Question

A sample of 7.60 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.

Part A
Assuming all the NH3 dissolves and that the volume of the solution remains at 0.450 L , calculate the pH of the resulting solution. Express your answer numerically to two decimal places.

Answer 1

First, get total amount og moles of NH3 bubbled

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Tmperature

R = ideal gas constant

n = PV/(RT)

P gas must be corrected

Pgas = Ptotal - Pvapr = 735-19.8 = 715.2 torr

T = 22°C = 22+273 K = 295 K

n = (715.2 )(0.45)/(62.4*295 )

n = 0.0174 moles

now,

[NH3] = mol of NH3 / V slution = 0.0174 / 0.45 = 0.0386666 M

now, get pH:

NH3 + H2O = NH4+ + OH-

Kb = [NH4+][OH-]/[NH3]

initially

[NH4+] = [OH-] = 0

[NH3] = 0.0386666

in equilibrium:

[NH4+] = [OH-] = 0 + x

[NH3] = 0.0386666 - x

substitute in Kb

1.8*10^-5 = x*x/(0.0386666 -x)

x^2 + 1.8*10^-5 - 0.0386666 * (1.8*10^-5)

x = [OH-] = 0.000825

pOH = -log(0.000825) = 3.083

pH = 14-pOH = 14-3.083 = 10.917

pH = 10.92