The ionization ofbenzoic acid is represented by this equation.

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\left(\mathrm{aq}\right)$

If a 0.045 M solution of benzoic acid has an $\left[{\mathrm{H}}^{+}\right]=1.7\times {10}^{-3}$, what is the ${\mathrm{K}}_{\mathrm{a}}$of benzoic acid?

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\leftrightarrow {\mathrm{H}}^{+}+{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}$

$K_a=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right]}$

$\left[{\mathrm{H}}^{+}\right]=1.7\times {10}^{-3}$

Using mole ratio $\left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\right]=1.7\times {10}^{-3}$

Final $\left[{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\right]=0.045-1.7\times {10}^{-3}=0.0433$

${\mathrm{k}}_{\mathrm{a}}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{\mathrm{C}}$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[1.7\times {10}^{-3}\right]\left[1.7\times {10}^{-3}\right]}{0.0433}=6.67\times {10}^{-5}$

So the ${\mathrm{K}}_{\mathrm{a}}$ of acid is $6.4\times {10}^{-5}\mathrm{M}$.