072 M of Hydrazoic acid HN3 Ka 24 x 105 is placed in a solu

0.72 M of Hydrazoic acid (HN3, Ka = 2.4 x 10-5) is placed in a solution with 0.62 M of sodium azide (NaN3). What is the pH of the solution at equilibrium?

Solution

From relation:

pH = pKa + [salt] / [acid] ----------(1)

given [Hydrazoic acid] = 0.72M and [sodium azide] = 0.62M and Ka = 2.4 ×10⁵

pKa = -logKa

pKa = -log2.4 ×10⁵

pKa = -(-4.62)

pKa = 4.62

put all the values in eq (1)

pH = 4.62 + log[0.62] / [0.72]

pH = 4.62 + log0.8611

pH = 4.62 - 0.065

pH = 4.55