Answer
$1L $ of $0.20M$ $NaOH$ and 0.30 moles of acetic acid are capable of making a buffer, because the solution has a weak conjugate acid-base pair in adequate amounts.
Work Step by Step
- There is a weak acid and a strong base, so we have to consider their reaction:
$CH_3COOH(aq)+NaOH(aq)\\ -->CH_3COO^−(aq)+Na^+(aq)+H_2O(l)$
- The limiting reactant is $NaOH$ (0.20 mol), so these are the final concentrations:
$CH_3COOH$: 0.30 − 0.20=0.10 mol
$NaOH $: 0.20 − 0.20=0 mol
$CH3COO^−$: 0 + 0.20= 0.20mol
- Since this solution has a conjugate acid-base pair ($CH_3COOH$ and $CH_3COO^-$), in adequate proportions, it is considered a buffer.
