Question

25.0 mL of a 0.0450 M butylamine solution is extracted with 105 mL of solvent. The partition coefficient for the reaction is 2.00 and the pKa of the protonated form of butylamine is 10.640. Calculate the concentration of butylamine remaining in the aqueous phase at pH = 9.80 and pH = 12.1.

Answer 1

At pH = 9.80

pKa = pH + log(salt/base)

10.64 = 9.80 + log(salt/base)

salt/base = 6.92

Total butylamine = 1 + 6.92 = 7.92

percent of compound in base form remained = 1 x 100/7.92 = 12.62%

concentration of butylamine remaining in aqueous phase = 0.045 x 25 x 73.14 x 0.1262 = 0.010 g

2 = (x/105)/[(0.010 - x)/25]

x = 0.01 g

amount remain in aqueous phase = 0.0104 - 0.01 = 0.0004 g

concentration of butylamine in aqueous phase = 0.004/73.14 x 0.025 = 0.0001 M

At pH = 12.1

pKa = pH + log(salt/base)

10.64 = 12.1 + log(salt/base)

salt/base = 0.034

Total butylamine = 1 + 0.034 = 1.034

percent of compound in base form remained = 1 x 100/1.034 = 96.7%

concentration of butylamine remaining in aqueous phase = 0.045 x25 x 73.14 x 96.7 = 0.0795 g

2 = (x/105)/[(0.0795 - x)/25]

x = 0.071 g

amount remain in aqueous phase = 0.0795 - 0.071 = 0.0085 g

concentration of butylamine in aqueous phase = 0.0085/73.14 x 0.025 = 0.0029 M