Answer to Question #170670 in Chemistry for Hyeri

Question

If a monoprotic acid is 3.8 percent ionized is a 0.1 F solution calculate its ionization constant what is the percentage ionazation of this acid in a 0. 2 F solution?

Answer 1

Ionization constant cam be calculated using the formula:

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

As 3.8 % of acid is ionized, the concentration of ionized products equals:

[H+] = [A-] = 3.8 % × [HA] / 100 % = 0.038 × 0.1 F = 0.0038 F

From here:

Ka = [H+] [A-] / [HA] = 0.0038 F × 0.0038 F / (0.1 F - 0.0038 F) = 1.5 × 10^-4

The concentration of ionized products in 0.2 F solution equals:

[H+]² = Ka × [HA] = 1.5 × 10^-4 × 0.2 F = 3 × 10^-5 F²

[H+] = 0.0055 F

The percentage ionization of the acid:

%i = ([H+] / [HA]) × 100% = (0.0055 / 0.2) × 100% = 2.75%

Answer: 1.5 × 10^-4; 2.75 %