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 $\pu{1.25*10^{-2}}$.


2 Answers 2


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


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 $\ce{CHCl2COOH}$.


\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}


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