JEE MAIN - Chemistry (2007 - No. 6)
The first and second dissociation constants of an acid H2A are 1.0 $$\times$$ 10−5
and 5.0 $$\times$$ 10−10 respectively. The overall dissociation constant of the acid will be :
5.0 $$\times$$ 10−5
5.0 $$\times$$ 1015
5.0 $$\times$$ 10−15
5.0 $$\times$$ 105
Explanation
$${H_2}A\,\rightleftharpoons\,{H^ + }\,\, + \,\,H{A^ - }$$
$$\therefore$$ $$\,\,\,{K_1} = 1.0 \times {10^{ - 5}} = {{\left[ {{H^ + }} \right]\left[ {H{A^ - }} \right]} \over {\left[ {{H_2}A} \right]}}$$ (Given)
$$H{A^ - } \to {H^ + } + {A^ - }$$
$$\therefore$$ $$\,\,\,{K_2} = 5.0 \times {10^{ - 10}}$$
$$ = {{\left[ {{H^ + }} \right]\left[ {{A^{ - - }}} \right]} \over {\left[ {H{A^ - }} \right]}}$$ (Given)
$$K = {{{{\left[ {{H^ + }} \right]}^2}\left[ {{A^{2 - }}} \right]} \over {\left[ {{H_2}A} \right]}}$$
$$ = {K_1} \times {K_2}$$
$$ = \left( {1.0 \times {{10}^{ - 5}}} \right) \times \left( {5 \times {{10}^{ - 10}}} \right)$$
$$ = 5 \times {10^{ - 15}}$$
$$\therefore$$ $$\,\,\,{K_1} = 1.0 \times {10^{ - 5}} = {{\left[ {{H^ + }} \right]\left[ {H{A^ - }} \right]} \over {\left[ {{H_2}A} \right]}}$$ (Given)
$$H{A^ - } \to {H^ + } + {A^ - }$$
$$\therefore$$ $$\,\,\,{K_2} = 5.0 \times {10^{ - 10}}$$
$$ = {{\left[ {{H^ + }} \right]\left[ {{A^{ - - }}} \right]} \over {\left[ {H{A^ - }} \right]}}$$ (Given)
$$K = {{{{\left[ {{H^ + }} \right]}^2}\left[ {{A^{2 - }}} \right]} \over {\left[ {{H_2}A} \right]}}$$
$$ = {K_1} \times {K_2}$$
$$ = \left( {1.0 \times {{10}^{ - 5}}} \right) \times \left( {5 \times {{10}^{ - 10}}} \right)$$
$$ = 5 \times {10^{ - 15}}$$
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