# 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, $\ce{HA}$, $K_a = 1.0$ x $10^{-4}$, is titrated with $\ce{NaOH}$. At the equivalence point, the concentration of $\ce{NaA}$ at the equivalence point is $0.010$ molar. What is the 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, which is fairly easy to guess, as the pH needs to be greater than 7, but 11 is much to 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 $\ce{[HA]}$.

Any help would be greatly appreciated.

• A clue: you know what would be starting pH? – Mithoron Aug 20 '15 at 22:04

• 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, $x=1.0\times10^{-6}\mathrm{M}$.

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

• I have one question: why do we know that $c = K_b$? – coloratura Aug 21 '15 at 17:55
• I don't understand the question! – Yomen Atassi Aug 21 '15 at 18:10
• Well, in your top equation, it seems that you are calculating the p$K_a$ of the hydrolysis of $A%^-$, which I think you do by subtracting the p$K_b$, from the p$K_w$, which means that $c$ is the p$K_b$, but please correct me if I'm wrong. – coloratura Aug 21 '15 at 18:32
• p$K_b$ is different from $c$. The chemistry underlying the top equation is proved in the next part of my answer. – Yomen Atassi Aug 21 '15 at 19:00