Question

Find the pH and percent ionization of a 0.100 M solution of a weak monoprotic acid having the given Ka values.

(a) Ka = 1.9 10-5

(b) Ka = 1.9 10-3

(c) Ka = 1.9 10-1

Answer 1

We know that the equilibrium of a weak monoprotic acid is as follows:

HA <-> H⁺ + A^-

We have 0.1 M HA, so using the constant definition:

Ka = [H⁺][A^-] / [HA]

Making an ICE table:

	HA	<->	H+	+	A-
I	0.1 M		0 M		0 M
C	-x		+x		+x
E	0.1 - x		x		x

For letter a):

1.9 x 10^-5 = x² / (0.1 - x)

Isolating for x:

x = [H⁺] = 0.00136894

pH = -log (0.00136894) = 2.8636

% ionization = (0.00136894/0.1) * 100 = 1.3689%

For letter b):

1.9 x 10^-3 = x² / (0.1 - x)

Isolating for x:

x = [H⁺] = 0.0128667

pH = -log (0.0128667) = 1.89

% ionization = (0.0128667/0.1) * 100 = 12.8667%

For letter c)

1.9 x 10^-1 = x² / (0.1 - x)

Isolating for x:

x = [H⁺] = 0.0724

pH = -log (0.0724) = 1.14

% ionization = (0.0724/0.1) * 100 = 72.4%