Question

a. Find the pH of a 0.100 M solution of a weak monoprotic acid having Ka= 1.1×10⁻⁵.

b.Find the percent dissociation of this solution.

c. Find the pH of a 0.100 M solution of a weak monoprotic acid having Ka= 1.8×10⁻³

^d. Find the percent dissociation of this solution.

e.Find the pH of a 0.100 M solution of a weak monoprotic acid having Ka= 0.16.

f.Find the percent dissociation of this solution.

Answer 1

a )

C = 0.100 , Ka = 1.1 x 10^-5

[H+] = sqrt (Ka x C) = sqrt (1.1 x 10^-5 x 0.1) = 1.05 x 10^-3 M

pH = -log [H+] = -log (1.05 x 10^-3)

pH = 2.98

b) percent dissociation = [H+] x 100 / C = 1.05 x 10^-3 x 100 / 0.1

percent dissociation = 1.05 %

c)

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

0.1-x x x ---------------> equilibrium

Ka = x^2 / 0.1-x

1.8 x 10^-3 = x^2 / 0.1-x

x^2 + 1.8 x 10^-3 x - 1.8 x 10^-4 = 0

x = 0.0125

[H+] = 0.0125 M

pH = 1.90

d)

percent dissocitation= ( x / C ) x 100

= (0.0125 / 0.1) x 100

= 12.5 %

percent dissocitation  = 12.5 %

e)

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

0.1-x x x ---------------> equilibrium

Ka = x^2 / 0.1-x

0.16 = x^2 / 0.1-x

x^2 + 0.16 x - 0.016 = 0

x = 0.0697

[H+] = 0.0697M

pH = 1.16

f )

percent dissocitation= ( x / C ) x 100

= (0.0697 / 0.1) x 100

= 69.7 %

percent dissocitation  = 69.7 %