What is the pH of a solution prepared by mixing 10 mL of 0.10M NH3 and 5 mL of 0.24M NH4+? Ka for NH4 = 5.56x10^-10?

After mixing:
Final volume = (10 + 5) mL = 15 mL
[NH₃] = (0.10 M) × (10/15) = 0.067 M
[NH₄⁺] = (0.24 M) × (5/15) = 0.08 M

Consider the dissociation of NH₄⁺ ions :
NH₄⁺(aq) ⇌ NH₃(aq) + H⁺(aq) …. Ka = 5.56 × 10⁻¹⁰

Henderson-Hasselbalch equation:
pH = pKa + log([NH₃]/[NH₄⁺]) = -log(5.56 × 10⁻¹⁰) + log(0.067/0.08) = 9.2