$0.10mol$ $NaOH$ would form a buffer with 1L of $0.20M$ $ CH_3COOH$.
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 $NaOH = 0.10$ $moles$ - 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 $NaOH$ (0.10 moles), so, after the reaction, we get: $NaOH = 0.10 - 0.10 = 0$ moles $CH_3COOH = 0.20 - 0.10 = 0.10$ moles and $CH_3COO^- = 0 + 0.10 = $ 0.10 moles. (Product of the last reaction.) - Therefore, we have this weak conjugate acid-base pair, with equal number of moles, which makes this a buffer solution.