Answer
$0.30mol$ $NaOH$ wouldn't form a buffer with 1L of $0.20M$ $ CH_3COOH$.
- Since this solution doesn't have any significant acid, it can't act as buffer, because it is not capable of neutralizing an added base.
Work Step by Step
1. Analyze the results of the reaction between the strong base $(NaOH)$ and the weak acid $(CH_3COOH)$.
- Initial: $CH_3COOH = 1L * 0.20M = 0.20$ moles
$NaOH = 0.30$ $moles$
- Since these are monoprotics acids, the ratio is 1 to 1, and $NaOH$ is strong:
$CH_3COOH(aq) + NaOH(aq) \lt -- \gt CH_3COO^-(aq) + Na^+(aq) + H_2O$
The limiting reactant is $CH_3COOH$ (0.20 moles), so, after the reaction, we got:
$NaOH = 0.30 - 0.20 = 0.10$ moles
$CH_3COOH = 0.20 - 0.20 = 0$ mol
and
$CH_3COO^- = 0 + 0.20 = $ 0.20 moles. (Product of the last reaction.)
- Since this solution doesn't have any significant acid, it can't act as buffer, because it is not capable of neutralizing an added base.
