Question

Calculate the percent ionization of hydrazoic acid (HN3) in solutions of each of the following concentrations (Ka=1.9x10^-5)

Part a) .412 M

Part b) .108 M

Part c) 4.01x10^-2

Answer 1

ANSWER:

K_a for HN₃ = 1.9 * 10^-5

Part (a):

The reaction can be written as:

K_a = {[H⁺] [N₃^-]} / {[HN₃]}

1.9 * 10^-5 = { [x] [x]} / {[0.412 - x]}

1.9 * 10^-5 * 0.412 = x² {x is very smaller than 0.412 so, it is negleted}

7.83 * 10^-6 = x²

x = 0.0028 M = [H⁺]

And,

percentage ionization = {[H⁺] * 100} / {[HF]}

= {0.0028 M * 100} / {0.412 M}

= 6.8 % ionized

Hence, % ionization of a 0.412 M HF solution is 6.8 %.

Part (b) :

The reaction can be written as:

K_a = {[H⁺] [N₃^-]} / {[HN₃]}

1.9 * 10^-5 = { [x] [x]} / {[0.108 - x]}

1.9 * 10^-5 * 0.108 = x² {x is very smaller than 0.108 so, it is negleted}

2.05 * 10^-6 = x²

x = 0.0014 M = [H⁺]

And,

percentage ionization = {[H⁺] * 100} / {[HF]}

= {0.0014 M * 100} / {0.108 M}

= 1.3 % ionized

Hence, % ionization of a 0.108 M HF solution is 1.3 %.

Part (c) :

The reaction can be written as:

K_a = {[H⁺] [N₃^-]} / {[HN₃]}

1.9 * 10^-5 = { [x] [x]} / {[4.01 * 10^-2 - x]}

1.9 * 10^-5 * 4.01 * 10^-2 = x² {x is very smaller than 4.01 * 10^-2 so, it is negleted}

7.62 * 10^-7 = x²

x = 0.00087 M = [H⁺]

And,

percentage ionization = {[H⁺] * 100} / {[HF]}

= {0.00087 M * 100} / {4.01 * 10^-2 M}

= 2.2 % ionized

Hence, % ionization of a 4.01 x 10^-2 M M HF solution is 2.2 %.