Question

For 260.0 mL of a buffer solution that is 0.3187 M in CH3CH2NH2 and 0.2887 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Kb=5.6⋅10−4).

Answer 1

c)
pKb of ethylamine = 3.37 ( check this value on your book)
pOH = 3.37 + log 0.2887/ 0.3187= 3.32
pH = 14 - pOH =10.68 ( initial pH)
moles ethylamine = 0.260 L x 0.318 M=0.08268
moles ethylammonium = 0.260 L x 0.2887 M=0.0750
CH3CH2NH3++ OH- = CH3CH2NH2 + H2O
moles CH3CH2NH3+ = 0.0750 - 0.0200 =0.055
moles CH3CH2NH2 = 0.0826 + 0.0200 =0.1026
concentration CH3CH2NH3+ = 0.0750/ 0.260=0.288 M
concentration CH3CH2NH2 = 0.0826/0.260 L=0.317 M
pOH = 3.37 + log 0.288 / 0.317 =3.32
pH = 14 - 3.32=10.68