Question

Aspirin (C9H8O4) is a weak acid with a Ka = 3.16 x 10–4. What is the pH of a solution that is 0.25 M in aspirin?

Please do not copy and paste the response from this question as it has previously been asked on this site. The work there didn't help me with the problem.

Answer 1

Lets suppose that aspirin has the following form:

HA, then the dissociation will be:

..................HA ==== H + A, then write the equilibrium charte (ICE)

I................0.25........0............0

C...............-x............x............x

E............0.25-x.......x.............x

The equilibrium constant is the relation ship between products to reactants, for acids it is called Ka, so:

Ka= [H] [X] / [HA]

x² / 0.25 - x = 3.16 x 10^–4

if you solve this equation for x, you will get a value of 0.00874

this is the concentration of H so

PH = - log H

PH = - log (0.00874) = 2.06

*hope it helps =)