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I have been working on this lab forever and have not been able to get a $K_\text{a}$ anywhere close to the accepted value. I’m sure I’m missing something really simple.

We titrated $\ce{CH3COOH}$ with $\ce{NaOH}$.

I used $10.00\ \mathrm{mL}$ of $\ce{CH3COOH}$. Volume of $\ce{NaOH}$ at the equivalence point was $10.14\ \mathrm{mL}$, and the concentration of $\ce{NaOH}$ was $0.1530\ \mathrm{mol \cdot L^{-1}}$.

First I found the concentration of $\ce{CH_3COOH}$ by using $c_\text{a} \cdot V_\text{a} = c_\text{b} \cdot V_\text{b}$, so $c_\text{a} \cdot 10.00\ \mathrm{mL} = 0.1530\ \mathrm{mol \cdot L^{-1}} \cdot 10.14\ \mathrm{mL}$
$c_\text{a} = 0.1551\ \mathrm{mol \cdot L^{-1}}$

Then I used the equation: $\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$ and the pH (at the midpoint of $5.07\ \mathrm{mL}\ \ce{NaOH}$) which was $4.87$

$$K_\text{a}= \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}] [\ce{H3O+}]}=[\ce{CH3COO-}]= 10^{-4.87}= 1.36 \times 10^5 $$ $$ K_\text{a}= \frac{(1.36 \times 10^5)^2}{ 0.1551} = 1.19 \times 10^{-9} $$ Any help on what I am doing wrong would be greatly appreciated!

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I’m not sure if I’m missing your question, but I thought that perhaps you could use the relation $\mathrm{pH}=\mathrm{p}K_\text{a}$ at the half equivalence point? The answer would be still incorrect but closer?

Otherwise, you could probably use stoichiometry to solve for $K_\text{a}$ at the beginning or the end of the titration by constructing an ICE table or otherwise. By considering the deprotonation at the start, so the concentration of protons would be equivalent to concentration of conjugate base. Then, instead of inserting the pH at the half equivalence point, insert the pH before the titration.

If you need help doing the algebra, I could try...

But I think what you did wrong was inserting the concentration at the half equivalence point, rather than before the titration.

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  • $\begingroup$ Thank you for the help, I believe using the relation of pH=pKa at the half equivalence point was the way to go. I was just trying to make it more difficult than it needed to be. $\endgroup$ – Amber Apr 11 '15 at 21:51

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