Answer
a. $HBr(aq) + H_2O(l) \longrightarrow Br^-(aq) + H_3O^+(aq)$
b. $C_2H_5COOH(aq) + H_2O(l) \leftrightharpoons {C_2H_5COO}^-(aq) + H_3O^+(aq)$
c. $H_2S(aq) + H_2O(l) \leftrightharpoons HS^-(aq) + H_3O^+(aq)$
d. $HCN(aq) + H_2O(l) \leftrightharpoons CN^-(aq) + H_3O^+(aq)$
e. $HF(aq) + H_2O(l) \leftrightharpoons F^-(aq) + H_3O^+(aq)$
f. $HClO_4(aq) + H_2O(l) \longrightarrow ClO{_4}^-(aq) + H_3O^+(aq)$
Work Step by Step
The pattern is:
$$Acid(aq) + H_2O(l) \longrightarrow Conjugate \space base + H_3O^+(aq)$$
If the acid is a strong acid, we use the single right arrow, if the acid is weak, we use the double arrow.
To get the conjugate base, remove a hydrogen and 1 from the charge of the molecule/ion.
