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Please show detailed work. Thanks. A 10.00 mL aliquot of a F neutral ethylenedia

ID: 592381 • Letter: P

Question

Please show detailed work. Thanks.

A 10.00 mL aliquot of a 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. The titrant is made up to the same concentration as the ethylenediamine solution. What is/are the equivalence point volume(s)? What is the initial pH of the ethylenediamine solution (before adding any titrant)? What is the pH after adding 2.50 mL of the titrant (titrant concentration is the same as your ethylenediamine concentration)? What is the pH after adding 5.00 mL of the titrant? What is the pH after adding 10.00 mL of the titrant? What is the pH after adding 30.00 mL of the titrant? 4. 5. 6. 7. 8.

Explanation / Answer

Titration of 10 ml of ethylenediamine

1. HCl will be used as the titrant.

2. two equivalent points are there in the titration.

3. If the concentration of both solutions are same, lets say 0.1 M, then equivalence point volume would be,

first equivalence point HCl = 10 ml

second equivalence point HCl = 20 ml.

4. initial pH (assuming 0.1 M ethylenediamine solution)

base + H2O <==> baseH+ + OH-

Kb1 = 8.5 x 10^-5 = x^2/0.1

x = [OH-] = 2.91 x 10^-3 M

pOH = -log[OH-] = -log(2.91 x 10^-3) = 2.53

pH = 14 - pOH = 11.47

5. after 2.5 ml HCl added

[base] remains = 0.1 M x 10 ml - 0.1 M x 2.5 ml = 0.75 mmol

[baseH+] formed = 0.1 M x 2.5 ml = 0.25 mmol

Using Hendersen-Hasselbalck equation,

pH = pKa2 + log(base/acid)

     = 9.93 + log(0.75/0.25) = 10.41

6. pH after 5 ml HCl added

moles [base] remains = moles [baseH+] formed

pH = pKa2 = 9.93

7. pH after adding 10 ml HCl

Ist equivalence point

pH = 1/2(pKa1 + pKa2) = 1/2(6.43 + 9.93) = 8.18

8. pH after 30 ml HCl added

this is past IInd equivalence point

excess [H+] = 0.1 M x 10 ml/40 ml = 0.025 M

pH = -log[H+] = 1.60

4.

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