Answer:
molar solubility in water = 2.412 * 10^-4 mol/L
molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L
Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH
Explanation:
The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.
The equation for the solubilization reaction of Mg(OH)2 can be given as:
Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)
Ksp can then be given as followed:
Ksp = [Mg^2+][OH^–]²
<u>Step 2:</u> Calculate the solubility in water
Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)
The mole ratio Mg^2+ with OH- is 1:2
So there will react X of Mg^2+ and 2X of OH-
The concentration at equilibrium will be XM Mg^2+ and 2X OH-
Ksp = [Mg^2+][OH^–]²
5.61*10^-11 = X * (2X)² = X *4X² = 4X³
X = <u>2.412 * 10^-4 mol/L = solubility in water</u>
<u>Step 3</u>: Calculate solubility in 0.130 M NaOH
The initial concentration of Mg^2+ = 0 M
The initial concentration of OH- = 0.130 M
The mole ratio Mg^2+ with OH- is 1:2
So there will react X of Mg^2+ and 2X +0.130 for OH-
The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-
The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .
Let's consider:
[Mg+2] = X
[OH] = 0.130
Ksp = [Mg^2+][OH^–]²
5.61*10^-11 = X *(0.130)²
5.61*10^-11 = X * (0.130)^2
X = <u>3.32*10^-9 = solubility in 0.130 M NaOH
</u>
<u>Step 4:</u> Calculate how many times Mg(OH)2 is better soluble in pure water.
(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5
Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH
