Question

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

(a) $\mathbf{\text{AgCl}}$;

(b) ${\mathbf{\text{SrCO}}}_{\mathbf{\text{3}}}$.

Short answer

1. The solubility of AgCl is not influenced by pH changes. Since ${\mathrm{Cl}}^{-}$is the anion of a strong acid.

   $\mathrm{AgCl}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+{\mathrm{CI}}^{-}\left(\mathrm{aq}\right)$

2. When the pH decreases, ${\text{SrCO}}_3$ become more soluble and the reactions are –

   $$\begin{gathered} {\text{SrCO}}_3\text{(s)}\rightleftharpoons {\text{Sr}}^{\text{2 +}}{\text{(aq) + CO}}_{\text{3}}^{\text{2 -}}\text{(aq)}l \\ {\text{CO}}_{\text{3}}^{\text{2 -}}{\text{+ 2H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq)}\rightleftharpoons {\text{CO}}_{\text{2}}{\text{(aq) +H}}_{\text{2}}{\text{O(l) + 2OH}}^{\text{-}}\text{(aq)} \end{gathered}$$
   

   
   

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 $\mathrm{AgCl}$ –

$\mathrm{AgCl}\text{(s)}\rightleftharpoons {\text{Ag}}^{\text{+}}{\text{(aq) + CI}}^{\text{-}}\text{(aq)}$

The anion ${\mathrm{Cl}}^{-}$ is formed from a strong acid, HCl. So it won’t readily react with the protons of the strong acid.

Thus, the solubility of AgCl is not influenced by the changes in pH.

Subpart (b)

First, write the dissolution equation for ${\text{SrCO}}_3$–

${\text{SrCO}}_3\text{(s)}\rightleftharpoons {\text{Sr}}^{\text{2 +}}{\text{(aq) + CO}}_{\text{3}}^{\text{2 -}}\text{(aq)}$

The formed ${\text{CO}}_{\text{3}}^{\text{2 -}}$ion reacts further with the present water to form carbonic acid –

data-custom-editor="chemistry" ${\text{CO}}_{\text{3}}^{\text{2 -}}{\text{+ 2H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{(aq) + 2OH}}^{\text{-}}\text{(aq)}\rightleftharpoons {\text{CO}}_{\text{2}}{\text{(aq) +H}}_{\text{2}}{\text{O(l) + 2OH}}^{\text{-}}\text{(aq)}$

When ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ions are added, the pH value of solution decreases. This addition decreases the concentration of ${\mathrm{OH}}^{-}$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 towards the products.

Therefore, the decrease of pH value makes ${\text{SrCO}}_3$ more soluble.