Question

The percentage ionizations of 0.01M HCN (K_a=4.9x10^-10) and 0.01 M NH₃( K_b=1.8x10^-5), are respectively:

a) 4.24%, 2.2%        b) 2.2%, 4.24%          c) 2.2x10^-6%, 4.24x10^-4%      d) 6.0%, 7.0%   e)7.0%, 6.0%

Answer 1

Correct answer is : (b) 2.2 % , 4.24 %

Explanation

For HCN

[H⁺] = (K_a * C)^1/2

[H⁺] = (4.9 x 10^-10 * 0.01 M)^1/2

[H⁺] = (4.9 x 10^-12 M²)^1/2

[H⁺] = 2.2 x 10^-6 M

Percentage ionization = ([H⁺] / C) * 100

Percentage ionization = (2.2 x 10^-6 M / 0.01 M) * 100

Percentage ionization = 0.022 %

For NH₃

[OH^-] = (K_b * C)^1/2

[OH^-] = (1.8 x 10^-5 * 0.01 M)

[OH^-] = (1.8 x 10^-7 M²)^1/2

[OH^-] = 4.24 x 10^-4 M

percent ionization = ([OH^-] / C) * 100

percent ionization = (4.24 x 10^-4 M / 0.01 M) * 100

percent ionization = 4.24 %