60.0 mL of a 0.0400 M triethanolamine solution is extracted with 115 mL of solve
ID: 1060169 • Letter: 6
Question
60.0 mL of a 0.0400 M triethanolamine solution is extracted with 115 mL of solvent. The partition coefficient for the reaction is 4.00 and the pKa of the protonated form of triethanolamine is 7.762. Calculate the concentration of triethanolamine remaining in the aqueous phase at pH = 6.50 and pH = 8.50.
60.0 mL of a 0.0400 M triethanolamine solution is extracted with 115 mL of solvent. The partition coefficient for the reaction is 4.00 and the pKa of the protonated form of triethanolamine is 7.762. Calculate the concentration of triethanolamine remaining in the aqueous phase at pH 6.50 and pH 8.50. Number M Triethanolamine at pH 6.50 Number Triethanolamine at pH 8.50 MExplanation / Answer
Let B denote the amine. The partition of B in aqueous and organic phases can be denoted as
B (aq) <=====> B (organic)
K = [B]org/[B]aq
Let D be the distribution co-efficient so that D is defined as
D = total concentration of B in organic phase/total concentration of B in aqueous phase = [B]org/[B]aq + [B-H+]aq ……(1)
Now, since B is a weak base, we must have
B (aq) + H+ -----> B-H+ (aq)
The reverse reaction is
BH+ (aq) ------> B (aq) + H+ (aq)
Ka = [B]aq[H+]aq/[BH+]aq
=====> [BH+]aq = [B]aq[H+]aq/Ka …..(2)
Put (2) in (1) and obtain
D = [B]org/[B]aq + [B]aq[H+]aq/Ka = [B]org/[B]aq(1 + [H+]aq/Ka) = [B]org/[B]aq*(1/1 + [H+]aq/Ka)
= K*(1/1 + [H+]aq/Ka) = K.Ka/(Ka + [H+]) …… (3)
Again, let we have m moles of the base out of which fraction q remains in the aqueous phase at equilibrium. Therefore,
K = (1 – q)m/Vorg/qm/Vaq = (1 – q)m.Vaq/q.m.Vorg
====> K.q.m.Vorg = m.Vaq – q.m.Vaq
====> q.m (K.Vorg + Vaq) = m.Vaq
====> q = Vaq/Vaq + KVorg …..(4)
At pH = 6.50, [H+] = 10-6.50 = 3.16*10-7.
Given pKa = 7.762, Ka = 10-7.762 = 1.7298*10-8
Given K = 4.00, find D at pH 6.50 as D = (4.00)*(1.7298*10-8)/(1.7298*10-8 + 3.16*10-7) = 0.207.
Substitute K by D in expression 3 and obtain
q = (60.0 mL)/(60.0 mL) + (0.207)*(115.0 mL) = 0.7159 0.72 = 72% left in water
Therefore, concentration of triethanolamine left in water is [Triethanolamine] at pH 6.50 = 0.72*0.0400 M = 0.0288 M (ans).
At pH = 8.50, [H+] = 10-8.50 = 3.16*10-9 M
D = (4.00)*(1.7298*10-8)/(1.7298*10-8 + 3.16*10-9) = 3.382
q = (60.0 mL)/(60.0 mL + 3.382*115.0 mL) = 0.1336 0.13 = 13% left in water.
Therefore, [Triethanolamine] at pH 8.50 = 0.13*0.0400 M = 0.0052 M (ans).
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