Question

In: Chemistry

A sample of 7.70 L of NH3 (ammonia) gas at 22 ∘C and 735 torr is...

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.

Solutions

Expert Solution

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

Kb 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 = pKb + log [ BH+ / B]

Thus,

pOH = pKb + 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


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