Question

What is the molar solubility of Mg(OH)2 in a basic solution with a pH of 12.50? Ksp for Mg(OH)2 is 5.6 × 10-12. What is the molar solubility of Mg(OH)2 in a basic solution with a pH of 12.50? Ksp for Mg(OH)2 is 5.6 × 10-12. 1.1 × 10-4 M 5.6 × 10-9 M 2.4 × 10-6 M 1.8 × 10-10 M

Answer 1

Answer:

$S=5.6\times 10^{-9}\ M$

Explanation:

$Mg(OH)_2$ will form its respective ions in the solution as:

Consider the ICE take for $Mg(OH)_2$ as:

                                     Mg(OH)₂         ⇄        Mg²⁺ + 2OH⁻

At t =equilibrium                                              S            2S            

Where, S is the molar solubility

The expression for Solubility product of $Mg(OH)_2$ is:

$K_{sp}=[Mg^{2+}][OH^-]^2$

Given that:

pH = 12.50

Also, pH + pOH = 14  

So, pOH = 14 - 12.50 = 1.5

$pOH=-log(OH^-)$

$[OH^-]=10^{(-1.5)}=0.03162$

So, Total [OH⁻ ] = S + 0.03162

$K_{sp}=5.6\times 10^{-12}$

So,

$5.6\times 10^{-12}=S\times {(S + 0.03162)}^2$

Since, S is very small, So, S + 0.03162 ≅ 0.03162

Thus,

$5.6\times \:10^{-12}=S\times \left(\:0.03162\right)^2$

$S=\frac{5.6}{10^{12}\times \:0.0009998244}$

$S=5.6\times 10^{-9}\ M$

The molar solubility of $Mg(OH)_2$ in a basic solution with a pH of 12.50 is:- $5.6\times 10^{-9}\ M$