Question

Determine the Cd2+ concentration when 0.01 mol CdSO4 and 0.1 mol KCN are added to 1 L of 0.1 M ammonia.

______ M

Answer 1

The formation constants K formation cadium with ammonia and cadmium with CN^- are given as

Cd2+ + 6 NH3 <----------> [Cd(NH3)6]2+

2.6 x 105

Cd2+ + 4 CN- <----------> [Cd(CN)4]2-

7.7 x 1016

From this it is clear in the said reaction we should only form K₂[Cd(CN)₄] under the conditions given in the question.

The reaction for that will be

CdSO₄+ 4 KCN K₂[Cd(CN)₄] +   K₂SO₄

We have 0.01 mol CdSO4 and 0.1 mol KCN

In 1 L Ammonia so concentration of CN^- is 0.1mol in 1L which is 0.1M

So from the formation constant we have all the Cd²⁺ first gets converted to [Cd(CN)₄]^2-

Then as per the equilibrium Cd²⁺ is released

Kf = [[Cd(CN)₄]^2-]/[Cd²⁺][CN^-]⁴

7.7 x 10¹⁶ = 0.01/[Cd²⁺] x (0.1)⁴

[Cd²⁺] = 0.01/7.7 x 10¹⁶ x (0.1)⁴

[Cd²⁺] = 1.29 x 10^-15M

This will be the concentration of Cd²⁺