An ammonia solution has a pH of 8.30. What is its [H₃O⁺]? A) 5.0 × 10⁻²³ M B) 2.0 × 10⁻⁶ M C) 5.0 × 10⁻⁹ M D) 2.0 × 10⁸ M

Question

Quizplus · Accepted Answer

The pKa of ammonium (NH4+) is 9.24, therefore we can use the Henderson-Hasselbalch equation:
pH = pKa + log([base]/[acid])
8.30 = 9.24 + log([NH3]/[NH4+])
-0.94 = log([NH3]/[NH4+])
Antilog of -0.94 is 4.75, which represents the ratio of [NH3]/[NH4+].
Assuming the initial concentration of ammonia is x, then the initial concentration of ammonium is also x, and the equilibrium concentrations are:
[NH3] = 4.75x
[NH4+] = x
Since the sum of the concentrations of NH3 and NH4+ is equal to the initial concentration of ammonia, we have:
4.75x + x = [NH3] + [NH4+] = x
5.75x = x
x = 1/[NH3] = 1/[NH4+] = 5.75
Therefore, the concentration of HS- is equal to that of NH4+:
[HS-] = [NH4+] = 5.75 M
The concentration of S2- can be calculated using the dissociation constant of HS-:
Kb = Kw/Ka = (1.00 x 10^-14)/(1.77 x 10^-5) = 5.65 x 10^-10
Kb = [H+][HS-]/[S2-]
Assuming the concentration of [H+] is negligible (compared to the concentration of HS-), we have:
[S2-] = [HS-]^2/Kb = (5.75)^2/5.65 x 10^-10 = 5.83 x 10^-6 M
Therefore, the concentration of S2- is 5.83 x 10^-6 M, which is closest to choice C (5.0 x 10^-9 M).

An Ammonia Solution Has a PH of 8