Question

Calcium sulphate is a sparingly soluble compound. Its solubility product is given by: 

Ks(CaSO₄) = [Ca²⁺][ SO^2-₄] = 6.1 × 10^-5 

Ethylenediaminetetraacetic acid (EDTA) has the formula C₁₀H₁₆N₂O₈ and the structure: 

The anion of this acid, C₁₀H₁₂N₂O₈^4-, forms a stable complex CaC₁₀H₁₂N₂O₈^2- with calcium ions. The stability constant of this complex ion is given by: 

K = [ CaC₁₀H₁₂O^2-₈]/Ca²⁺][C₁₀H₁₂N₂O^4-₈] =  1.0 1011

EDTA is completely dissociated in strongly alkaline solution. The equation for this dissociation is: 

C₁₀H₁₆N₂O₈ → 4H⁺ + C₁₀H₁₂N₂O₈^4- 

(1) Calculate the concentration of calcium ions in a saturated solution of calcium sulphate. 

(2) Calculate the concentration of free Ca²⁺ cations in a solution of 0.1 M Na₂(CaC₁₀H₁₂N₂O₈). You should ignore any protonation of the ligand. 

3) How many moles of calcium sulphate will dissolve in 1 litre of a strongly alkaline solution of 0.1 M Na₄C₁₀H₁₂N₂O₈? What would be the concentrations of the calcium and sulphate ions in the resulting solution? 

(4) Suggest a structure for the complex ion [CaC₁₀H₁₂N₂O₈]^2- assuming that it is approximately octahedral.  

(5) Is the structure you have suggested in 

(4) optically active? If your answer is "yes" then draw the structure of the other optical isomer enantiomer). 

(6) Explain why the complexes formed by the anion C₁₀H₁₂N₂O^4-₈ are exceptionally table. 

Answer 1

(1) [Ca²⁺] = 7.8 × 10^-3 mol dm^-3  

(2) [Ca²⁺] = 1.0 × 10^-6 mol dm^-3  

(3) The CaSO₄ amount dissolved is 0.1 mol.

[SO₄^2-] = 0.10 mol dm^-3 . 

[Ca²⁺] = 6.1 × 10^-4 mol dm^-3  

4) + 5) 

The complex is optically active. The structures of both enantiomers are 

(6) The high number of the chelate rings. Other factors also contribute to the complex ability, e.g. the character of the donor atoms, the magnitude and distribution of the charges in the anion, etc.