## Chemistry: Atoms First (2nd Edition)

$H_3C_6H_5O_7(aq) + H_2O(l) \lt -- \gt H_2C_6H_5{O_7}^-(aq) + H_3O^+(aq)$ $H_2C_6H_5{O_7}^-(aq) + H_2O(l) \lt -- \gt HC_6H_5{O_7}^{2-}(aq) + H_3O^+(aq)$ $HC_6H_5{O_7}^{2-}(aq) + H_2O(l) \lt -- \gt C_6H_5{O_7}^{3-}(aq) + H_3O^+(aq)$
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. $HC_6H_5{O_7}^{2-}(aq) + H_2O(l) \lt -- \gt C_6H_5{O_7}^{3-}(aq) + H_3O^+(aq)$