Question

A sample of 7.60L of NH3 (ammonia) gas at 22 ?C and 735 torr is bubbled into a 0.300L solution of 0.400 M HCl (hydrochloric acid).

The Kb value for NH3 is 1.8

Answer 1

First of all let us find the moles of NH3 using the equation PV = nRT

P = 735 torr = 0.967 atm , V = 7.60 L , R = 0.0821 L-atm/mol-K , T = 273 + 22 = 295 K

n = PV/RT = 0.967 atm * 7.60 L / (0.0821 L-atm/mol-K * 295 K) = 0.303 mol

Moles of HCl = Molarity * Volume(in L) = 0.400 M * 0.300 L = 0.12 mol

Now,

NH3 +     HCl -------->     NH4Cl

0.303        0.12                   0         (Initial)

- 0.12       - 0.12               + 0.12    (Change)

0.183          0                      0.12     (Final)

So a basic buffer is formed between NH3 and NH4Cl

pH = 14 - pOH

pH = 14 - [ pKb + log( [NH4+] / [NH3] ) ]

pKb = - log(Kb) = - log( 1.8 x 10^-5) = 4.745

pH = 14 - [ 4.745 + log( 0.12 / 0.183) ]

pH = 14 - 4.562

pH = 9.438 = 9.44

So, pH of final solution = 9.44