Question

Calculate the ph of a solution that is 0.30 M in ammonia (NH3) and 0.20 M in ammonium chloride. Kb=1.76 x 10^-5

Then calculate the ph if you add 10 mL of 1.0 M HCl to 50 mL of your solution in the problem above.

Answer 1

kb   = 1.76*10^-5

Pkb = -logKb

       = log1.76*10^-5  

        = 4.754

POH   = Pkb + log[NH4Cl]/[NH3]

         = 4.754 + log0.2/0.3

         = 4.754-0.1760    = 4.578

PH   = 14-POH

         = 14-4.578   = 9.422 <<<<answer

no of moles of NH4Cl = molarity * volume in L

                                    = 0.2*0.05 = 0.01 moles

no of moles of NH3   = molarity * volume in L

                                  = 0.3*0.05   = 0.015 moles

no of moles of HCl = molarity *volume in L

                               = 1*0.01 = 0.01 moles

no of moles of NH4Cl after the addition of 0.01 moles of HCl = 0.01+0.01 =0.02 moles

no of moles of NH3 after the addition of 0.01 moles of HCl = 0.015-0.01   = 0.005 moles

POH = Pkb + log[NH4Cl]/[NH3]

         = 4.754 + log0.02/0.005

           = 4.754+0.602 = 5.356

PH   = 14-POH

        = 14-5.356   = 8.644 >>>answer