Question

Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61*10^(-11). It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. What is the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.200 NaOH solution?

Answer 1

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.

$Kps=5.61 \times 10^{-11} = [Mg^{2+} ].[OH^{-}]^{2} =S.(2S)^{2} =4S^{3} \\S=2.41 \times 10^{-4}$

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'

$Kps=5.61 \times 10^{-11} = [Mg^{2+} ].[OH^{-}]^{2}=(S').(0.200+2S')^{2}$

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.