Q19.83P - Chemistry: Molecular Nature Of Matter And Change

Question

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

(a)$\mathbf{CuBr}$; 

(b)${\mathbf{\text{Ca}}}_{\mathbf{\text{3}}}{\mathbf{\text{(PO}}}_{\mathbf{\text{4}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$.

Step-by-Step Solution

Answer

• The solubility of  $\mathrm{CuBr}$ is not affected by the  pH and the reaction is –

          $\mathrm{CuBr}\text{(s)}\rightleftharpoons {\text{Cu}}^{\text{ + }}{\text{(aq) + Br}}^{\text{ - }}\text{(aq)}$

• The solubility of calcium phosphate is not influenced by the changes in pH of the solution${\text{Ca}}_3{\text{(PO}}_4{\text{)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{3Ca}}^{\text{2 + }}{\text{(aq) + 2PO}}_{\text{4}}^{\text{3 - }}\text{(aq)}$

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 $\mathrm{CuBr}$  –

$\mathrm{CuBr}\text{(s)}\rightleftharpoons {\text{Cu}}^{\text{ + }}{\text{(aq) + Br}}^{\text{ - }}\text{(aq)}$ 

When data-custom-editor="chemistry" ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ions are added, the pH  value of solution decreases. This addition of hydronium ions doesn't affect the  ${\mathrm{Br}}^{-}$ ions, because they form a strong acid HBr which easily dissociates to H⁺ and Br^- .This addition also doesn't affect the concentration of   Cu⁺ ions. Thus, the concentrations of products and reactants do not change.

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

 Therefore, the pH  value doesn't affect solubility of  CuBr.

Subpart (b)

First, write the dissolution equation for ${\text{Ca}}_3{\text{(PO}}_4{\text{)}}_{\text{2}}$ –

 ${\text{Ca}}_3{\text{(PO}}_4{\text{)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{3Ca}}^{\text{2 + }}{\text{(aq) + 2PO}}_{\text{4}}^{\text{3 - }}\text{(aq)}$

The phosphate ion is the conjugate base of the strong acid that is ${\mathrm{H}}_3{\mathrm{PO}}_4$.

So, by altering the pH, the phosphate(weak conjugate base) ions remain intact in the solution and hence, there is no change in its solubility.