How to calculate the pH of a 0.325 M solution of pyridine hydrofloride?

How to calculate the $\mathrm{pH}$ of a $0.325~\mathrm{M}$ solution of $\ce{C5H5NHF}$? The $K\_\mathrm{b}$ for $\ce{C5H5N}$ and $\ce{C5H5NH+}$ is $1.7\times 10^{-9}$. 


Here's my work, but it's wrong (at least I tried):


$$\ce{C5H5NHF -> C5H5NH+ + F- }$$
$$K\_\mathrm{a}(\ce{HF})= 4.0\cdot10^{-4}$$
$$K=\frac{K\_\mathrm{w}}{K\_\mathrm{a}}$$
$$K=\frac{1\cdot10^{-14}}{4.0\cdot10^{-4}}=2.5\cdot10^{-11}$$


Construct an ice table:


$$
\begin{array}{|c|c|c|c|}\hline
&\ce{F-}&\ce{+H2O(l)->}&\ce{HF}&+\ce{OH-}\\\hline
\text{I}&\pu{0.325M}&&0&0 \\\hline
\text{C}&-x&&x&x\\\hline
\text{E}&\pu{0.325M}-x&&x&x\\\hline
\end{array}
$$


Set up the equation (small x-approximation):
$$2.5\cdot10^{-11} = \frac{x^2}{0.325-x}$$
Now, cross multiply:
$$x^2 = 5.87\cdot10^{-12}$$
Take the square root of both sides:
$$[\ce{OH-}] = x = 2.42\cdot10^{-6}$$
You want to get from $[\ce{OH-}]$ to $\mathrm{pOH}$:
$$\mathrm{pOH} = -\log\left(2.42\cdot10^{-6}\right) = 5.616$$
Then, from $\mathrm{pOH}$ to $\mathrm{pH}$:
$$\mathrm{pH} = 14 - 5.616 = 8.384$$ 
Which is the wrong answer.


What am I doing wrong? How do I even use the $K\_\mathrm{b}$ value in my calculation? 




I think part of the issue is that you are looking at the wrong reaction. I believe the reaction should be $$\ce{C5H5NHF\_{(s)} ->C5H5NH+\_{(aq)} + F-\_{(aq)}->C5H5N\_{(aq)} + HF\_{(aq)}}$$ 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+} >\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](http://www.cengage.com/chemistry/discipline_content/owl/support/chapters/owlbook_ch17.pdf) has a chapter discussing acid/base mixtures) 


A good approximation for the pH can be found in this [paper](http://pubs.acs.org/doi/abs/10.1021/ed016p140#/doi/abs/10.1021/ed016p140), 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.


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