The Ka of hydrazoic acid (HN3) is 1.9 × 10-5 at 25°C. What is the pH of a 0.35-M aqueous solution
of HN3?
To find the pH of a solution of HN3, we need to determine the concentration of H+ ions in the solution. HN3 is a weak acid, so we can use the equilibrium expression to find the concentration of H+ ions.
The balanced chemical equation for the dissociation of HN3 in water is:
HN3 ⇌ H+ + N3-
The Ka expression for HN3 is:
Ka = [H+][N3-] / [HN3]
Since the concentration of H+ ions is very small compared to the concentration of HN3, we can assume that the concentration of H+ ions at equilibrium is approximately equal to the concentration of HN3 that dissociates. We can represent this using the equation:
[H+] = [HN3] * √(Ka)
Given that the concentration of HN3 is 0.35 M and the Ka of HN3 is 1.9 × 10^(-5), we can substitute these values into the equation to find the concentration of H+ ions:
[H+] = 0.35 * √(1.9 × 10^(-5))
Calculating this expression gives us:
[H+] ≈ 0.35 * 1.38 × 10^(-3) ≈ 4.83 × 10^(-4) M
Now, to find the pH of the solution, we can use the formula:
pH = -log[H+]
Substituting the concentration of H+ ions we found into this formula, we get:
pH = -log(4.83 × 10^(-4))
Using a calculator, we can determine that the pH of the 0.35 M aqueous solution of HN3 is approximately 3.32.