Question
If $${K_{sp}}\left( {PbS{O_4}} \right) = 1.8 \times {10^{ - 8}}$$ and $${K_a}\left( {HSO_4^ - } \right) = 1.0 \times {10^{ - 2}}$$ the equilibrium constant for the reaction. $$PbS{O_4}\left( s \right) + {H^ + }\left( {aq} \right) \rightleftharpoons $$ $$HSO_4^ - \left( {aq} \right) + P{b^{2 + }}\left( {aq} \right)$$ is
A.
$$1.8 \times {10^{ - 6}}$$
B.
$$1.8 \times {10^{ - 10}}$$
C.
$$2.8 \times {10^{ - 10}}$$
D.
$$1.0 \times {10^{ - 2}}$$
Answer :
$$1.8 \times {10^{ - 6}}$$
Solution :
$$PbS{O_4}\left( s \right) \rightleftharpoons $$ $$P{b^{2 + }}\left( {aq} \right) + SO_4^{2 - }\left( {aq} \right)\left( {{K_{sp}}} \right)...\left( {\text{i}} \right)$$
$$HSO_4^ - \left( {aq} \right) \rightleftharpoons $$ $${H^ + }\left( {aq} \right) + SO_4^{2 - }\left( {aq} \right)\,\left( {{K_a}} \right)...\left( {{\text{ii}}} \right)$$
$${\text{Subtracting equation (ii) from (i), then}}$$
$$PbS{O_4}\left( s \right) + {H^ + }\left( {aq} \right) \rightleftharpoons $$ $$P{b^{2 + }}\left( {aq} \right) + HSO_4^ - \left( {aq} \right)\left( {{K_{eq}}} \right)$$
$$\eqalign{
& \therefore \,\,{K_{eq}} = \frac{{{K_{sp}}}}{{{K_a}}} \cr
& = \frac{{1.8 \times {{10}^{ - 8}}}}{{1.0 \times {{10}^{ - 2}}}} \cr
& = 1.8 \times {10^{ - 6}} \cr} $$