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JEE Advanced · Chemistry · 8. Ionic Equilibrium

The dissociation constant of a substituted benzoic acid at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-4}\). The \(\mathrm{pH}\) of \(0.01 \mathrm{M}\) solution of its sodium salt is

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(D) 8

Step-by-step Solution

Detailed explanation

The hydrolysis reaction of conjugate base of acid is
\(A^{-}(a q)+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HO}^{-}+\mathrm{HA}\)
\(K_h=\frac{K_w}{K_a}=\frac{10^{-14}}{10^{-4}}=10^{-10}\)
Since, degree of hydrolysis is negligible;
\(\left[\mathrm{OH}^{-}\right]=\sqrt{K_h C}=10^{-6} \cdot p[\mathrm{OH}]=6\) and \(\mathrm{pH}=14-6=8\)
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