Hydrazoic acid, HN3, is a weak acid that has an equilibrium constant, Ka?

Question

, equal to 2.8 x 10-5 at 25°C. At 0.300 liter sample of 0.050 molar solution is prepared.

c. To 0.150 liter of this solution, 0.80 gram of sodium azide, NaN3, is added.

The salt dissolves completely. Calculate the pH of the resulting solution at 25°C if the volume of the solution remains unchanged.

d. To this remaining 0.150 liter of the original solution, 0.075 liter of 0.100 molar NaOH solution is added. Calculate the [OH-] for the resulting solution.

Answer 1

pH = pKa + log([N3-]/[HN3])

pKa = -log(2.8x10^-5) = 4.55

0.150 L x 0.050 M = 0.0075 mol HN3

0.80 g NaN3/ 65 g/mol = 0.0123 mol NaN3

c) pH = 4.55 + log(0.0123/0.0075) = 4.55 + 1.64 = 6.19

d) 0.075 L x 0.100 = 0.0075 mol NaOH

mol HN3 = 0.0075 - 0.0075 = 0

mol NaN3 = 0.0075

volume = 0.225 L

NaN3 molarity = 0.0075/0.225 = 0.0333

N3- + H2O --> HN3 + OH-

Kb = Kw/Ka = 1.0x10^-14/2.8x10^-5 = 3.57x10^-10

Kb = {HN3][OH-] / [N3-]

after hydrolysis of N3- :

let X = [HN3] = [OH-]

[N3-] = 0.033 - X

assume X << 0033

Kb = 3.57x10^-10 = X^2/0.033

X = 3.4x10^-6

pOH = 5.46

pH = 14 - 5.46 = 8.53