How can one calculate the pH of a solution? 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:

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

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?
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:  


Find the conjugated base of your acid  (for example $\small\ce{HCOOH}$ becomes $\small\ce{HCOO-}$ (correct me if i'm wrong here)  
Write the equation of the solution
Now you want to prepare an aqueous solution. 
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 $\ce{mol/l}$)
Then you write that:  

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

$$\ce{CH3COOH\ -&gt;\ CH3COO- \ +\ H+}$$
the $\small\ce{pH}$ concentration is given by:
$$\ce{pH\ =\ -\log(\small[H_3O+ ])}$$ 


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