Question

A sample of 7.70 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.

Part A:

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

1st find the mol of NH3

Given:

P = 735.0 torr

= (735.0/760) atm

= 0.9671 atm

V = 7.7 L

T = 22.0 oC

= (22.0+273) K

= 295 K

find number of moles using:

P * V = n*R*T

0.9671 atm * 7.7 L = n * 0.08206 atm.L/mol.K * 295 K

n = 0.3076 mol

Given:

M(HCl) = 0.4 M

V(HCl) = 0.35 L

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.4 M * 0.35 L = 0.14 mol

We have:

mol(HCl) = 0.14 mol

mol(NH3) = 0.3076 mol

0.14 mol of both will react

excess NH3 remaining = 0.1676 mol

Volume of Solution = 0.35 + 0 = 0.35 L

[NH3] = 0.1676 mol/0.35 L = 0.4789 M

[NH4+] = 0.14 mol/0.35 L = 0.4 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.4/0.4789}

= 4.667

use:

PH = 14 - pOH

= 14 - 4.6666

= 9.3334

Answer: 9.33