3
$\begingroup$

A solution contains $\pu{0.09 M}~\ce{HCl}, \pu{0.09 M}~\ce{CHCl2COOH}$ and $\pu{0.1 M}~\ce{CH3COOH}$. The $\mathrm{pH}$ of the solution is $1$. Given that the $K_\mathrm{a}$ of acetic acid is $10^{-5}$, calculate the $K_\mathrm{a}$ of $\ce{CHCl2COOH}$.

My attempt is as follows. We have $[\ce{H+}] \text{ from } \ce{HCl} = [\ce{HCl}] = \pu{0.09 M}$.

Then, for the other two acids:

$$\begin{align} [\ce{H+}] \text{ from } \ce{CH3COOH} &= x \\ \frac {10^{-2}\cdot x^{2}}{10^{-1}-x} &= K_\mathrm{a} \\ [\ce{H+}] \text{ from } \ce{CHCl2COOH} &= y \\ \mathrm{pH_\text{total}} &= 1 \\ -\log_{10}(0.09+x+y) &= 1 \\ 0.09+x+y &= 10^{-1} \\ \end{align}$$

But the answer key says the answer is $\pu{1.25*10^{-2}}$.

$\endgroup$
3
$\begingroup$

I just wanted to add an answer based on "no" assumptions, except (and that is occasionally a big assumption) that concentration equals activity. At first it does not seem very practical, as of its size, but I couldn't stop me from adding it here as I like to see the whole picture. In the end, it is actually not that hard.

The exact equation for the proton concentration is:

$$[\ce{H+}] = [\ce{OH-}] + [\ce{Cl-}] + [\ce{CHCl2COO-}] + [\ce{CH3COO-}]$$

which is the sum of the hydroxide ions from the autodissociation of water and the sum of the resulting acid residue ions from the three acids. As is shown here [1], this system expands to the following (and can be reduced again, if appropriate simplifications are taken into account for each acid, as is shown here [2]):

$$ [\ce{H+}] = [\ce{OH-}] + \frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}} + \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}} + \frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}} $$

where $K_{\mathrm a1}$, $K_{\mathrm a2}$, and $K_{\mathrm a3}$ are the $K_{\mathrm a}$'s of $\ce{HCl}$, $\ce{CHCl2COOH}$, and $\ce{CH3COOH}$ respectively.

Now what you need to find is $K_{\mathrm a2}$. Solving it at this stage for $K_{\mathrm a2}$ is probably not a good idea aka nothing nice to handle, as the system is quite large. But as you know all values except $K_{\mathrm a2}$, it's actually not that bad.

$$ \underbrace{[\ce{H+}]}_\pu{0.1 mol/l} = \underbrace{[\ce{OH-}]}_\pu{10^{-13} mol/l} + \underbrace{\frac{[\ce{HCl}]_0 K_{\mathrm a1}}{[\ce{H+}] + K_{\mathrm a1}}}_\pu{0.09 mol/l} + \frac{[\ce{CHCl2COOH}]_0 K_{\mathrm a2}}{[\ce{H+}] + K_{\mathrm a2}} + \underbrace{\frac{[\ce{CH3COOH}]_0 K_{\mathrm a3}}{[\ce{H+}] + K_{\mathrm a3}}}_\pu{10^{-5} mol/l} $$

$\ldots$ I'll leave units from here on (everything is in $\pu{mol/l}$). This simplifies quite much into: $$\begin{align} 0.1 &= 10^{-13} + 0.09 + \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}} + 10^{-5}\\ 0.1 - 10^{-13} - 0.09 - 10^{-5} &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\ 0.01^* &= \frac{0.09\,K_{\mathrm a2}}{0.1 + K_{\mathrm a2}}\\ 0.01\,(0.1 + K_{\mathrm a2}) &= 0.09\,K_{\mathrm a2}\\ 0.001 + 0.01 K_{\mathrm a2} &= 0.09 K_{\mathrm a2}\\ 0.001 &= 0.09 K_{\mathrm a2} - 0.01 K_{\mathrm a2}\\ 0.001 &= (0.09 - 0.01)\,K_{\mathrm a2}\\ 0.001 &= 0.08\,K_{\mathrm a2}\\ K_{\mathrm a2} &= \frac{0.001}{0.08}\\ &= 0.0125 \end{align}$$


* I rounded to $0.01$; it was actually $0.00999\ldots$ which would lead to $\pu{0.0124861... mol/l}$

[1] Equations 2.1 and 2.2
[2] Assumption 1 for HCl, Assumption 3 for acetic acid

| improve this answer | |
$\endgroup$
2
$\begingroup$

Let's use $K_1 = K_\mathrm{a}$ for $\ce{CH3COOH}$ and $K_2 = K_\mathrm{a}$ for $\ce{CHCl2COOH}$

\begin{align} K_1 &= \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} \\[3pt] K_2 &= \dfrac{[\ce{H+}][\ce{CHCl2COO-}]}{[\ce{CHCl2COOH}]} \end{align}

Given that the $\mathrm{pH}$ is $1$, we'll assume that that can be written to at least four decimal places $1.0000$, and that concentrations can be used instead of activities.

Thus for acetic acid

$$ \frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{K_1}{[\ce{H+}]} = \frac{10^{-5}}{0.1} = 10^{-4}$$

so for all practical purposes acetic acid doesn't contribute any significant amount of the $\ce{H+}$.

Since we know that $\pu{0.09 M}$ (we'll assume $\pu{0.0900 M}$) of the $\ce{H+}$ comes from the $\ce{HCl}$, that leaves $\pu{0.0100 M}$ of $\ce{H+}$ from the $\ce{CHCl2COOH}$.

Thus

\begin{align} [\ce{CHCl2COO-}] &= \pu{0.0100 M} \\ [\ce{CHCl2COOH}] &= \pu{0.0900 M} - [\ce{CHCl2COO-}] \\ &= \pu{0.0800 M} \\ K_2 &= \frac{[\ce{H+}][\ce{CHCl2COO-}]}{[\ce{CHCl2COOH}]} \\ &= \frac{(\pu{0.1 M})(\pu{0.01 M})}{\pu{0.0800 M}} \\ &= \pu{1.25\times10^{-2} M} \end{align}

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.