Answer
This is the net ionic reaction when $KOH$ is added to this buffer:
$OH^-(aq) + N{H_4}^+(aq) \lt -- \gt N{H_3}(aq) + H_2O(l)$
Work Step by Step
- Potassium hydroxide: $KOH$.
When a base is added to a buffer, its acid is going to try to neutralize the added compound:
$KOH(aq) + N{H_4}^+(aq) \lt -- \gt K^+(aq) + N{H_3}(aq) + H_2O(l)$
To write the net ionic equation, identify the totally ionizated/dissociated compounds: $HNO_3$, and break them into ions:
$K^+ (aq) + OH^-(aq) + N{H_4}^+(aq) \lt -- \gt K^+(aq) + N{H_3}(aq) + H_2O(l)$
- Remove the spectators ions:
$OH^-(aq) + N{H_4}^+(aq) \lt -- \gt N{H_3}(aq) + H_2O(l)$