How can I calculate pH of weak acid?

1 Answer
Dec 6, 2014

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.