Question

For 260.0 mL of a buffer solution that is 0.2922 M in CH3CH2NH2 and 0.2677 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Kb=5.6⋅10−4).

Answer 1

For 260.0 mL of a buffer solution that is 0.2922 M in CH3CH2NH2 and 0.2677 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0100 mol of NaOH(Kb=5.6⋅10−4).

this is a buffer so

CH3CH2NH3Cl -->CH3CH2NH3+

CH3CH2NH2

initial pH

Kb = 5.6*10^-4

pKb = -log(kb) = -log(5.6*10^-4 =3.2518

pOH = pKb + log(CH3CH2NH3+/CH3CH2NH2)

pOH = 3.2518+ log(0.2677/0.2922) = 3.213

pH = 14-pOH = 14-3.213= 10.787

after addition of NaOH

mmol of base initially = MV = 260*0.2922= 75.972 mmol of base

mmol of conjugate = MV = 260*0.2677= 69.602 mmol of conjugate

after adding HCl...

mmol = 0.01*10^3 = 10 mmol

conjugate will react , therefore decreases, and form base, which increases

then

mmol of base initially= 75.972 +10 = 85.972

mmol of conjugate = 69.602 -10 = 59.602

then

pOH = pKb + log(CH3CH2NH3+/CH3CH2NH2)

pOH = 3.2518+ log(59.602/85.972) = 3.0927

pH = 14-pOH = 14-3.0927= 10.9073