Question

A sample of 7.70 L of NH3 (ammonia) gas at 22 ∘C and 735 torr is bubbled into a 0.400 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 0.400 L , calculate the pH of the resulting solution. Express your answer numerically to two decimal places.

Answer 1

Given,

The volume of NH₃(g) = 7.70 L

Temperature(T) = 22 ^oC + 273.15 = 295.15 K

Pressure(P) = 735 Torr x ( 1 atm / 760 Torr) = 0.9671 atm

Volume of HCl solution = 0.400 L

Concentration of HCl solution = 0.400 M

K_b value for NH₃ = 1.8 x 10^-5

Firstly, calculating the number of moles of NH₃ gas that is bubbled into the HCl solution,

We know, the ideal gas equation,

PV = nRT

Rearranging the formula,

n = PV /RT

n = ( 0.9671 atm x 7.70 L) / (0.08206 L.atm/mol.K x  295.15 K)

n = 0.3074 mol of NH₃ gas

Now, Calculating the number of moles of HCl solution,

= 0.400 M x 0.400 L

= 0.16 mol of HCl

Now, The reaction between NH₃ and HCl is,

NH₃(g) + HCl(aq) NH₄⁺(aq)

Drawing an ICE chart,

	NH3(g)	HCl(aq)	NH4+(aq)
I(moles)	0.3074	0.16	0
C(moles)	-0.16	-0.16	+0.16
E(moles)	0.1474	0	0.16

Now, The new concentrations of [NH₃] and [NH₄⁺]

[NH₃] = 0.1474 mol / 0.400 L = 0.3686 M

[NH₄⁺] = 0.16 mol / 0.400 L = 0.4 M

Now, We know, the Henderson-Hesselbalch equation,

pOH = pK_b + log [ BH⁺ / B]

Thus,

pOH = pK_b + log [[NH₄⁺] / [NH₃]]

pOH = -log [ 1.8 x 10^-5] +  log [[0.4 / [0.3686]

pOH = 4.7447 + 0.0354

pOH = 4.78

Now, We know,

pH + pOH = 14

pH + 4.78 = 14

pH = 9.22