Answer
$H_3C_6H_5O_7(aq) + H_2O(l) \lt -- \gt H_2C_6H_5{O_7}^-(aq) + H_3O^+(aq)$
$H_2C_6H_5{O_7}^-(aq) + H_2O(l) \lt -- \gt HC_6H_5{O_7}^{2-}(aq) + H_3O^+(aq)$
$HC_6H_5{O_7}^{2-}(aq) + H_2O(l) \lt -- \gt C_6H_5{O_7}^{3-}(aq) + H_3O^+(aq)$
Work Step by Step
1. Write the reaction where $H_3C_6H_5O_7$ behaves as an acid in water.
$H_3C_6H_5O_7(aq) + H_2O(l) \lt -- \gt H_2C_6H_5{O_7}^-(aq) + H_3O^+(aq)$
- It will donate one proton to the water molecule, producing $H_3O^+$.
2. Write the reaction where the conjugate base acts as an acid:
$H_2C_6H_5{O_7}^-(aq) + H_2O(l) \lt -- \gt HC_6H_5{O_7}^{2-}(aq) + H_3O^+(aq)$
- Repeat step 2 until the conjugate base is incapable of donating protons.
$HC_6H_5{O_7}^{2-}(aq) + H_2O(l) \lt -- \gt C_6H_5{O_7}^{3-}(aq) + H_3O^+(aq)$
