Question

A solution containing a monoprotic weak acid (HA) that is 8.82% dissociated was found to have a pH of 2.91.

What is the concentration of A– at equilibrium?

What is the concentration of HA at equilibrium?

Calculate the pKa for this acid.

Answer 1

1)

use:

pH = -log [H+]

2.91 = -log [H+]

[H+] = 1.23*10^-3 M

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

C      0 0

C-x      x x

x = [H+] = 1.23*10^-3 M

So,

[A-] = x = 1.23*10^-3 M

Answer: 1.23*10^-3 M

2)

% dissociation = x*100/C

8.82 = (1.23*10^-3)*100/C

C = 0.0139 M

[HA] = C-x

= 0.0139 M - 1.23*10^-3 M

= 0.0127 M

Answer: 0.0127 M

3)

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

= (1.23*10^-3)*(1.23*10^-3) / (0.0127)

= 1.19*10^-4

use:

pKa = -log Ka

= -log (1.19*10^-4)

= 3.92

Answer: 3.92