Question

Calculate the pH of 4.0 × 10^-5 M phenol (K_a = 1.6 × 10^-10).

Short answer

Answer

pH of the given phenol solution comes out to be 9.8.

Step-by-step solution

 Calculation of [H+] for Hydrolysis

Reaction involved:

Phenol dissolves in water to from phenoxide ion and hydronium ion.

$$\begin{gathered} H{C}_6{H}_5{O}_{\left(aq\right)}+H_{2}O\iff C_8{H}_5{O^{-}}_{\left(aq\right)}+H_3{O^{+}}_{\left(aq\right)} \\ H{C}_6{H}_5{O}_{\left(aq\right)}+C{H}_{2\iff }{C}_8{H}_5{O^{-}}_{\left(aq\right)}+H_3{O^{+}}_{\left(aq\right)} \\ 4\times {10}^{-5}OO \\ 4\times {10}^{-5}-xXX \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[\mathrm{X}\right]\left[\mathrm{X}\right]}{\left[4\times {10}^{-5}-\mathrm{X}\right]}=1.6\times {10}^{-10} \\ {\mathrm{X}}^2=4\times {10}^{-5}\times 1.6\times {10}^{-10}-1.6\times {10}^{-10} \\ {\mathrm{X}}^2=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2a} \end{gathered}$$

$$\begin{gathered} \left[H^{+}\right]=-\frac{\left(1,6\times {10}^{-10}\pm \sqrt{{\left(1.6\times {10}^{-10}\right)}^2-4\left(1\right)-\left(6.4\times {10}^{-15}\right)}\right)}{2\left(1\right)} \\ \left[H^{+}\right]=1.6\times {10}^{-10} \end{gathered}$$

Calculation of pH for Phenol

$$\begin{gathered} pH=-\log \left[H^{+}\right] \\ pH=-\log \left(1.6\times {10}^{-10}\right) \\ pH=9.8 \end{gathered}$$