Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

c) For 220.0 mL of a buffer solution that is 0.285 M in CH3CH2NH2 and 0.255 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.010 mol of NaOH.

Answer 1

C2H5NH2 has pkb of 3.193

pOH = pkb + log [C2H5NH3Cl]/[C2H5NH2]

pOH = 3.193 + log ( 0.255/0.285)

      = 3.145

pH = 14-3.145 = 10.86

OH- moles added = NaOH moles = 0.01

Initial C2H5NH2 moles = M x V = 0.285 x 220/1000 = 0.0627

Initial C2H5NH3Cl moles= M x V = 0.255 x 220/1000 = 0.0561

OH- reacts with C2H5NH3Cl and we get C2N5NH2

hence after OH- addition C2H5NH3Cl moles = 0.0561-0.01 = 0.0461

C2H5NH2 moles = 0.0627+ 0.01 = 0.0727

pOH = 3.193 + log ( 0.0461/0.22) / ( 0.0727/0.22)

   = 3

pH = 14-3 = 11