Problem: Sodium azide, NaN3 is an ionic compound consisting of Na+ ions and N3- ions. A 0.50 M solution of NaN3 has a pH of 9.21. Calculate Ka for hydrazoic acid, HN3.A. 1.9 x 10-5B. 3.5 x 10-6C. 2.4 x 10-8D. 5.3 x 10-10E. 6.2 x 10-9

FREE Expert Solution

We are being asked to calculate the Ka for hydrazoic acid HN3 if a 0.50 M solution of NaN3 has a pH of 9.21.


NaN3 is a salt and it will break up in the solution:

NaN3(aq) → Na+(aq) + N3-(aq)

1 mol of NaN3 produces 1 mol N3-

[NaN3] = [N3-] = 0.50 M


N3- is the conjugate base of a weak base HN3:

       HN3(aq)    +   H2O(l)    →    N3-(aq)    +    H3O+(aq)
     (weak base)                     (conjugate acid)


We're going to calculate for the Ka of HN3 using the following steps:

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Problem Details

Sodium azide, NaN3 is an ionic compound consisting of Na+ ions and N3- ions. A 0.50 M solution of NaN3 has a pH of 9.21. Calculate Ka for hydrazoic acid, HN3.

A. 1.9 x 10-5

B. 3.5 x 10-6

C. 2.4 x 10-8

D. 5.3 x 10-10

E. 6.2 x 10-9

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