Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 1.2×105.

ID: 1009968 • Letter: F

Question

Find the percent ionization of a 0.120 M solution of a weak monoprotic acid having Ka= 1.2×105

Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 1.2×103.

Find the percent ionization of a 0.120 M solution of a weak monoprotic acid having Ka= 1.2×103.

Find the pH of a 0.120 M solution of a weak monoprotic acid having Ka= 0.18.

Find the percent ionization of a 0.120 M solution of a weak monoprotic acid having Ka= 0.18.

A ) For monoprotic acid, let x amount has dissociated,

Ka = [H3O+][A-]/[HA]

1.2 x 10^-5 = x^2/0.120

x = [H3O+] = 1.2 x 10^-3 M

pH = -log[H3O+] = 2.921

B ) Percent ionization = 1.2 x 10^-3 x 100/0.120 = 1.00 %

C ) For monoprotic acid with x amount of dissociation,

Ka = [H3O+][A-]/[HA]

1.2 x 10^-3 = x^2/0.120

x = [H3O+] = 0.012 M

pH = -log[H3O+] = 1.921

D ) Percent ionization = 0.012 x 100/0.12 = 10.00 %

E ) For monoprotic acid with x amount of dissociation,

Ka = [H3O+][A-]/[HA]

0.18 = x^2/(0.120 - x)

x^2 + 0.18x - 0.0216 = 0

x = [H3O+] = 0.0823 M

pH = -log[H3O+] = 1.084

F ) Percent ionization = 0.0823 x 100/0.12 = 68.58 %

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