Answer
c. $0.2M HNO_3$ and $0.4M NaF$ will produce a buffer.
Work Step by Step
Number of moles:
$HNO_3: $ $\frac{0.2mol}{1L} \times 1L = 0.2mol$
$NaF: $ $\frac{0.4mol}{1L} \times 1L = 0.4mol$
In $NaF$: $Na^+$ is the conjugate pair of a strong base, so it doen't affect the pH of the water. And $F^-$ is the conjugate base of a weak acid, so it will act as a weak base.
Since there are a weak base and a strong acid in the solution, this reaction will occur:
$HNO_3(aq) + F^-(aq) -- \gt HF(aq) + NO_3^-(aq)$
$HNO_3$ is the limiting reactant, because it has the lower number of moles (0.2 mol). So, the reaction will consume 0.2 mol of $HNO_3$ and $F^-$, and produce $0.2mol$ of $HF$ and $NO_3^-$
Therefore, these are the final amounts in moles:
$HNO_3 : 0.2mol - 0.2mol = 0$
$F^-: 0.4mol -0.2mol = 0.2mol (F^-)$
$HF: 0 + 0.2 mol (produced) = 0.2mol(HF)$
As we can see, we have $HF$ and $F^-$ in equal amounts, so, this is a buffer.
