Question

Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 1.0×10−5.

Find the percent dissociation of this solution.

Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 1.6×10⁻³.

Find the percent dissociation of this solution.

Answer 1

1 and 2)
for simplicity lets write weak acid as HA

HA          ----->     H+   +   A-
0.120                 0         0
0.120-x               x         x

Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((1.0E-5)*0.120) = 1.10E-3

pH = -log [H+] = -log (1.10E-3) = 2.96

percent dissociation = x*100/c = 1.10E-3 * 100/0.120 = 0.917 %

3 and 4)
for simplicity lets write weak acid as HA

HA          ----->     H+   +   A-
0.120                 0         0
0.120-x               x         x

Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((0.0016)*0.120) = 1.39E-2

pH = -log [H+] = -log (1.39E-2) = 1.86

percent dissociation = x*100/c = 1.39E-2 * 100/0.120 = 11.6 %