Question

Part B How many grams of dry NH4Cl need to be added to 2.40 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.66? Kb for ammonia is 1.8×10−5. Express your answer with the appropriate units.

mass of NH4Cl = ?

Answer 1

Pkb = -logKb

        = -log1.8*10^-5

         = 4.75

PH   = 8.66

POH = 14-PH

           = 14-8.66   = 5.34

no of moles of NH3 = molarity * volume in L

                                  = 0.8*2.4 = 1.92 moles of NH3

POH    = Pkb + log[NH4Cl]/[NH3]

5.34   = 4.75 + log[NH4Cl]/1.92

log[NH4Cl]/1.92    = 5.34-4.75

log[NH4Cl]/1.92   = 0.59

[NH4Cl]/1.92     = 10^0.59

[NH4Cl]/1.92     = 3.8904

[NH4Cl]            = 3.8904*1.92   = 7.4695 moles

mass of NH4Cl   = no of moles * gram molar mass

                           = 7.4695*53.5   = 399.62g >>>> answer