Answer
$0.10mol$ $CH_3COOH$ wouldn't form a buffer with 1L of $0.20M$ $ NaOH$.
- Since this solution doesn't have any significant acid, it can't act as a buffer, because, it can't neutralize an added base.
Work Step by Step
- There is a weak acid and a strong base, so we have to consider their reaction:
$CH_3COOH(aq) + NaOH(aq) \lt -- \gt CH_3COO^-(aq) + Na^+(aq) + H_2O(l)$
- The limiting reactant is $CH_3COOH$ (0.10 mol), so these are the final concentrations:
$CH_3COOH: 0.10 - 0.10 = 0$
$NaOH: 0.20 - 0.10 = 0.10 moles$
$CH_3COO^-: 0 + 0.10 = 0.10 moles$
- Since this solution doesn't have any significant acid, it can't act as a buffer, because, it can't neutralize an added base.
