# Finding the Ka of a weak acid after addition of a strong base

When 50.00 mL of an unknown 0.2461 M weak acid are titrated with 0.1968 M NaOH, it was found after the addition of exactly 16.00 mL of base that the pH of the solution was 3.79. From this information, calculate the Ka of the acid.

I have an idea of how to do this question, but looking in my book, I seem to be arriving at a wrong answer.

$$50 \text{mL of }0.2461\text{M weak acid} = 0.2461 \cdot 0.05 = 0.012305\text{ mol weak acid}$$ $$16 \text{mL of }0.1968\text{M NaOH} = 0.2461 \cdot 0.013 = 0.0031488\text{ mol NaOH}$$

$$\begin{array}{|c|cccc|}\hline & \ce{HA} & \ce{NaOH} & \ce{A^{-}} & \ce{H_{2}O} \\ \hline \mathrm I & 0.0123 & 0.0032 & 0 & -\\ \mathrm C & -0.0032 & -0.0032 & +0.0032 & - \\ \mathrm E & 0.0091 & 0 & 0.0032 & -\\ \hline \end{array}$$

$$\text{molarity } \ce{A^-} = \frac{0.0032~\textrm{mol}}{0.066~\textrm L} = 0.0485~\textrm M$$

from here, I work on the assumption that the weak conjugate base will associate in water.

$$\mathrm p\ce{OH} = 14 - \mathrm p\ce H = 14 - 3.79 = 10.21$$ $$[\ce{OH^-}] = 10^{-\mathrm p\ce{OH}} = 10^{-10.21} = 6.17\times 10^{-11}~\mathrm M$$

$$\begin{array}{|c|cccc|}\hline & \ce{A^-} & \ce{H_2O} & \ce{HA} & \ce{OH^-} \\ \hline \mathrm I & 0.0485 & - & 0 & 0\\ \mathrm C & -6.17\times 10^{-11} & - & +6.17\times 10^{-11} & +6.17\times 10^{-11} \\ \mathrm E & 0.0485 \text{ (negligible change)} & - & 6.17\times 10^{-11} & 6.17\times 10^{-11}\\ \hline\end{array}$$

\begin{align}K_\mathrm {b} &= \frac{(6.17\times 10^{-11})^{2}}{0.0485} = 7.85\times 10^{-20}\\ & = \frac{K_\mathrm W}{K_\mathrm b}\\ & = \frac{1.0\times 10^{-14}}{7.85\times 10^{-20}} = 127388.54 \leftarrow\text{This is obviously wrong}\;.\end{align}

Where did I go wrong? Any help is much appreciated!

• Ka of an acid doesn't change with addition of a strong base. – Dissenter Nov 6 '14 at 16:01

• I had been trying to get the proper answer ($5.61\cdot10^{-5}$) and for whatever reason I kept getting the wrong answer. Maybe I was using the moles of $HA$ and $A^-$ instead of their molarities in 66 mL of solution. Thanks a lot though! Can you think of situations where you would need to set up a second ICE table for its association in water: situations where doing just one ICE table isn't enough? – Kestrel Nov 6 '14 at 16:09