Question

A sample of 7.80 L of NH3 (ammonia) gas at 22 ∘C and 735 torr is bubbled into a 0.500 Lsolution 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 0.500 L , calculate the pH of the resulting solution.

Answer 1

Given,

The volume of NH₃ = 7.80 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 = 0.500 L

Concentration of HCl solution = 0.400 M

K_b value for NH₃ = 1.8 x 10^-5

Calculating the number of moles of NH₃ and number of moles of HCl,

We know, the ideal gas equation,

PV = nRT

Rearranging the formula,

n = PV/RT

Substituting the known values,

n = (0.9671 atm x 7.80 L) /( 0.08206 L.atm/mol. K x 295.15 K)

n = 0.3114 mol of NH₃

Now,

= 0.500 L x 0.400 M

= 0.2 mol of HCl

Now, the reaction between NH₃ and HCl is,

NH₃(aq) + HCl(aq) NH₄⁺(aq)

Drawing an ICE chart.

	NH3(aq)	HCl(aq)	NH4+(aq)
I(moles)	0.3114	0.2	0
C(moles)	-0.2	-0.2	+0.2
E(moles)	0.1114	0	0.2

Now, the new concentrations of NH₃ and NH₄⁺ are,

[NH₃] = 0.1114 mol / 0.5 L = 0.223 M

[NH₄⁺] = 0.2 mol / 0.5 L = 0.4 M

Now, the equilibrium reaction for NH₃ is,

NH₃(aq) + H₂O(l) NH₄⁺(aq) + OH^-(aq)

Drawing an ICE chart,

	NH3(aq)	NH4+(aq)	OH-(aq)
I(M)	0.223	0.4	0
C(M)	-x	+x	+x
E(M)	0.223-x	0.4+x	x

Now, the K_b expression is,

K_b = [NH₄⁺] [OH^-] / [NH₃]

1.8 x 10^-5 = [0.4+x] [x] / [0.223-x]

1.8 x 10^-5 = [0.4] [x] / [0.223] -------Here, (0.4+x] 0.4 and [0.223-x] 0.223 since, x<<0.4 and 0.223

x = 1.00 x 10^-5

Thus, [OH^-] = x = 1.0 x 10^-5 M

We know,

pOH = -log [OH^-]

pOH = -log [1.0 x 10^-5]

pOH = 4.9987

Now,

pH + pOH = 14

pH + 4.9987 = 14

pH = 9.00