Percent Ionization

Question

A.)Calculate the percent ionization of 7.5×10⁻³ butanoic acid 

B.)Calculate the percent ionization of 7.5×10⁻³ butanoic acid in a solution containing 7.5×10⁻² sodium butanoate.

Answer 1

HA = butanoic acid, CH3CH2CH2COOH

A^- = butanoate, CH3CH2CH2COO^-

Ka = 1.5x10^-5 (from the internet)

HA ==> H⁺ + A^-

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

1.5x10^-5 = (x)(x) /(7.5x10^-3 - x) and if x is small, 7.5x10^-3 - x = 7.5x10^-3

x² = 1.125x10^-7

x = 3.35x10^-4 and this is ~ 4.5% of 7.5x10^-3 so neglecting it in the above calculation is justified

%ionization = 3.35x10^-4/7.5x10^-3 x (100%) = 4.47%

(b) Common ion effect:

HA ====> H⁺ + A^-

7.5x10^-3.....0.........7.5x10^-2 .....Initial

-x...............+x.........+x.............Change

0.0075-x......x.........0.075+x......Equilibrium

1.5x10^-5 = (0.075 + x)(x) / 0.0075 - x and again assuming x is small, we have...

1.5x10^-5 = (0.075 )(x) / 0.0075

0.075x = 1.125x10^-7

x = 1.5x10^-6 M

%ionization = 1.5x10^-6 / 7.5x10^-3 (x100%) = 0.02%

NOTE the huge difference in % ionization when a common ion is present.