Question

Calcium oxalate, CaC₂O₄.H₂O, is a sparingly soluble salt of analytical and physiological importance.The solubility product is 2.1 × 10^-9 at 25°C.Oxalate ions can protolyse to form hydrogen oxalate ions and oxalic acid. The pKa values at 25°C are 1.23 (H₂C₂O₄) and 4.28 (HC₂O^-₄ ). At 25°C the ionic product of water is 1.0 × 10^-14 . 

(a) State those expressions for the equilibrium conditions which are of interest for the calculation of the solubility of calcium oxalate monohydrate. 

(b) State the concentration conditions which are necessary for the calculation of the solubility s (in mol dm^-3) of calcium oxalate in a strong acid of concentration C. 

(c) Calculate the solubility (in g dm^-3) of calcium oxalate monohydrate in a plant cell in which the buffer system regulates the pH to 6.5. 

(d) Calculate the solubility (in g dm^-3) of calcium oxalate monohydrate in hydrochloric acid with a concentration of 0.010 mol dm^-3.Give the concentration of hydrogen ions in the solution. 

(e) Calculate the equilibrium concentrations of all other species in solution d). 

Answer 1

(c) The solubility of calcium oxalate monohydrate is 6.7 × 10^-3.(Calculated according to equation (8) 

(d) Elimination of the concentrations of oxalate species using equations (1), (3), and (4) yields the following expressions for (5) and (6). (The concentration of hydroxide ions can be neglected.) 

Elimination of s from (8) and (9) results in 4th order equation.For this reason, an iterative method is to be preferred. The first approximation is [H⁺] = C . This value of 

[H⁺] can be used to calculate:

(i) solubility s from (8), 

(ii) the last two terms in (9), which are corrections. Now a new value for [H⁺] obtained from (9) may be used as a starting value for the next approximation. Two repeated operations give the following value for s: 

s = 6.6 × 10^-4 mol dm^-3 = 9.6 × 10^-2 g dm^-3

[H⁺] = 9.3 × 10^-3 mol dm^-3