If $pK_a = 4.76$, then $K_a = 1.74 \times 10^{-5}$
Then approximating that $K_a$ can be written in terms of concentration rather than the true definition which is in terms of activities:
$K_a = 1.74 \times 10^{-5}M = \frac{[A-][H+]}{[HA]} = \frac{[H+]^2}{c-[H+]} $
$1.74 \times 10^{-5}M = \frac{[H+]^2}{c-[H+]} $
where $c$ is the molar concentration of the vinegar
If you want pH explicitly in the equation, then approximating pH = -log[H+], then:
$1.74 \times 10^{-5}M = \frac{(10^{-pH}M)^2}{c-(10^{-pH}M)} $
So, for example if pH is 3, $10^{-3}$ is 0.001, so
$1.74 \times 10^{-5} = \frac{10^{-6}M}{c-(10^{-3}M)}$
$1.74 \times 10^{-5}c -1.74 \times 10^{-8}M = 10^{-6}M$
$1.74 \times 10^{-5}c = 1.02 \times 10^{-6}M$
$c = 0.059M$