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

1 Answer
Dec 31, 2015

Answer:

#C_6H_5CO_2H(aq) + H_2O(l) rightleftharpoons C_6H_5CO_2^(-) + H_3O^+#

Explanation:

#K_a# #=# #[[C_6H_5CO_2^-][H_3O^+]]/[[C_6H_5CO_2H(aq)]]#

The #pK_a# of benzoic acid is #4.2#, whereas the #pK_a# of acetic acid is #4.7#. Can you account for why acetic acid is a weaker acid than benzoic acid?

Note #pK_a = -log_(10)K_a#.