The first and second dissociation constants o | vedclass

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Question

$H_{2} A ightleftharpoons H^{+}+H A$

$	herefore K_{1}=1.0 	imes 10^{-5}=\frac{\left[H^{+} \| H A^{-}ight]}{\left[H_{2} Aight]}(	ext { Give

Vedclass · Accepted Answer

$5.0 	imes 10^{-15}$

Vedclass · Answer

$H_{2} A ightleftharpoons H^{+}+H A$

$	herefore K_{1}=1.0 	imes 10^{-5}=\frac{\left[H^{+} \| H A^{-}ight]}{\left[H_{2} Aight]}(	ext { Given })$

$H A^{-} ightarrow H^{+}+A$

$	herefore K_{2}=5.0 	imes 10^{-10}$

$=\frac{| H^{+} \| A^{-1}}{\left|H A^{-}ight|}(	ext {Given })$

$K=\frac{\left[H^{+}ight]^{2}\left[A^{2}ight]}{\left[H_{2} Aight]}$

$=K_{1} 	imes K_{2}$

$=\left(1.0 	imes 10^{-5}ight) 	imes\left(5 	imes 10^{-10}ight)$

$=5 	imes 10^{-15}$

The first and second dissociation constants of an acid $H_2A$ are $1.0 \times 10^{-5}$ and $5.0 \times 10^{-10}$ respectively. The overall dissociation constant of the acid will be

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