Question

A weak acid has a dissociation constant of 3.0 x 10-5. Calculate the pH of a 2.90 M solution of this acid.

Q 13

Answer 1

Lets write the dissociation equation of HA

HA -----> H+ + A-

2.9 0 0

2.9-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((3*10^-5)*2.9) = 9.327*10^-3

since c is much greater than x, our assumption is correct

so, x = 9.327*10^-3 M

So, [H+] = x = 9.327*10^-3 M

we have below equation to be used:

pH = -log [H+]

= -log (9.327*10^-3)

= 2.03

Answer: 2.03