Understanding how to calculate the pH of a buffer with ice tables 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:

$$\ce{CH3COOH + H2O &lt;=&gt; H3O + CH3COO-}$$

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

calculate the pH:

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


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 &amp; \ce{CH_3COOH} &amp; \ce{H_3O^+} &amp; \ce{CH_3COO^-}\\ \hline \mathrm I &amp; 0.1 &amp; &amp; 0.1\\ \Delta&amp; -x &amp; x &amp; x\\ \mathrm E &amp;0.1-x &amp; x &amp; 0.1 +x\\ \hline\end{array}

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