Question

What is the pH of a 0.084 M solution of hydrazoic acid. HN3?

Answer 1

its a week acid we need to construct the ICE table

since water is exceaa we do not to consider the concentration of H2O   

HN₃ + H₂O ---> H₃O⁺ + N3^-

I 0.084 0 0

C -x +x +x

E 0.084-x +x +x

now write the equilibrium expression or dissociation constant

Ka = [N3^- ] [H₃O⁺] / [HN₃ ]

Ka value for the hydrozoic acid is 2.5 x 10^-5 (from the text book standard value)

2.5 x 10^-5 = [x][x] / [0.084-x]

x² +  2.5 x 10^-5x - 2.1x10^-6 = 0

solve the quadratic equation

x = 0.001436 = [H3O+]

pH = -log[H3O+]

pH = -log[0.001436]

pH = 2.84