Question
The first and second dissociation constants of an acid $${H_2}A$$ are $$1.0 \times {10^{ - 5}}$$ and $$5.0 \times {10^{ - 10}}$$ respectively. The overall dissociation constant of the acid will be
A.
$$0.2 \times {10^5}$$
B.
$$5.0 \times {10^{ - 5}}$$
C.
$$5.0 \times {10^{15}}$$
D.
$$5.0 \times {10^{ - 15}}$$
Answer :
$$5.0 \times {10^{ - 15}}$$
Solution :
$$\eqalign{
& {H_2}A \rightleftharpoons {H^ + } + H{A^ - } \cr
& \therefore \,\,{K_1} = 1.0 \times {10^{ - 5}} \cr
& = \frac{{\left[ {{H^ + }} \right]\left[ {H{A^ - }} \right]}}{{\left[ {{H_2}A} \right]}}\,\,{\text{(Given)}} \cr
& H{A^ - } \to {H^ + } + {A^{2 - }} \cr
& \therefore \,\,{K_2} = 5.0 \times {10^{ - 10}} \cr
& = \frac{{\left[ {{H^ + }} \right]\left[ {{A^{2 - }}} \right]}}{{\left[ {H{A^ - }} \right]}}\,{\text{(Given)}} \cr
& K = \frac{{{{\left[ {{H^ + }} \right]}^2}\left[ {{A^{2 - }}} \right]}}{{\left[ {{H_2}A} \right]}} \cr
& = {K_1} \times {K_2} \cr
& = \left( {1.0 \times {{10}^{ - 5}}} \right) \times \left( {5 \times {{10}^{ - 10}}} \right) \cr
& = 5 \times {10^{ - 15}} \cr} $$