How to calculate the pH of a 0.325 M solution of pyridine hydrofloride?
I think part of the issue is that you are looking at the wrong reaction. I believe the reaction should be $$\ce{C5H5NHF_{(s)} -&gt;C5H5NH+_{(aq)} + F-_{(aq)}-&gt;C5H5N_{(aq)} + HF_{(aq)}}$$ When looking at this, we can start by noting that $\ce{HF}$ is a stronger acid than $\ce{C5H5NH+}$, so the reaction will favor the reactants. Then, we can compare the reactants and note that $\ce{C5H5NH+}$ has a greater acid strength than the base strength of $\ce{F-}$  (ie $\mathrm{K_a} $ of $\ce{C5H5NH+} &gt;\mathrm{K_b} \text{ of } \ce{F-} $) so the solution will be acidic. Its always useful to have this sort of sanity to check to make sure your answer makes sense. (For more details on this portion, Cengage has a chapter discussing acid/base mixtures) 

A good approximation for the pH can be found in this paper, as shown here: $$[\ce{H3O+}]=\mathrm{K_a}*\sqrt{\frac{\mathrm{K_w}}{\mathrm{K_a * K_b}}}$$ where $\mathrm{K_a}$ is for $\ce{C5H5NH+}$ and $\mathrm{K_b}$ is for 
$\ce{F-}$. This leads to $[\ce{H3O+}]=4.85*10^{-5}\to\mathrm{pK_a}=4.314$

As a comparison, the $\text{p}\ce{H}$ of $.325$M pyridinium is $2.86$, which it makes sense that the $\text{p}\ce{H}$ is higher when it is mixed with a base.