Question

A sample of 7.50 L of NH3 (ammonia) gas at 22 ∘C and 735 torr is bubbled into a 0.450 L solution of 0.400 M HCl (hydrochloric acid). The Kb value for NH3 is 1.8×10−5.

Assuming all the NH3 dissolves and that the volume of the solution remains at 0.450 L , calculate the pH of the resulting solution.

Answer 1

Moles PV = nRT
(735 Torr)(7.50 Litres) = n (62.4 Torr-Litres/mol-K)(295 K)
0.29946 moles of NH3
find moles
0.450L solution of 0.400 M HCL = 0.180 moles of HCl
0.29946 moles of NH3 reacts with 0.180 moles of HCl producing 0.160 moles of NH4+ ion,
and leaving 0.1435 moles of NH3 behind
find molarities
0.160 moles of NH4+ / 0.400 L = 0.400 M NH4+
0.1435 moles of NH3 / 0.400 L = 0.359 M NH3
NH4OH --> NH4+ & OH-
Kb = [ NH4+] [OH-] / [NH4OH]
1.8 e-5 = [ 0.400] [OH-] / [ 0.359]
[OH-] = 1.61 e-5
pOH = 4.79
pH = 9.21

your answer is
pH = 9.21