Question

10 ml of 0.1 M triprotic acid $H_{3}A$ is titrated with $0.1MNaOH$ solution. If the ratio of $\frac{\left[H_{3}A\right]}{\left[A^{3-}\right]}$ at $2^{nd}$ equivalence point is expressed as ${10}^{-p}$, then $p$ is :

[Given : $K_1={10}^{-4};K_2={10}^{-8};K_3={10}^{-12}$ for $H_{3}A$]

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

The correct option is B 6

At $2^{nd}$ equivalence point, solution contains $H{A}^{2-}$.
$\left[H^{+}\right]=\sqrt{K_2\times K_3}={10}^{-10}M$
$\frac{H_{3}A}{A^{3-}}=\frac{H_{3}A}{A^{3-}}\times \frac{H{A}^{2-}}{H{A}^{2-}}\times \frac{H_2{A}^{-}}{H_2{A}^{-}}\times \frac{\left[H^{+}\right]\left[H^{+}\right]\left[H^{+}\right]}{\left[H^{+}\right]\left[H^{+}\right]\left[H^{+}\right]}$
$⟹\frac{[H^{+}{]}^3}{K_1{K}_2{K}_3}=\frac{{10}^{-30}}{{10}^{-24}}={10}^{-6}$
$p=6$