, which is about 2.42. Exercise 6.42 reverses the calculation: a pH of 3.76 for vinegar means the hydrogen ion concentration is 10^-3.76, which is 1.7 x 10^-4 M.

Exercise 6.46 asks for the degree of dissociation and pH of 0.05 M acetic acid with Ka equal to 1.74 x 10^-5. The hydrogen ion concentration is the square root of Ka times c, about 9.3 x 10^-4 M, the pH is about 3.03, and the degree of dissociation is the hydrogen ion concentration divided by c, about 0.019 or 1.9 percent.

Exercise 6.52 asks for the pH of 0.001 M aniline using its Kb. The hydroxide concentration is the square root of Kb times c, the pOH follows, and the pH is 14 minus the pOH. Exercise 6.54 does the same for dimethylamine and then asks how the percentage ionised falls when sodium hydroxide is added, which is the common ion effect.

Exercise 6.55 converts the pH of biological fluids into hydrogen ion concentrations. Muscle fluid at pH 6.83 gives 1.5 x 10^-7 M, stomach fluid at pH 1.2 gives 6.3 x 10^-2 M, blood at pH 7.38 gives 4.2 x 10^-8 M and saliva at pH 6.4 gives 4.0 x 10^-7 M. Exercise 6.57 and 6.58 work from a mass of base to its pH through molarity and the hydroxide concentration.

6.5 Group D: buffers, hydrolysis and solubility product

Exercise 6.64 asks for two pH values, one for a 0.1 M chloroacetic acid solution and one for a 0.1 M solution of its sodium salt. The acid uses the square root formula, while the salt hydrolyses with Kh equal to Kw/Ka and gives a basic solution whose pH comes from the hydrolysis relation.

Exercise 6.65 uses the ionic product of water at 310 K, 2.7 x 10^-14, to find the pH of neutral water at that temperature. In neutral water the hydrogen and hydroxide concentrations are equal, so each is the square root of Kw, and the pH is about 6.78 rather than 7. The neutral point of the pH scale therefore depends on temperature.

Exercise 6.66 mixes strong acid and strong base solutions and asks for the pH of the resulting mixture. Work in moles of hydrogen and hydroxide ions, find which is in excess after mixing, divide the excess by the total volume, and calculate pH or pOH from the excess concentration.

Exercise 6.67 asks for the solubilities of several salts from their Ksp values. Write the expression from each dissociation equation, substitute the solubility, and solve. Silver chromate gives two silver ions and one chromate ion, so its Ksp is 4s cubed, while silver bromide gives one of each and its Ksp is s squared.

Exercise 6.68 finds the ratio of the molarities of saturated silver chromate and silver bromide solutions from their Ksp values. Solve for each solubility and divide, which gives the answer in terms of a cube root and a square root.

Exercise 6.69 mixes sodium iodate and cupric chlorate and asks whether copper iodate precipitates. Calculate the ion product after mixing, compare it with Ksp, and answer accordingly. Exercise 6.71 uses the same comparison to find the maximum concentrations that avoid precipitation, and exercise 6.73 tests the precipitation of sulphides in acidic solution, which is the basis of the separation of groups in qualitative analysis.