# Calculating pH in titration of monoprotic weak acid with strong base

$25.0\:\mathrm{ml}$ of $0.10\:\mathrm{M}$ $\ce{CH3COOH}$ is titrated with $0.10\:\mathrm{M}$ $\ce{NaOH}$. What is the $\mathrm{pH}$ after the addition of $30.0\:\mathrm{ml}$ $\ce{NaOH}$?

In broad strokes, as I understand it, the sodium hydroxide in this case would completely consume the acetic acid, leaving a solution of acetate ions, $\ce{Na+}$ and $\ce{OH-}$.

What makes sense to me then is to figure that the $\mathrm{pH}$ is now dominated by the concentration of hydroxide ions, so the $\mathrm{pH}$ would be given by

$\mathrm{pH} = 14-(-\log([\ce{OH-}])= 11.96$

where $[\ce{OH-}] = \frac{[(30.0 - 25.0) \times 10^{-4}]\:\mathrm{mol}}{(30.0+25.0) \times 10^{-3}\:\mathrm{l}}$

i.e., the remaining moles of hydroxide ($0.0005$) per liter of total volume ($55\:\mathrm{ml}$)

but is this correct? Do I need to take into account the conjugate base’s contribution to deprotonation of water? If so, how is that done?

I’d like to be clever and use the lack of $K_\mathrm{a}$ or related constants in the problem to indicate that I’m on the right track, but that would be giving too much credit to my professor’s design of these questions (and the prior question in this series does give $\mathrm{p}K_\mathrm{a}$ for acetic acid)

• The strong base contributes so much more (in magnitude) that the conjugate base (weak) can be neglected for a reasonable approximation. Thus, for the scope of the question, I believe you did it correctly -- many questions are designed for such approximations. – Andy Jun 11 '15 at 22:39

If you calculate the equilibrium constant of the neutralization reaction $$\ce{CH3COOH + OH- <=> CH3COO- + H2O}$$ $$K= \frac{K_a}{K_w}=\frac{10^{-4.75}}{10^{-14}}= 10^{9.25}$$
Now, if you write of reaction of acetate ion with water:$$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$ It's the opposite of the neutralization reaction. Its equilibrium constant is $$10^{-9.25}$$.
• The K value is actually $10^{9.25}$. – Ed V Feb 19 at 0:09