Question

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 600. mL of 0.953-M solution of NH3 in order to prepare a pH = 9.65 buffer?

Answer 1

no of moles of NH3 = molarity * volume in L

                                 = 0.953*0.6   = 0.5718moles

POH = 14-PH

          = 14-9.65

         = 4.35

Pkb of NH3 = 4.75

POH   = Pkb + log[NH4Cl]/[NH3]

4.35    = 4.75 + log[NH4Cl]/0.5718

log[NH4Cl]/0.5718    = 4.35-4.75

log[NH4Cl]/0.5718    = -0.4

[NH4Cl]/0.5718         = 10^-0.4

[NH4Cl]/0.5718         = 0.3981

[NH4Cl]                  = 0.3981*0.5718

[NH4Cl]                = 0.2276 moles

no of moles = W/G.M.Wt

mass of NH4Cl    = no of moles * G.M.Wt

                            = 0.2276*53.49   = 12.17g >>>>answer