Question

The pH of a 0.10 M solution of hydrazoic acid, HN₃, is 2.86. Calculate the Ka value for hydrazoic acid.

A. 1.4 x 10-4
B. 1.9 x 10-5
C. 1.9 x 10-3
D. 1.4 x 10-5
E. 2.6 x 10-3

Answer 1

Solution:- HN₃(hydrazoic acid) is a weak monoprotic acid means one mole of it gives 1 mole of H⁺ in the solution. The equation could be written as...

HN₃(aq) + H₂O <--------> H₃O⁺(aq) + N₃^-(aq)

Let's make the ice table for this weak acid.

      HN₃(aq) + H₂O <--------> H₃O⁺(aq) + N₃^-(aq)

I       0.10                                0                      0

C         -X                               +X                    +X

E       0.10 - X                         X                        X

Ka = [H₃O⁺] [N₃^-]/[HN₃]

we know that, pH = - log [H₃O⁺]

on taking antilog...

[H₃O⁺] = 10^-pH

pH is given as 2.86.

[H₃O⁺] = 10^-2.86 = 0.00138

From ice table, [H₃O⁺] = [N₃^-] = 0.00138

Let's plug in the values in Ka expression..

Ka = [(0.00138)²](0.10 - 0.00138)

Ka = 1.93 x 10^-5

So, the corect choice is (B) 1.9 x 10^-5.