Question

For 300.0 mL of a buffer solution that is 0.2604 M in CH3CH2NH2 and 0.2404 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

Using Hendersen-Hasselbalck equation,

pH = pKa + log([CH3CH2NH2]/[CH3CH2NH3Cl])

Initial pH

pKb = -log[Kb] = -log(5.6 x 10^-4) = 3.25

pKa = 14 - pKb = 10.75

pH = 10.75 + log(0.2604/0.2404)

      = 10.78

After addition of NaOH

moles of NaOH added = 0.02 mol

New [CH3CH2NH2] = (0.2604 M x 0.3 L + 0.02 mol)/0.3 L = 0.327 M

New [CH3CH2NH3Cl] = (0.2404 M x 0.3 L - 0.02 mol)/0.3 L = 0.174 M

final pH would be thus,

pH = 10.75 + log(0.327/0.174)

      = 11.01