Question

The K, of a monoprotic weak acid is 0.00812. What is the percent ionization of a 0.138 M solution of this acid? percent ionization:

Answer 1

Assume weak mono protic acid HA. Consider a dissociation of weak acid HA in water.

HA + H₂O H₃O ⁺ + A ^-

For above reaction, Ka = [H₃O ⁺] [A ^-] / [HA] = 0.00812

To calculate [H₃O ⁺], [A ^-], [HA] use ICE table.

Concentration (M)	HA	H3O +	A -
Initial	0.138
Change	-X	+X	+X
Equilibrium	0.138 - X	X	X

Hence we can write, Ka = X ² / 0.138 - X = 0.00812

X ² = (0.138 - X) 0.00812  = 1.120 x 10 ^-03 - 0.00812 X

X ² + 0.00812 X - 1.120 x 10 ^-03 = 0

Comparing above equation with a X ² + b X + c = 0, we get a = 1, b= 0.00812 and c = -1.120 x 10 ^-04

We can solve above equation for value of X.

X= - b +/- (b²-4ac) / 2a

X = - 0.00812 +/- (0.00812)2 - 4 x 1 x (-1.120 x 10 ^-04) / 2 x 1

X = - 0.00812 +/- 6.593 x 10 ^-05 + 4.480 x 10 ^-04 / 2

X = -0.00812 +/- 0.02267 / 2

Therefore, X = - 0.00812 + 0.02267 /2 or X = -0.00812 - 0.02267 / 2

Therefore, X= 0.007275 or X = - 0.01540

Acceptable value of X is 0.007275

Therefore, X = 0.007275 M = [H₃O ⁺]

% Ionization of HA = [H₃O ⁺] / [HA] x 100

% Ionization of HA = (0.007275 / 0.138) x 100

% Ionization of HA = 5.27 %