Question

A $\mathbf{0}\mathbf{.}\mathbf{0450}\mathbf{M}$solution of benzoic acid has a pH of$\mathbf{2}\mathbf{.}\mathbf{78}$. Calculate${\mathbf{pK}}_{\mathbf{a}}$for this acid

Short answer

The ${\mathrm{pK}}_{\mathrm{a}}$ for benzoic acid is$=4.20$

Step-by-step solution

Define pKa

${\mathrm{pK}}_{\mathrm{a}}$is the number that describes the acidity of a particular molecule. The formula to find the${\mathrm{pK}}_{\mathrm{a}}$ is

${\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}$

Reaction of dissociation

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CO}}_2\mathrm{H}\stackrel{}{\rightleftharpoons }{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CO}}_2^{-}+{\mathrm{H}}^{+} \\ \left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CO}}_2\mathrm{H}\right]=\mathrm{F}-{10}^{-2.78} \\ \left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CO}}_2^{-}\right]=\left[{\mathrm{H}}^{+}\right]={10}^{-278} \end{gathered}$$

Calculating the Ka of benzoic acid

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CO}}_2^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CO}}_2\mathrm{H}\right]} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{{\left({10}^{-2.78}\right)}^2}{0.0450-{10}^{-2.78}}=6.35\times {10}^{-5} \end{gathered}$$

Calculating pKa of benzoic acid

Substituting the value of $K_{\mathrm{a}}$in the formula ${\mathrm{pK}}_{\mathrm{a}}$ we get

${\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}=-\log \left(6.35\times {10}^{-5}\right)=4.20$