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Let's say you are asked to calculate the pH of a solution with:

  • $\mathrm{0.1~M}~\ce{HCl}$
  • $\mathrm{0.2~M}~\ce{CH3COOH}$.
  • $K_\mathrm a = 2\times 10^{-5}$

This is how I would do it:

$$\ce{CH3COOH + H2O <=> H3O + CH3COO-}$$

The corresponding ICE table: \begin{array}{|c|cccc|}\hline ~ & \ce{CH_3COOH}& \ce{H_2O}& \ce{H_3O^+}& \ce{CH_3COO^-}\\ \hline \mathrm I & 0.2 & & & 0.1\\\Delta & -x & & x & x\\ \mathrm E& 0.2-x & & x & 0.1 +x\\\hline\end{array}

calculate the pH:

$$\begin{align} K_\mathrm a &= \frac{\ce{[H3O+][CH3COO^{-}]}}{[\ce{CH3COOH}]}\\ &= \frac{x \cdot (0.1+ x)}{(0.2-x)} \end{align}$$ as $x$ is very small, we leave it out: $$\begin{align} \implies 2\times 10^{-5} &= \frac{0.1x}{0.2}\\ \implies 2\times 10^{-5} &= \frac{0.1x}{0.2}\\ \implies~~~~~~~~~~~~ x &= 4\times 10^{-5}\\ \implies -\log(x) &= 4.40~= \mathrm{pH}\\ \end{align}$$

  • First of all. Is this correct?
  • Where do you include $\ce{HCl}$? I didn't write it anywhere. And I see most of the people not writing the second molecule. How/where is it included?
  • What is the molecule in the first column of the ice table? What does it represent? Do you always write the acid in that column?

I made that exercise up. The original exercise has $0.1$ molar $\ce{HCl}$ and $0.1$ molar $\ce{CH3COOH}$. I changed it in order to clearly see which value is which one at which place in stead of all having the same ones. This is the original exercise:

\begin{array}{|c|ccc|}\hline & \ce{CH_3COOH} & \ce{H_3O^+} & \ce{CH_3COO^-}\\ \hline \mathrm I & 0.1 & & 0.1\\ \Delta& -x & x & x\\ \mathrm E &0.1-x & x & 0.1 +x\\ \hline\end{array}

\begin{align}K_\mathrm a &= \frac{x\cdot (0.1 +x)}{(0.1-x)}\\ &\approx \frac{x\cdot 0.1}{0.1}\\ \implies x&= 2\times 10^{-5}\\\therefore -\log(x) &= 4.70\;. \end{align}

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  • $\begingroup$ You seem to have some command of LaTeX formatting please review the edits to understand that you don't need '\ce{}' to make the the equations do subscripts and multiplication. $\endgroup$ – A.K. Jan 12 '16 at 20:40
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    $\begingroup$ @A.K. Please make sure when you edit, that you don't accidentally convert multiplication sings to decimal delimiters. $\endgroup$ – Martin - マーチン Jan 13 '16 at 4:32
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If ...

[...] you are asked to calculate the pH of a solution with:

  • 0.1 M HCl
  • 0.2 M AcOH ($K_a = 2 \cdot 10^{−5}$)

it would be a tremendous waste of time to assume it being a buffer solution and think about how you can get everything to work.

Hydrochloric acid is at least 1000000 times stronger than acetic acid which tells you that the resulting pH will be the pH of your 0.1 M HCl which then again is a very easy question (namely what is the pH of a 0.1 M HCl solution?).


Even if the question would have been to calculate the pH of a solution of

  • 0.1 M HCl
  • 0.2 M NaOAc

the question would have been anything else than hard to solve, too.

Adding 0.1 mol of HCl to 0.2 mol of NaOAc will give you 0.1 mol acetic acid and 0.1 mol acetate which only needs to be entered into the Henderson-Hasselbalch equation.

$$pH = pK_a + \log_{10} \frac{[A^-]}{[HA]}$$

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The answer given is totally wrong. With 0.1 molar HCl, then the pH would be 1.0 from the HCl and virtually none of the acetic acid would be dissociated.

REVISED PROBLEM

I made that exercice up. The original exercice has $0.1$ molar $\ce{HCl}$ and $0.1$ molar $\ce{CH3COOH}$.

This makes no sense.

The work shown in the second image is for 0.1 molar $\ce{CH3COOH}$ and 0.1 molar $\ce{CH3COO^{-}}$. That work gets the right answer, but the wrong way. At equilibrium: $$[\ce{CH3COOH}] = 0.1 - x$$ but $x$ is so small that $$[\ce{CH3COOH}] \approx 0.1$$ and $$[\ce{CH3COO^{-}}] \approx 0.1$$ which simplifies the mathematics considerably.

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  • $\begingroup$ ok I made that exercice up. The original exercice has 0.1 molar HCl and 0.1 molar $\ce{CH3COOH}$. I changed it in order to clearly see which value is which one at which place in stead of all having the same ones. This is the original exercice: imgur.com/d3URXIk . Could you please tell what is wrong? (I've seen some people even use two ice tables to solve it) $\endgroup$ – ohiliouh Jan 12 '16 at 20:32
  • $\begingroup$ Or could you please show/explain how to solve it correctly? $\endgroup$ – ohiliouh Jan 12 '16 at 20:49

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