# Calculating the dissociation constant of one acid in a mixture

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

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 , 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 ):

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

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 $$\pu{0.0124861... mol/l}$$

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}