What is the ph of a 0.015 m aqueous solution of hydrazoic acid (hn3) (ka = 1.9 x 10–5) at 25â°c?

Answer is: pH value of aqueous solution of hydrazoic acid is 4,72.
Chemical reaction: N₃H + H₂O ⇄ N₃⁻ + H₃O⁺.
c(N₃H) = 0,015 M.
Ka(N₃H) = 1,9·10⁻⁵.
[N₃⁻] = [H₃O⁺] = x.
[N₃H)] = 0,015 M - x.
Ka = [N₃⁻] · [H₃O⁺] / [N₃H].
1,9·10⁻⁵= x² / 0,015 M -x.
Solve quadratic equation: x = [H₃O⁺] = 0,000019 M.
pH = -log [H₃O⁺] = 4,72.

The kinetic energy of the liquid increases. hope this !