Question

REPORT FORM ; ACID TITRATION CURVE.

Name of acid (or acid salt) NH4Cl

mass of solid acid = 0.160g NH4Cl

volume of water used to dissolve acid =40ml

NaOH(titrant)=30ml

Data analysis:

PH at equivalent point=11.25, pH at half equivalent point =9.49

Ka of =3.2*10^-10

Questions: calculate pH at beginning of titration, pH at equivqlent point, pH after 20.0ml of titrant were added.

Thank you.kindly assist

Answer 1

no of mol of NH4Cl(acid) = 0.16/53.5 = 0.003 mol

concentration of NH4Cl = n/v in L

        = 0.003/0.04

       = 0.075 M

initial pH = 7-1/2(pkB+logC)

   C = concentration of NH4Cl = 0.075 M

pka = -log(3.2*10^-10) = 9.49

PKB = 14-9.49 =

pH = 7-1/2(4.51+log0.075)

     = 5.3

pH at equivalence point ,

no of mol of NH4Cl(acid) = 0.16/53.5 = 0.003 mol

no of mol of NaOH = 0.003 mol

concentration of NaOH solution = n/v = 0.003/0.03 = 0.1

At equivalence point, total NH4Cl converts in to NH3,H2O,NaCl

concentration of NH3 produced = 0.003/(40+30)*1000 = 0.0428 M

pH of NH3 = 14 - 1/2(pkb-logC)

          = 14-1/2(4.51-log0.0428)

      pH = 11.06

after addition of 20 ml titrant

no of mol of NH4Cl(acid) = 0.16/53.5 = 0.003 mol

no of mol of NaOH = 20*0.1/1000 = 0.002 mol

pH = pka + log(base/acid)

   = 9.49+log(0.002/(0.003-0.002))

   = 9.79