Question

What is the equivalence point pH for the titration of 25ml 0.80M diethyleneamine (kb = 5.9x10-4) titrated by 0.60M strong acid? A. 3.74 B. 5.62 C.6.55 D. 8.62

Answer 1

If the base is represented by B then the initial reaction before the equivalence point :

B + H2O ==> BH^+ + OH-

initial moles of base present = 0.80 M *0.025 L = 0.02 moles

	B	BH^+	OH-
inital	0.02	0	0
change	-x	+x	+x
equilibrium	0.02-x	+x	+x

Kb = x^2/(0.02-x)

or, 5.9x10-4 = x^2/(0.02-x)

or, x =0.0032 moles

So, moles of acid requied to neutralize this weak base = 0.0032 moles

volume of acid required = 1000mL *0.0032/0.6 = 5.3 mL

At the equivalence point all the base is converted to BH^+. Now the equilibrium will be as follows:

BH^+ + H2O ==> B + H3O ^+

Ka of BH^+ is : Ka = Kw/Kb = 1*10^-14/5.9*10^-4 = 0.17*10^-10

total volume at this point = 25 +5.3 = 30.3 mL

	BH^+	B	H3O^+
inital	0.0032	0	0
change	-x	+x	+x
equilibrium	0.0032-x	+x	+x

Ka = x^2/(0.0032-x)

or, 0.17*10^-10 =x^2/(0.0032-x)

or, 5.44*10^-14 - 0.17*10^-10x - x^2 = 0

or, x = 2.33 *10^-7

moles of H3O^+ = 2.33*10^-7 moles

Molarity = 2.33*10^-7moles*1000mL/30.3 mL = 0.077 *10^-4 M

pH = -log[H3O^+] = -log[0.077 *10^-4] = 5.11

pH of the equivalence point is 5.11

Answer =B