The ionization constants for ethylenediamine are K b1 = 8.47 x 10-5and K b2 = 7.

Question

The ionization constants for ethylenediamine are Kb1 = 8.47 x 10-5and Kb2 = 7.05 x 10-8.

What analytical concentration in M of an HCl solution is required if the second equivalence point is reached when 764.5 mL are titrated with 137.5 mL of a NH2CH2CH2NH2 solution with an analytical concentration of 0.1333 M in order to reach the second equivalence point?

B. The ionization constants for ethylenediamine are Kb1 = 8.47 x 10-5and Kb2 = 7.05 x 10-8.

What is the pH of the solution when the second equivalence point has been reached in the above titration?

Explanation / Answer

A)

we know that

volume at second equivalence point = 2 x volume at first equivalence point

so

764.5 = 2 x volume at first equivalence point

so

volume at first equivalence point = 382.25 ml

now

at first equivalence point

moles of acid = moles of base

also

moles = molarity x volume (L)

so

Ma x Va = Mb x Vb

Ma x 382.25 = 0.1333 x 137.5

Ma = 0.048

so

the concentration of HCl is 0.048 M

B)

now at second equivalence point

moles of HCl reacted = 0.048 x 0.7645

moles of HCl reacted = 0.036696

so moles of salt formed = 0.036696

now

final volume = 764.5 + 137.5 = 902

so

conc of salt = 0.036696 / 0.902

conc of salt = 0.04

now

the salt is a weak acid

so

[H+] = sqrt ( Ka x C)

also

Ka = Kw / Kb2

so

Ka = 10-14 / 7.05 x 10-8

Ka = 1.418 x 10-7

so

[H+] = sqrt ( 1.418 x 10-7 x 0.04)

[H+] = 7.53 x 10-5

now

pH = -log [H+]

pH = -log 7.53 x 10-5

pH = 4.123

now

so

the pH at second equivalence point is 4.123

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