Question

A 10.00 mL aliquot of a 0.02000 F neutral ethylenediamine (H2NCH2CH2NH2) solution is to be titrated.

1. What titrant should be used: NaOH or HCl?

2. How many equivalence points are there in this titration?

3. If the titrant is 0.0500 F, what is/are the equivalence point volume(s)?

4. What is the initial pH of the ethylenediamine solution (before adding any titrant)?

5. What is the pH after adding 2.00 mL of the titrant?

6. What is the pH after adding 4.00 mL of the titrant?

7. What fraction of the ethylenediamine is in the “intermediate” form at this point?

8. What is the pH after adding 6.00 mL of the titrant?

9. What is the pH after adding 8.00 mL of the titrant?

10. What is the pH after adding 15.00 mL of the titrant?

Ka value = 1.18 * 10^-10 pka value = 9.92

Please show calculations

thank you

Answer 1

1)

HCl will be used as the titrant

2)

It is diprotic weak base-strong acid titration.Since ethylene diamine is diprotic , two equivalence points are seen in in the curve.

3)

Given data,

V_en = 10 mL , M_en = 0.02 F

M_HCl = 0.05 F

First equivalence point , V_HCl = ( M_en x V_en) / M_HCl

= (0.02 F x 10 mL) / 0.05 F

= 4 mL

The second equivalence point is reached after adding an additional 4 mL of HCl. for a total volume of 10 mL

4)

H2NCH2CH2NH2 -----> OH- + H3NCH2CH2NH2

0.02 - x x x

Consider,

K_b = [OH-] [H_en] / [en]

8.47 x 10^-5 = x * x / 0.02- x

On solving we get ,

[OH-] = x = 1.3 x 10^-4

p^OH = -log [OH-]

= - log [1.3 x 10^-4]

= 3.88

We know , pH + p^OH = 14

pH = 14 - p^OH

= 14 - 3.88

pH = 10.1