$\mathrm{C_6H_5NH_2}+\mathrm{H_2O}$ ⇌ $\mathrm{C_6H_5NH_3^+}+\mathrm{OH^-}$ $K_b$ = $\frac{[\mathrm{C_6H_5NH_3^+}][\mathrm{OH^-}]}{[\mathrm{C_6H_5NH_2}]}$ = $\frac{[\mathrm{OH^-}]^2}{[\mathrm{C_6H_5NH_2}]}$ $[\mathrm{OH^-}]$ = $\sqrt{K_b[\mathrm{C_6H_5NH_2}]}$ = $\sqrt{(4.27\times 10^{-10})(10^{-3})}$ = 6.534 × $10^{-7}\,\mathrm{M}$ pOH = - log (6.534 × $10^{-7})$ = 6.18 ∴ pH = 14 - 6.18 = 7.82 $\mathrm{C_6H_5NH_2}+\mathrm{H_2O}$ ⇌ $\mathrm{C_6H_5NH_3^+}+\mathrm{OH^-}$ $\begin{array}{lccc}\text{Initial} & C & 0 & 0 \\ \text{At eqm.} & C-C\alpha & C\alpha & C\alpha \end{array}$ $K_b$ = $\frac{C\alpha \cdot C\alpha}{C(1-\alpha)}$ = $\frac{C\alpha^2}{1-\alpha}$ = $C\alpha^2$ (Since 1 ⋙ α) ∴ α = $\sqrt{\frac{K_b}{C}}$ = $\sqrt{\frac{4.27\times 10^{-10}}{10^{-3}}}$ = 6.53 × $10^{-4}$ $pK_b$ = –log(4.27 × $10^{-10}$) = 9.37 $pK_a + pK_b$ = 14 (for a pair of conjugate acid and base) ∴ $pK_a$ = 14 – 9.37 = 4.63 i.e. –log $K_a$ = 4.63 or, log $K_a$ = –4.63 or, $K_a$ = antilog(–4.63) = 2.34 × $10^{-5}$ pH = 7.82, a = 6.53 × $10^{-4}$, $K_a$ = 2.34 × $10^{-5}$