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6-2.Equilibrium-II (Ionic Equilibrium)
medium
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
A
$0.2 \times 10^5$
B
$5.0 \times 10^{-5}$
C
$5.0 \times 10^{15}$
D
$5.0 \times 10^{-15}$
(AIEEE-2007)
Solution
$H_{2} A \rightleftharpoons H^{+}+H A$
$\therefore K_{1}=1.0 \times 10^{-5}=\frac{\left[H^{+} \| H A^{-}\right]}{\left[H_{2} A\right]}(\text { Given })$
$H A^{-} \rightarrow H^{+}+A$
$\therefore K_{2}=5.0 \times 10^{-10}$
$=\frac{| H^{+} \| A^{-1}}{\left|H A^{-}\right|}(\text {Given })$
$K=\frac{\left[H^{+}\right]^{2}\left[A^{2}\right]}{\left[H_{2} A\right]}$
$=K_{1} \times K_{2}$
$=\left(1.0 \times 10^{-5}\right) \times\left(5 \times 10^{-10}\right)$
$=5 \times 10^{-15}$
Standard 11
Chemistry
