Question

A sample of 7.80 L of NH3 (ammonia) gas at 22 ∘C and 735 torr is bubbled into a 0.250 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 0.250 L , calculate the pH of the resulting solution.

Express your answer numerically to two decimal places.

Answer 1

T = 22⁰C   = 22 + 273 = 295K

P   = 735torr = 735/760   = 0.967atm

V   = 7.8L

PV = nRT

n    = PV/RT

        = 0.967*7.8/0.0821*295   = 0.311 moles

no of moles of HCl = molarity * volume in L

                                = 0.4*0.25 = 0.1 moles

   NH3    +   HCl -------------> NH4Cl

I      0.311       0.1                       0

C    -0.1         -0.1                      0.1

E     0.211        0                        0.1

    Pkb = -logKb = -log1.8*10^-5    = 4.75

no of moles of NH3 = 0.211moles

no of moles of NH4Cl = 0.1 moles

     POH = PKb + log[NH4Cl]/[NH3]

               = 4.75 + log0.1/0.211

               = 4.75 -0.3242   = 4.4258

PH      = 14-POH

          = 14-4.4258   = 9.6 >>>>>answer