Answer
(c) $0.20M HF$ and $0.10M NaOH$
Work Step by Step
- Since we have a strong base and a weak acid, a neutralization reaction will occur:
$NaOH(aq) + HF(aq) \lt -- \gt Na^+(aq) + F^-(aq) + H_2O(l)$
$NaOH$ is the limiting reactant (because at equal volumes, it has the lower concentration $(0.10M)$)
Therefore:
$[NaOH]_{final} = 0.10M (Initial) - 0.10M = 0$
$[HF]_{final} = 0.20M (Initial) - 0.10M = 0.10M$
$[F^-] = 0 (Initial) + 0.10M = 0.10M$
Notice: For reactants, the concentration decreases, because the compound is being consumed. And for the products, the compound is being produced, so, the concentration increases.
Since this solution has $HF$ and $F^-$ in equal amounts, it can act as a buffer.
