How would you write the ionization equation for benzoic acid (C6H5COOH)?

Dec 31, 2015

${C}_{6} {H}_{5} C {O}_{2} H \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {C}_{6} {H}_{5} C {O}_{2}^{-} + {H}_{3} {O}^{+}$
${K}_{a}$ $=$ $\frac{\left[{C}_{6} {H}_{5} C {O}_{2}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[{C}_{6} {H}_{5} C {O}_{2} H \left(a q\right)\right]}$
The $p {K}_{a}$ of benzoic acid is $4.2$, whereas the $p {K}_{a}$ of acetic acid is $4.7$. Can you account for why acetic acid is a weaker acid than benzoic acid?
Note $p {K}_{a} = - {\log}_{10} {K}_{a}$.