Question

A sample of 7.5 L of NH3 gas at 22 ∘C and 735 torr is bubbled into a 0.50-L solution of 0.40 M HCl. Assuming that all the NH3 dissolves and that the volume of the solution remains 0.50 L, calculate the pH of the resulting solution.

Answer 1

Given,

The volume of NH₃ gas = 7.5 L

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

Pressure of NH₃ gas = 735 Torr x ( 1 atm /760 Torr) = 0.9671 atm

Volume of HCl solution = 0.50 L

Concentration of HCl solution = 0.40 M

Calculating the number of moles of NH₃ gas and HCl,

We know,

K_b for NH₃ = 1.8 x 10^-5

We know, the ideal gas equation,

PV = nRT

Rearranging the formula,

n = PV /RT

Substituting the known values and calculating the moles of NH₃ gas,

n = (0.9671 atm x 7.5 L) /(0.08206 L.atm/mol.K x 295.15 K)

n = 0.2995 mol of NH₃ gas

Now, calculating the number of moles of HCl solution,

= 0.50 L x 0.40 M

= 0.2 mol of HCl

Now, the reaction between NH₃ and HCl is,

NH₃(aq) + H⁺(aq) NH₄⁺(aq)

Drawing an ICE chart,

	NH3(aq)	H+(aq)	NH4+(aq)
I(moles)	0.2995	0.2	0
C(moles)	-0.2	-0.2	+0.2
E(moles)	0.0995	0	0.2

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

[NH₃] = 0.0995 mol / 0.5 L = 0.1989 M

[NH₄⁺] = 0.2 mol/0.5 L = 0.4 M

Now, We know the Henderson-Hasselbalch equation,

pOH = pK_b + log [C.acid /base]

pOH = -logK_b + log [NH₄⁺ /NH₃]

pOH = -log(1.8 x10^-5) + log [0.4 /0.1989]

pOH = 4.74 + 0.303

pOH = 5.048

Now, We know,

pH + pOH = 14

Rearranging the formula,

pH = 14 - pOH

pH = 14 - 5.048

pH = 8.95 Or 8.9 [ 2S.F]