Question

Write equations to show whether the solubility of either of the following is affected by pH:

(a) Fe(OH)₂;

(b) CuS.

Short answer

(a) The solubility of Fe(OH)₂increases on decreasing the pH.

$\mathrm{Fe}(\mathrm{OH}{)}_2\text{(s)}\rightleftharpoons {\text{Fe}}^{\text{2 +}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq)}$

(b) When the pH decreases, CuS become more soluble and the reaction is –

$\text{CuS(s)}\rightleftharpoons {\text{Cu}}^{\text{2 +}}{\text{(aq) +S}}^{\text{2 -}}\text{(aq)}$

Step-by-step solution

Le Chatelier’s Principle

According to Le Chatelier's principle, when there is increased concentration of products, the equilibrium changes and shifts toward the reactants and vice versa, when there is increased concentration of reactants, the equilibrium shifts toward the products.

Subpart (a)

First, write the dissolution equation for$\mathrm{Fe}(\mathrm{OH}{)}_2$–

$\mathrm{Fe}(\mathrm{OH}{)}_2\text{(s)}\rightleftharpoons {\text{Fe}}^{\text{2 +}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq)}$

When H₃O⁺ions are added, the pH value of solution decreases. This addition decreases the concentration of OH^- ions, because they easily react with hydronium ions to form water.

According to Le Chatelier's principle, this decrease in product concentration moves the reaction toward the products.

Therefore, decrease of pH value makes Fe(OH)₂ more soluble.

Subpart (b)

First, write the dissolution equation for CuS–

$\text{CuS(s)}\rightleftharpoons {\text{Cu}}^{\text{2 +}}{\text{(aq) +S}}^{\text{2 -}}\text{(aq)}$

When H₃O⁺ions are added, the pH value of solution decreases. This addition of hydronium ions decreases the concentration of S^2- ions, because they easily react to form a weak acid H₂S.

${\text{S}}^{\text{2 -}}{\text{(aq) + 2H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\rightleftharpoons {\text{H}}_2{\text{S(aq) + 2H}}_{\text{2}}\text{O(l)}$

According to Le Chatelier's principle, this decrease in product concentration moves the reaction toward the products.

Therefore, decrease of pH value makes CuS more soluble.