equilibrium - What's the pH of vinegar containg 5% acetic acid?

Vinegar generally contains 5% acetic acid. We would expect the pH of vinegar to be approximately:
a. 0
b. 3
c. 7
d. 9
e. 12

I don't have the key for this question, so I just want to make sure that I got it right and what I got doesn't seem to be one of the answer choices.

My answer was 1.3, so I rounded it to 1 and there wasn't an answer choice. Here is what I did:

$$\log(.05/1) = -1.3010$$

How else would you do it because you were already given the final concentration from the original concentration?

Am I missing something here?

Edit: The professor said to use this formula for calculating it:

$$K_\mathrm{a} = \frac{(.05x)^2}{x}$$

and got 3, which made no sense, any way this is possible?

Answer

You are not given a concentration, but a percentage value. Assuming its mass percent (wt%), you first have to calculate the actual concentration of the solution like this:

5% Acetic acid means in a $\pu{1 kg}$ solution, you can find $0.05 \times \pu{1000 g} = \pu{50 g}$ acetic acid. With a molar mass of $\pu{60.05 g/mol}$, this is similar to

$$n = \frac{\pu{50 g}}{\pu{60.05 g/mol}} = \pu{0.83 mol}$$

The concentration $C = n/V$, therefore is

$$C = \frac{\pu{0.83 mol}}{\pu{0.95 L}} = \pu{0.874 mol/L}$$

Acetic acid is a weak acid. This means it does not dissociate completely and $\mathrm{p}K_\mathrm{a}$ has to be taken into account to calculate the pH. The $\mathrm{p}K_\mathrm{a}$ of acetic acid is about $4.76$. The pH is then defined as:

$$\mathrm{pH} = 0.5 (\mathrm{p}K_\mathrm{a} - \log C_\mathrm{a}) = 0.5 (4.76 - \log 0.874) = 2.4$$