Question

The first and second dissociation constant 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. $0.2\times {10}^5$

Answer 1

The correct option is C $5.0\times {10}^{-15}$

$H_{2}A\rightleftharpoons H{A}^{-}+H^{+}\therefore K_1=1.0\times {10}^{-5}=\frac{\left[H^{+}\right]\left[H{A}^{-}\right]}{\left[H_{2}A\right]}$ $H{A}^{-}\to H^{+}+A^{2-}\therefore K_2=5.0\times {10}^{-10}=\frac{\left[H^{+}\right]\left[A^{2-}\right]}{\left[H{A}^{-}\right]}$
Overall dissociation constant $K=\frac{\left[H^{+}{]}^2\right[A^{2-}]}{\left[H_{2}A\right]}=K_1\times K_2$
$K_1\times K_2=1.0\times {10}^{-5}\times 5.0\times {10}^{-10}$
$K=K_1\times K_2=5.0\times {10}^{-15}$