Question

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

(a) PbI₂;

(b) Hg₂(CN₂).

Short answer

(a) The solubility of ${\mathrm{PbI}}_2$is not affected by the role="math" localid="1663414071710" $\mathrm{pH}$and the reaction is –

${\text{PbI}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{Pb}}^{\text{2 +}}{\text{(aq) + 2I}}^{\text{-}}\text{(aq)}$

(b) When the data-custom-editor="chemistry" $\mathrm{pH}$decreases, ${\mathrm{Hg}}_2\left({\mathrm{CN}}_2\right)$become more soluble and the reaction is –

${\text{Hg}}_{\text{2}}{\text{(CN)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{2Hg}}^{\text{+}}{\text{(aq) + 2CN}}^{\text{-}}\text{(aq)}$

Step-by-step solution

Introduction

When the anion of a weak acid is present on the left side of the solubility equilibrium then, the decrease in pH of that salt solution increases the solubility of that particular salt.

Subpart (a)

First, write the dissolution equation for ${\mathrm{PbI}}_2$–

${\text{PbI}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{Pb}}^{\text{2 +}}{\text{(aq) + 2I}}^{\text{-}}\text{(aq)}$

When ${\mathrm{H}}_3\mathrm{O}+$ions are added, the $\mathrm{pH}$value of solution decreases. This addition of hydronium ions doesn't affect the ions, because they form a strong acid$\left(\mathrm{HI}\right)$ which easily dissociates to ${\mathrm{H}}^{+}\mathrm{and}{\mathrm{I}}^{-}$.

According to Le Chatelier's principle, there is no change in equilibrium position, because there is no change in concentration of products or reactants.

Therefore, the $\mathrm{pH}$value doesn't affect solubility .

Subpart (b)

First, write the dissolution equation for–

${\text{Hg}}_{\text{2}}{\text{(CN)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{2Hg}}^{\text{+}}{\text{(aq) + 2CN}}^{\text{-}}\text{(aq)}$

When ${\mathrm{H}}_3{\mathrm{O}}^{+}$ions are added, the $\mathrm{pH}$value of solution decreases. This addition of hydronium ions decreases the concentration of ${\mathrm{CN}}^{-}$ions, because they easily react to form a weak acid $\mathrm{HCN}$.

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

Therefore, decrease of $\mathrm{pH}$value makes ${\mathrm{Hg}}_2\left({\mathrm{CN}}_2\right)$more soluble.