Let's call the acid HA.
As it's a weak acid it will dissociate according to :
HArightleftharpoonsH^(+)+A^-
So we can write:
K_a=([H]^(+)[A^(-)] )/([HA])
You need to know its initial concentration and K_a value so here's an example:
What is the pH of a 0.01 mol/litre solution of ethanoic acid?
K_(a)=1.7xx10^(-5)mol.dm^(-3)
CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+
Initially there are 0.01 mol CH_3COOH
When equilibrium is established we can say that x moles dissociate so we must be left with (0.01-x) moles of CH_3COOH and x moles of CH_3COO^- and x moles H^+
So
K_(a)=(x.x)/((0.01-x))
At this point we make an assumption that the amount x is so small compared to 0.01 that we can assume (0.01-x)rarr0.01
So x^2=K_(a)xx0.01
x^(2)=1.7xx10^(-5)xx0.01=1.7xx10^(-7)
x = 4.12xx10^(-4)=[H^+]
pH = -log[H^+] = -log4.12xx10^(-4)=3.4
This is at 298K.