Answer
a) 3.26
Work Step by Step
Here, the weak acid is HF and the conjugate base is F$^{-}$. The following equations show how the weak acid dissociates in H$_{2}$O (water):
HF$_{(aq)}$ + H$_{2}$O$_{(l)}$ <=> H$_{3}$O$^{+}$$_{(aq)}$ + F$^{-}$$_{(aq)}$
NaF$_{aq}$ --> Na$^{+}$$_{(aq)}$ + F$^{-}$$_{(aq)}$
Note that, here, Na$^{+}$$_{(aq)}$ is simply a spectator ion, and we can ignore it in this buffer solution. Also, NaF$_{aq}$ fully dissociates because NaF is a soluble ionic compund.
pH = pKa + log($\frac{[conjugate base]}{[weak acid]}$)
pH = pKa + log($\frac{[F^{-}]}{[HF]}$)
Since, HF is a very weak acid, it does not dissociate very much. Hence, the F$^{-}$$_{(aq)}$ its dissociation provides in the buffer solution is negligible and can be ignored. Hence, the [HF] = 0.76M = 0.76 molL$^{-1}$, and [F$^{-}$] = [NaF] = 0.98M = 0.98 molL$^{-1}$.
Hence, pH = pKa + log($\frac{0.98}{0.76}$)
From the little blue table at the bottom of Page-787, we know that pKa of HF = 3.15
=> pH = 3.15 + log($\frac{0.98}{0.76}$)
=> pH = 3.26 (3sf)
Hence, answer is a) 3.26.
