Question

A sample of 7.50 L of NH3 (ammonia) gas at 22 ∘C and 735 torr is bubbled into a 0.350 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 0.350 L , calculate the pH of the resulting solution.

Answer 1

Ans :- pH = 9.32

Explanation :-

K_b of NH₃ = 1.8×10⁻⁵

pK_b = - log K_b

pK_b = - log 1.8 x 10^-5

pK_b = 4.74

From Ideal gas equation, we have

PV = nRT

No. of moles of NH₃ = n = PV/RT = (735atm/760)(7.50 L)/(0.0821 L atm K^-1mol^-1)(273+22) = 0.2995 mol

Moles of HCl added = Molarity x Volume = 0.400 M x 0.350 L = 0.140 mol

ICF table is :

.....................NH₃..............+....................HCl -------------------------> NH₄Cl

Initial (I)........0.2995 mol.........................0.140 mol...........................0.0 mol

Change (C)...-0.140 mol.........................-0.140 mol..........................+0.140 mol

Final (F).........0.1595 mol........................0.0 mol................................0.140 mol

Using Henderson-Hasselbalch equation, we have

pOH = pK_b + log [NH₄Cl]/[NH₃]

pOH = 4.74 + log 0.140/0.1595

pOH = 4.74 - 0.0566

pOH = 4.68

pH = 14-pOH

pH = 14-4.68

pH = 9.32