Question

How many grams of dry NH4Cl need to be added to 2.40 L of a 0.300 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.65? Kb for ammonia is 1.8×10−5.

Answer 1

no of moles of NH3 = molarity * volume in L

                                  = 0.3*2.4   = 0.72 moles

PH = 8.65

POH = 14-PH

           = 14-8.65    = 5.35

Kb    = 1.8*10^-5

PKb   = -logKb

          = -log1.8*10^-5

           = 4.75

POH   = Pkb + log[NH4Cl]/[NH3]

5.35   = 4.75 + log[NH4Cl]/0.72

5.35-4.75   = log[NH4Cl]/0.72

log[NH4Cl]/0.72   = 0.6

[NH4Cl]/0.72         = 10^0.6

[NH4Cl]/0.72         = 3.981

[NH4Cl]                 = 3.981*0.72   = 2.866

no of moles of NH4Cl   = 2.866

mass of NH4Cl            = no of moles * gram molar mass

                                    = 2.866*53.5   = 153.33g of NH4Cl >>>>>answer