Answer:
% ionization for benzoic acid = 0.08%
% ionization for sodium benzoate = 2.5%
The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions
Explanation:
Ka for pure water = 1.0 * 10-⁷
Ka for sodium benzoate = 6.5*10-⁵
1. For benzoic acid (C6H5COOH)
C6H5COOH ==== C6H5COO‐ + H+
0.15M 0 0
0.15-x x x
Ka = [C6H5COO-] [H+] / [C6H5COOH]
Ka = [X] [X] / 0.15 - X
1.0*10-⁷ = [X]² / 0.15 - x
But x is negligible compared to 0.15,
(1.0*10-⁷)*0.15 = x²
Take square root of both sides,
X = 1.22 * 10-⁴
% ionization = ( [H+] / [C6H5COOH] ) * 100
% ionization = (1.22*10-⁷ / 0.15) * 100
% ionization = 0.08%
2. For C6H5COONa
Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions
Ka for C6H5COONa = 6.5*10-⁵
6.5*10-⁵ = [X]² / (0.10 - X)
Cross-multiply both sides;
(6.5*10-⁵ * 0.10) = X²
Take square root of both side,
X= 2.5*10-³
% ionization = (2.5*10-³ / 0.10) *100
% ionization = 2.5%
