To solve this problem, we need to use the concept of weak acid-base equilibrium and the Henderson-Hasselbalch equation. The reaction between NH3 and HNO3 can be written as:
NH3 + HNO3 → NH4+ + NO3-
Before any HNO3 is added, the NH3 solution is a weak base with the following equilibrium equation:
NH3 + H2O ⇌ NH4+ + OH-
where the Kb of NH3 is 1.8 × 10-5. At the start, we have 0.10 M NH3, and the concentration of OH- can be calculated using the Kb expression:
Kb = [NH4+][OH-]/[NH3]
[OH-] = Kb[NH3]/[NH4+] = 1.8 × 10^-5 × 0.10 / x, where x is the concentration of NH4+.
At the equivalence point, the moles of HNO3 added equals the moles of NH3 initially present, and the solution contains only NH4+ and NO3-. Therefore, the concentration of NH4+ is:
x = 0.10 - 0.05 = 0.05 M
At this point, all the OH- ions have been consumed by the HNO3, so the pH of the solution depends on the concentration of NH4+. The Henderson-Hasselbalch equation for a weak acid-base system is:
pH = pKa + log([base]/[acid])
where pKa is the negative logarithm of the acid dissociation constant (Ka) and [base]/[acid] is the ratio of the concentrations of the weak base and its conjugate acid.
For NH3, the conjugate acid is NH4+ and the pKa can be calculated as:
pKa = -log(Ka) = -log(1.8 × 10^-5) = 4.74
Plugging in the values, we get:
pH = 4.74 + log(0.05/0.05) = 4.74
Therefore, the pH of the solution after the addition of 50.0 mL of HNO3 is 4.74.