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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?

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  • 2
    $\begingroup$ You didn't take into account the ionization of $\ce{C5H5NH^{+}}$ anywhere. That equilibrium will affect the final pH. You have a problem asking about the pH of a salt solution made of both a weak acid and a weak base. This will make it trickier than usual. $\endgroup$ – Nicolau Saker Neto Jun 1 '15 at 1:17
  • $\begingroup$ So, my K equation should actually look like 1/Ka * 1/Kb * Kw to take in the account of the Kb value? $\endgroup$ – Ewi Jun 1 '15 at 1:57
  • $\begingroup$ The fact that the pyridinium cation $\ce{C5H5NH+}$ is more acidic (pKa = 5.25) than $\ce{F-}$ is basic (pKb = 10.83) somewhat supports this idea that the dissociation of $\ce{F-}$ is supposed to be ignored. Although I definitely think that the question should have given a non-basic anion such as $\ce{Cl-}$ in this case. $\endgroup$ – orthocresol Jun 2 '15 at 11:26
  • $\begingroup$ The intro of this old ACS paper might answer your question. $\endgroup$ – Tyberius Apr 1 '17 at 21:01
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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 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.

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