We are being asked to calculate the K_{a} for hydrazoic acid HN_{3} if a 0.50 M solution of NaN_{3} has a pH of 9.21.

NaN_{3} is a salt and it will break up in the solution:

**NaN _{3}_{(aq)}**

1 mol of NaN_{3} produces 1 mol N_{3}^{-}

[NaN_{3}] = **[N _{3}^{-}] = 0.50 M**

N_{3}^{-} is the conjugate base of a weak base HN_{3}:

** HN _{3(aq)}**

(weak base) (conjugate acid)

**We're going to calculate for the Ka of HN _{3} using the following steps:**

Sodium azide, NaN_{3} is an ionic compound consisting of Na^{+} ions and N3^{-} ions. A 0.50 M solution of NaN_{3} has a pH of 9.21. Calculate K_{a} for hydrazoic acid, HN_{3}.

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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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.