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So, I have 1 litre of acetate buffer, and 0.1 mol of oxonium ions/$\ce{H3O+}$ is added. The task is to calculate the pH of this solution.

What I was thinking:

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$ $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \mathrm{p}K_\mathrm{a} + \log\frac{n(\ce{Ac-})}{n(\ce{HAc})}$$ $$\mathrm{pH} = 4.74 + \log\frac{n(\ce{Ac-}) - 0.1\ \mathrm{mol}}{n(\ce{HAc}) + 0.1\ \mathrm{mol}}$$

But neither $c(\ce{Ac-})$ nor $c(\ce{HAc})$ (which should, in theorey, be equal) is given. The only thing I know (or I think I know) is this: $$\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \frac{1}{1} = 1$$

So, is it possible to calculate the pH from that? Am I missing something?

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A buffer is most effective when conjugate acid and base are in a 1:1 ratio, but that is not always the case (particularly after we start adding acid to it.) Do you know the concentration of buffering compounds prior to the addition of acid? If not, the question can't be answered - the pH after adding the acid could vary widely depending on the buffering capacity of the solution. – Jason Patterson Jul 5 '15 at 3:35

I'm tring to write the methodology
If anything is wrong please point it out to me i'll be happy to correct!

To find $\small\ce{pH} $ of a solution you can:

  1. Find the conjugated base of your acid (for example $\small\ce{HCOOH}$ becomes $\small\ce{HCOO-}$ (correct me if i'm wrong here)

  2. Write the equation of the solution
    Now you want to prepare an aqueous solution.

  3. Find the concentration of $\small\ce{[CH_3COOH]}$ you use the formula to determine the concentration in solution of acid

    $$ \ce{[CH_3COOH]\ =\ C\ -\ [CH3COO- ]} $$

    ($\ce{C}$ is the concentration of brought acid in $\ce{mol/l}$)

  4. Then you write that:

    $$\qquad\ce{[H+ ]\ =\ [CH3COO- ]}$$ as both dissociate from the same acid as

    $$\ce{CH3COOH\ ->\ CH3COO- \ +\ H+}$$

  5. the $\small\ce{pH}$ concentration is given by:
    $$\ce{pH\ =\ -\log(\small[H_3O+ ])}$$

$pH$ here you are! (without any unit)

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