Answer:
1.72 × 10⁵
Explanation:
Let's consider the reaction for the solubilization of Mg(OH)₂.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
To find its solubility in pure water we will use an ICE Map. We recognize 3 stages: Initial, Change and Equilibrium and complete each row with the concentration or change in concentration.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
We can find the value of the solubility (S) from the solubility product Kps.
[tex]Kps=5.61 \times 10^{-11} = [Mg^{2+} ].[OH^{-}]^{2} =S.(2S)^{2} =4S^{3} \\S=2.41 \times 10^{-4}[/tex]
To calculate the solubility (S') of Mg(OH)₂ in a 0.200 M NaOH solution, we need to take into account the common ion OH⁻ that comes from NaOH. NaOH is a strong electrolyte.
NaOH(aq) ⇒ Na⁺(aq) + OH⁻(aq)
I 0.200 0 0
C -0.200 +0.200 +0.200
E 0 0.200 0.200
The initial concentration of OH⁻ for the solubilization of Mg(OH)₂ will be 0.200 M.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0.200
C +S' +2S'
E S' 0.200 + 2S'
[tex]Kps=5.61 \times 10^{-11} = [Mg^{2+} ].[OH^{-}]^{2}=(S').(0.200+2S')^{2} \\[/tex]
In the term (0.200 + 2S'), 2S' is very small so it can be omitted to simplify calculations. Then, S' = 1.40 × 10⁻⁹
The ratio S/S' is 2.41 × 10⁻⁴/1.40 × 10⁻⁹ = 1.72 × 10⁵.
