# Calculating pH: Weak Acid, Strong Base

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

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 $$\ce{[HA]}$$.

• 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: $$K=\frac{K_w}{K_a}= 10^{-10}$$
$$K=\frac{x^2}{c-x}=\frac{x^2}{1.0\times10^{-2}-x}\approx\frac{x^2}{1.0\times10^{-2}} =1.0\times10^{-10}$$ Where $x$ is the concentration of ion hydroxide and the weak acid $\ce{H}A$.
We have, $x=1.0\times10^{-6}\mathrm{M}$.
$$[\ce{H+_\mathrm{(aq)}}]=\frac{10^{-14}}{1.0\times10^{-6}}\mathrm{M}$$ $$\mathrm{pH}=8.0$$