Question

We have a solution that contains 0.30 M Zn⁺² ion and 0.50 M Mn⁺² ion. If you add NaOH to make the pH of the solution 10, what concentrations (M) of Zinc and Manganese will still be in solution? What percentage of the first metal was removed when the second metal starts to fall?

Using the equations if nessecary:

ΔG° = -nFE°

ln K = (nE°) / 0.0257

E = E° - (0.0257 / n) ln Q

Charge (coulombs) = Amps x time (seconds)

Answer 1

The ionizations of Zn(OH)₂ and Mn(OH)₂ are as below.

Zn(OH)₂ (s) <========> Zn²⁺ (aq) + 2 OH^- (aq)                      K_sp = 3*10^-17

Mn(OH)₂ (s) <========> Mn²⁺ (aq) + 2 OH^- (aq)                    K_sp = 2*10^-13

The pH of the solution is 10.

It is known that

pH + pOH = 14

======> 10 + pOH = 14

======> pOH = 14 – 10 = 4

Further,

pOH = -log [OH^-]

where [OH^-] determines the molar concentration of OH^- in solution.

Therefore,

-log [OH^-] = 4

======> [OH^-] = antilog (-4) M

======> [OH^-] = 1*10^-4 M.

Plug [OH^-] in the K_sp expressions and determine the concentrations of Zn²⁺ and Mn²⁺ in solution when the pH of the solution is 10.

K_sp [Zn(OH)₂] = [Zn²⁺][OH^-]²

======> 3*10^-17 = [Zn²⁺]*(1*10^-4)² (ignore units here since K_sp is dimensionless)

======> [Zn²⁺] = (3*10^-17)/(1*10^-4)² = 3*10^-9

The concentration of Zn²⁺ ion remaining in solution at pH = 10 is 3*10^-9 M (ans).

K_sp [Mn(OH)₂] = [Mn²⁺][OH^-]²

======> 2*10^-13 = [Mn²⁺]*(1*10^-4)² (ignore units here since K_sp is dimensionless)

======> [Mn²⁺] = (2*10^-13)/(1*10^-4) = 2*10^-5

The concentration of Mn²⁺ ion remaining in solution at pH = 10 is 2*10^-5 M (ans).

Since Zn(OH)₂ has a lower K_sp than Mn(OH)₂, hence, Zn(OH)₂ precipitates first. The question requires to determine [Zn²⁺] when Mn(OH)₂ starts to precipitate.

Determine [OH^-] when Mn(OH)₂ starts to precipitate. Plug in [Mn²⁺] = 0.50 M in the K_sp expression and obtain [OH^-].

K_sp = [Mn²⁺][OH^-]²

======> 2*10^-13 = (0.50)*[OH^-]²

======> [OH^-]² = (2*10^-13)/(0.50) = 4*10^-13

======> [OH^-] = 6.324*10^-7

Put [OH^-] = 6.324*10^-7 M in the K_sp expression for Zn(OH)₂ and obtain [Zn²⁺] when Mn(OH)₂ begins to precipitate.

3*10^-17 = [Zn²⁺][OH^-]²

======> 3*10^-17 = [Zn²⁺]*(6.324*10^-7)²

======> [Zn²⁺] = (3*10^-17)/(6.324*10^-7)² = 7.50*10^-5

The concentration of free Zn²⁺ in the solution = 7.50*10^-5 M.

The concentration of Zn²⁺ removed = (0.30 – 7.50*10^-5) M = 0.299925 M.

Percentage of Zn²⁺ ion removed = (0.299925 M)/(0.30 M)*100

= 99.975%

≈ 99.97% (ans).