Question

(a) What happens to the concentration of Cadmium when the pH of a solution increases from 7 to 12? Calculate the concentration of cadmium in each case. Assume that the solubility of cadmium is controlled only by hydroxide. The solubility product constant is as follows:

Cd(OH)_{2 (s)} <--> Cd²⁺ + 2OH^- K_sp= 2x10^-14

The national groundwater drinking water standard for cadmium is 0.005 mg/L. (b) What pH will ensure that any Cadmium in excess of 0.005 mg/L will precipitate out and not be in solution?

Answer 1

Ans. Calculating Ksp

Ksp expression is equivalent to reaction quotient when a solid form is being converted into aqueous form.

# Solubility at pH 7

At pH 7-

            pOH = 14 – 7 = 7

            [OH^-] = antilog (-7) = 10^-7 M

Now,

Ksp = [Cd²⁺] [OH^-]²   

Or, 2.0 x 10^-14 = [Cd²⁺] (10^-7)²

Or, [Cd²⁺] = 2.0 x 10^-14 / 10^-14 = 2.0

1 mol Cd²⁺ is donated by 1 mol Cd(OH)₂.

Hence, solubility at pH 7 = 2.0 M

# Solubility at pH 12

At pH 12-

            pOH = 14 – 12 = 2

            [OH^-] = antilog (-2) = 0.01 M

Now,

Ksp = [Cd²⁺] [OH^-]²   

Or, 2.0 x 10^-14 = [Cd²⁺] (10^-2)²

Or, [Cd²⁺] = 2.0 x 10^-10

Hence, solubility at pH 7 = 2.0 x 10^-10 M

Conclusion: Solubility of Cd(OH)₂decreases when pH is increased from 7 to 12.

#b. Maximum permitted solubility = 0.005 mg/ L

                                                            = 5.0 x 10^-6 g/ L                    ; [1 g = 10^-3 mg]

                                                            = (5.0 x 10^-6 g L^-1) / (112.411 g/ mol)

                                                            = 4.448 x 10^-8 mol / L

                                                            = 4.448 x 10^-8 M

So, [Cd²⁺] in specified solution = 4.448 x 10^-8 M

Now,

Putting the values in Ksp expression to calculate [OH^-]-

2.0 x 10^-14 = (4.448 x 10^-8) [OH^-]²

Or, (2.0 x 10^-14) / (4.448 x 10^-8) = [OH^-]

Or, [OH^-] = (4.496 x 10^-7)^½ = 6.706 x 10^-4

Hence, [OH^-] = 6.706 x 10^-4 M

Now,

            pOH = -log [OH^-] = -log (6.706 x 10^-4) = 3.174

And,    pH = 14 – pOH = 14 – 3.174 = 10.826