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Q. A solution contains $\pu{0.09M}~\ce{HCl}, \pu{0.09M}~\ce{CHCl2COOH}$ and $\pu{0.1M}~\ce{CH3COOH}$. The pH of the solution is 1. If $\ce{K_a}$ for acetic acid is $10^{-5}$, then calculate $\ce{K_a}$ for $\ce{CHCl2COOH}$.

Attempt- $$[\ce{H+}] \text{ of } \ce{HCl} = [\ce{HCl}] = \pu{0.09M}$$ $$[\ce{H+}] \text{ of } \ce{CH3COOH} = x$$ $$\frac {10^{-2}\cdot x^{2}}{10^{-1}-x} = \ce{K_a}$$ $$[\ce{H+}] \text{ of } \ce{CHCl2COOH}=y$$ $$\mathrm{pH_{total}}=1$$ $$-\text{log}[0.09+x+y]=1$$ $$0.09+x+y=10^{-1}$$

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

Edit- I realised that the concentrations were not taken in equilibrium. Have made changes to reflect.

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  • $\begingroup$ Please have a look at some hints on formatting :) $\endgroup$ – Klaus-Dieter Warzecha Nov 14 '16 at 8:05
  • $\begingroup$ Sorry, I had trouble using the \space identifier it is now fixed $\endgroup$ – Akshat Batra Nov 14 '16 at 8:06
  • $\begingroup$ No need to be sorry for fighting! Three remarks: Chemical equations look much better if you use $\ce{...}$ for them. MathJax in titles is discouraged. I'd simply write acid dissociation constant instead. Instead of *, you can use \cdot or \times in scientific notation. $\endgroup$ – Klaus-Dieter Warzecha Nov 14 '16 at 8:11
  • $\begingroup$ I migrated here from Math SE, and thought the Latex would be same. Apparently not! I changed the title and would remember your advice in the future. $\endgroup$ – Akshat Batra Nov 14 '16 at 8:15
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Let's use $K_1 = K_a$ for $\ce{CH3COOH}$ and $K_2 = K_a$ for $\ce{CHCl2COOH}$

$K_1 = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$

$K_2 = \dfrac{[\ce{H+}][\ce{CHCl2COO-}]}{[\ce{CHCl2COOH}]}$

Given that the pH is 1, we'll assume 1.000 and that concentrations can be used instead of activities.

Thus for acetic acid

$ \dfrac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \dfrac{K_1}{[\ce{H+}]} = \dfrac{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 0.09 (we'll assume 0.0900) of the $\ce{H+}$ comes from the $\ce{HCl}$, that leaves 0.0100 of $\ce{H+}$ from the $\ce{CHCl2COOH}$.

Thus

$\ce{[CHCl2COO-] = 0.0100}$

$\ce{[CHCl2COOH] = 0.0900 - [CHCl2COO-] = 0.0800}$

$K_2 = \dfrac{[\ce{H+}][\ce{CHCl2COO-}]}{[\ce{CHCl2COOH}]} = \dfrac{(0.1)(0.01)}{0.0800} = 1.25\times10^{-2}$

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I guess the mistake you are doing is in the equation for $K_a$ for $\ce{CH3COOH}$. See if the following makes sense.

$$\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}=K_a(\ce{CH3COOH})$$ Let $x$ be the $[\ce{H+}]$ from acetic acid, and $y$ from the other organic acid. $$\frac{10^{-1}x}{0.1-x}=10^{-5}$$ $$0.09+x+y=0.1 \text{ (Balancing for H+)}$$ $$\frac{10^{-1}y}{[\ce{CHCl2COOH}]}=K_a(\ce{CHCl2COOH})$$

This should give you the answer.

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  • $\begingroup$ I see you took the ion of the acid to be of the same concentration of the acid, shouldn't it be equal to the concentration of $H^{+}$ $\endgroup$ – Akshat Batra Nov 14 '16 at 11:09
  • $\begingroup$ The equation for Ka does not discriminate between the H+ obtained from different sources. It is an equilibrium constant that just requires the products of the concentrations (with proper exponents) to be equal to some constant. $\endgroup$ – Satwik Pasani Nov 14 '16 at 13:53
  • $\begingroup$ I confirmed with my chem teacher, the equilibrium concentration of $[H^{+}]$ would be equal to $[CH_{3}COOH]$. Thus, $10^{-5}= \frac {x^{2}}{0.1-x}$ $\endgroup$ – Akshat Batra Nov 14 '16 at 14:19
  • $\begingroup$ I think the final pH value is what tells us the equilibrium H+ concentration. And in the equation for Ka, why do you use $x^2$? The concentration of H+ and that of the acetate ion will not be the same. $\endgroup$ – Satwik Pasani Nov 15 '16 at 2:15
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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: $$\ce{[H+] = [OH-] + [Cl-] + [CHCl2COO-] + [CH3COO-]}$$ 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): $$ \ce{[H+] = [OH-] + \frac{[HCl]0 k_{a1}}{[H+] + k_{a1}} + \frac{[CHCl2COOH]0 k_{a2}}{[H+] + k_{a2}} + \frac{[CH3COOH]0 k_{a3}}{[H+] + k_{a3}} }$$

Now what you need to find is $\mathrm k_{a2}$. Solving it at this stage for $\mathrm k_{a2}$ is probably not a good idea aka nothing nice to handle, as the system is quite large. But as you know all values except $\mathrm k_{a2}$ its actually not that bad. $$ \ce{\underbrace{[H+]}_{0.1 mol/l} = \underbrace{[OH-]}_{10^{-13} mol/l} + \underbrace{\frac{[HCl]0 k_{a1}}{[H+] + k_{a1}}}_{0.09 mol/l} + \frac{[CHCl2COOH]0 k_{a2}}{[H+] + k_{a2}} + \underbrace{\frac{[CH3COOH]0 k_{a3}}{[H+] + k_{a3}}}_{10^{-5} mol/l} }$$ $\ldots$ I'll leave units from here on. This simplifies quite much into: $$\begin{align} 0.1 &= 10^{-13} + 0.09 + \frac{0.09\,\mathrm k_{a2}}{0.1 + \mathrm k_{a2}} + 10^{-5}\\ 0.1 - 10^{-13} - 0.09 - 10^{-5} &= \frac{0.09\,\mathrm k_{a2}}{0.1 + \mathrm k_{a2}}\\ 0.01^* &= \frac{0.09\,\mathrm k_{a2}}{0.1 + \mathrm k_{a2}}\\ 0.01\,(0.1 + \mathrm k_{a2}) &= 0.09\,\mathrm k_{a2}\\ 0.001 + 0.01 \mathrm k_{a2} &= 0.09 \mathrm k_{a2}\\ 0.001 &= 0.09 \mathrm k_{a2} - 0.01 \mathrm k_{a2}\\ 0.001 &= (0.09 - 0.01)\,\mathrm k_{a2}\\ 0.001 &= 0.08\,\mathrm k_{a2}\\ \mathrm k_{a2} &= \frac{0.001}{0.08}\\ &= 0.0125 \end{align}$$


* I rounded to $0.01$; it was actually $0.00999\ldots$ which would lead to 0.0124861... mol/l
[1] Equations 2.1 and 2.2
[2] Assumption 1 for HCl, Assumption 3 for acetic acid

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