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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!

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  • $\begingroup$ Ka of an acid doesn't change with addition of a strong base. $\endgroup$ – Dissenter Nov 6 '14 at 16:01
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When you used the ICE method the second time, you forgot that HA does not have an initial concentration of 0. Instead, the initial concentration of HA is 0.0091 as you calculated previously. For simplicity's sake, you can just use Henderson-Hasselbalch equation after your first ICE method to determine the answer.

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  • $\begingroup$ 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? $\endgroup$ – Kestrel Nov 6 '14 at 16:09
  • $\begingroup$ I guess the issue is not so much about whether you use one or two or even three ICE tables, but more so on looking at the net change in concentration of each specie in solution. $\endgroup$ – user2759975 Nov 6 '14 at 16:24
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Here is a different approach:

Initially we have: 50,00 *0,2461 = 12,305 mmole of acid A. This amount of the acid is being titrated with 16,00*0,1968 M = 3,1488 mmole of NaOH. By the titration, 3,1488 mmole of the acid A will be converted to its corresponding base B.

pH = pka + log [B]/[A] => 3,79 = pka + log [3,1488/(12,305-3,1488)] = pka - 0,46357 => pka = 4,25 => ka = 10-4,25 = 5,62*10-5

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