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.