What is the solubility of Mg(OH)₂ at a pH of 12.00? (Ksp Mg(OH)₂ is 1.6 × 10⁻¹³)

Answer 1

Hi Joe S.!

This problem requires us set up an ICE table in order to figure the molar solubility. Additionally, we are given that the pH of the solution is equal to 12. This is also a common ion effect scenario because we have some other species contributing hydroxide ions to solution. This will affect the behavior of our Mg(OH)₂ species. So, let's have a look!

Step 1: Find the [OH^-]

We are given the pH of the solution as pH = 12. We can use this to find the pOH and subsequently the [OH^-]. In this problem, I will solve straight for the [OH^-]:

pH + pOH =14 => pOH = 14 - pH

=> pOH = -log[OH^-]

-log[OH^-] = 14 - pH

log[OH^-] = pH - 14

[OH^-] = 10^{pH - 14}

<=> 10^{12 - 14}

[OH^-] = 10^-2 M = 0.01 M

Step 2: ICE Table

We will take the [OH^-] found in the previous step and plug this into our initial [OH^-]:

Mg(OH)₂ <--------------> Mg²⁺ + 2OH^-

I ----- 0 0.01M

Δ ----- +s +2s

Eq ----- s 0.01M + 2s

Step 3: Equilibrium Expression

We can now substitute our equilibrium concentrations into our equilibrium expression, Ksp:

Ksp = [Mg²⁺][OH^-]² = 1.6x10^-13

Ksp = (s)*(0.01M + 2s)² = 1.6x10^-13

=> s << 0.01M

NOTE: This is the affect that we discussed at the top. We assume that the amount of OH^- contributed by the dissolution of Mg(OH)₂ is negligible compared with that already present. Consequently, we can ignore it.

Ksp = (s)*(0.01M)² = 1.6x10^-13 => s = 1.6x10^-13/(0.01M)²

s = 1.6x10^-9 M

Therefore, the molar solubility of Mg(OH)₂ at pH = 12 is 1.6x10^-9 M.

Hope this helps!

Cheers