How can one calculate the pH of a solution?

Question

So, I have 1 litre of acetate buffer, and 0,1mol of oxonium ions/H3O+ is added. The task is to calculate the pH of this solution.

What I was thinking:

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

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?

Answer 1

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)