Le Chatelier's Principle Question 1

\ce{Mg(OH)2(s) <--> Mg^{2+}(aq) + 2 OH-(aq)}

The exothermic dissolution of $\ce{Mg(OH)2}$ in water is represented by the equation above. The $K_{sp}$ of $\ce{Mg(OH)2}$ is $1.8 \times 10^{-11}$ . Which of the following changes will increase the solubility of $\ce{Mg(OH)2}$ in an aqueous solution?

Decreasing the pH

Increasing the pH

Adding

\ce{NH3}

to the solution

Adding

\ce{Mg(NO3)2}

to the solution

Approach

The $K_{sp}$ of magnesium hydroxide is low. The reactants are favored in the reaction. To increase the solubility, we need $\ce{Mg(OH)2}$ to dissociate. In other words, we want the equilibrium position to shift to the right.

Decreasing the pH involves adding acid. The hydroxide would be consumed through neutralization with the acid. Since products are being removed, the equilibrium position will shift to the right.

Increasing the pH would likely have an opposite effect and would shift the equilibrium position to the left, decreasing solubility.

$\ce{NH3}$ , a weak base, would react with water and produce some hydroxide ions, resulting in a shift in equilibrium position to the left.

Adding $\ce{Mg(NO3)2}$ would result in increased $\ce{Mg^{2+}}$ , shifting the equilibrium position to the left.