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:
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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