Question

Iodine is soluble to a certain extent in pure water. It is, however, more soluble in solutions containing iodide ions.By studying the total solubility of iodine as a function of iodide concentration, the equilibrium constants of the following reactions can be determined:

| Equation | Equilibrium constants |
| I2()⇋ I2(aq) | k1 | (1) |
| I2() + I– (aq)⇋ I−3 (aq) | k2 | (2) |
| I2(aq) + I– (aq)⇋ I-3 (aq) | k3 | (3) |

a) Give the equilibrium equations for (1) –

(3). Solutions of known potassium iodide concentration [I– ]tot were equilibrated with solid iodine. Subsequent titration with sodium thiosulphate solution served to determine the total solubility of iodine [I2]tot.The experiments yielded the following results:

| [I– ]tot / mmol dm-3 | 10.00 | 20.00 | 30.00 | 40.00 | 50.00 |
| [I– ]tot / mmol dm-3 | 5.85 | 10.53 | 15.11 | 19.96 | 24.82 |

(b) Plot [I2]tot versus [I– ]tot in a diagram.

(c) Derive a suitable algebraic expression relating [I2]tot and [I– ]tot.

(d) Use the graph to determine values of the equilibrium constants k1,k2, and k3.

Answer 1

(a) Equilibrium equations The following relations are valid for the concentrations of the aqueous solutions:

(b) See diagram on the next page.

(c) The relation between [I2]tot and [I– ]tot is as follows:

(These values are calculated by the least square method.)