Question

Calculate the percent ionization of formic acid solutions having the following concentrations:

1) 0.520 M

2) 0.140 M

3) 4.60 * 10 ^ -2 M

Answer 1

Formic acid dissociation:

HCO₂H <-> H⁺ + HCO₂^-

With Ka = 1.8 x 10^-4

We make also our BCA table, for number 1)

	HCO2H	<->	H+	+	HCO2-
B	0.52		0		0
C	-x		+x		+x
A	0.52 - x		x		x

1.8 x 10^-4 = [H⁺] [HCO₂^-] / [HCO₂H]

1.8 x 10^-4 = [x²] / [0.52 - x]

x = 0.00958

%ionization = 0.00958 / 0.52 = 1.84%

For number 2)

1.8 x 10^-4 = [x²] / [0.14 - x]

x = 0.00493

%ionization = 0.00493 / 0.14 = 3.52%

For number 3)

1.8 x 10^-4 = [x²] / [0.046 - x]

x = 0.002789

%ionization = 0.002789 / 0.046 = 6.063%