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. $5.0\times 10^{-5}$
• B. $5.0\times 10^{15}$
• C. $5.0\times 10^{-15}$
• D. $0.0\times 10^5$

Answer 1

Answer: (C)

$H_{2}A\rightleftharpoons H{A}^{-}+H^{+}{K}_1=\frac{\left[H{A}^{-}\right]\left[H^{+}\right]}{\left[H_{2}A\right]}$
$H{A}^{-}\rightleftharpoons H^{+}+A^{2-}{K}_2=\frac{\left[H^{+}\right]\left[A^{2-}\right]}{\left[H{A}^{-}\right]}$
For the reaction
$H_{2}A\rightleftharpoons 2H^{+}+A^{2-}$
$K=\frac{{\left[H^{+}\right]}^2\left[A^{2-}\right]}{\left[H_{2}A\right]}=K_1\times K_2$
$=1\times 10^{-5}\times 5\times 10^{-10}$
$=5\times 10^{-15}$
Hence, (3) is correct.