Question

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

Short 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.

${\text{HC}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}C}}_{\text{6}}{\text{H}}_{\text{5}}{{\text{O}}^{\text{-}}}_{\text{(aq)}}{\text{+\hspace{0.17em}H}}_3{{\text{O}}^{+}}_{\text{(aq)}}$

$\begin{array}{l}{\text{HC}}_{\text{6}}{\text{H}}_{\text{5}}{\text{O}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}C}}_{\text{6}}{\text{H}}_{\text{5}}{{\text{O}}^{\text{-}}}_{\text{(aq)}}{\text{+\hspace{0.17em}H}}_{\text{3}}{{\text{O}}^{\text{+}}}_{\text{(aq)}}\\ {\text{4×10}}^{\text{-5}}\text{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em}0\hspace{0.05em}\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}\hspace{1em}0}\\ {\text{4×10}}^{\text{-5}}\text{-x\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{[x][x]}}{{\text{[4×10}}^{\text{-5}}\text{-x]}}{\text{\hspace{0.17em}=\hspace{0.17em}1.6×10}}^{\text{-10}}\\ {\text{x}}^{\text{2}}{\text{\hspace{0.17em}=\hspace{0.17em}4×10}}^{\text{-5}}{\text{×1.6×10}}^{\text{-10}}{\text{-1.6×10}}^{\text{-10}}\text{x}\end{array}$

${\text{x}}^{\text{2}}{\text{+1.6×10}}^{\text{-10}}{\text{x-6.4×10}}^{\text{-15}}\text{=0}$

$\begin{array}{l}\text{x}=\frac{-\text{b}\pm \sqrt{{\text{b}}^2-4\text{ac}}}{2a}\\ {\text{[H}}^{+}\text{]}=\frac{-(1.6\times {10}^{-10})\pm \sqrt{{(1.6\times {10}^{-10})}^2-4\left(1\right)(\text{-6.4}\times {\text{10}}^{\text{-15}})}}{2\left(1\right)}\\ {\text{[H}}^{+}\text{]}=1.6\times {10}^{-10}\end{array}$

Calculation of pH for Phenol 

$\begin{array}{l}{\text{pH=-log[H}}^{\text{+}}]\\ \text{pH}=-\log (1.6\times {10}^{-10})\\ \text{pH}=9.8\end{array}$