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This is a question that had me puzzled for quite a while. I feel that information is missing.

A weak acid, $\ce{HA}$, $K_\textrm{a} = \pu{1.0E-4 M}$, is titrated with $\ce{NaOH}$. The concentration of $\ce{NaA}$ at the equivalence point is $\pu{0.010 M}$. What is the $\mathrm{pH}$ at the equivalence point?
a) 4.0
b) 6.0
c) 7.0
d) 8.0
e) 11.0

The correct answer is :

d) 8.0

The answer is fairly easy to guess, as the $\mathrm{pH}$ needs to be greater than 7, but 11 is much too basic. However, I'm not sure how I would be able to figure that out from the information given? I've tried using the Henderson-Hasselbalch equation, but I can't figure out the relationship between $\ce{[NaA]}$ or $\ce{[A-]}$ and $\ce{[HA]}$.

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  • At the equivalence point, you have an aqueous solution of weak base $\ce{Na}A$ at the concentration of $c$. The pH of the solution is given by the equation: $$\mathrm{pH}=\frac{1}{2}(\mathrm{p}K_a +\mathrm{p}K_w -\mathrm{p}c)$$ In your case: $\mathrm{pH}=\frac{1}{2}(4.0 +14 -2.0)=8.0$
  • Now, if you are not familiar with the above equation, you can write the reaction that could happen in the solution; the reaction of $\ce{Na}A$, completely dissociated, with water: $$A^- +\ce{ H2O <=> OH- +H}A$$ The constant of this equilibrium is: $$K=\frac{K_w}{K_a}= 10^{-10}$$

    On the other hand, you have:

$$K=\frac{x^2}{c-x}=\frac{x^2}{1.0\times10^{-2}-x}\approx\frac{x^2}{1.0\times10^{-2}} =1.0\times10^{-10}$$ Where $x$ is the concentration of ion hydroxide and the weak acid $\ce{H}A$.

We have, $x=1.0\times10^{-6}\mathrm{M}$.

$$[\ce{H+_\mathrm{(aq)}}]=\frac{10^{-14}}{1.0\times10^{-6}}\mathrm{M}$$ $$\mathrm{pH}=8.0$$

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