Question

Construct a curve for the titration of 50.00 mL of a 0.1000 M solution of ethylenediamine with a 0.2000 M solution of HCl in the following table. For each titration, calculate the pH after the addition of 0.00, 12.50, 20.00, 24.00, 25.00, 26.00, 37.50, 45.00, 49.00, 50.00, 51.00, and 60.00 mL of HCl.

Answer 1

The titration of ethylenediamine, which will abbreviate as en, with HCl is an example of a

weak base/strong acid titration curve. Because en is dibasic, there are two equivalence points. The first equivalence point occurs when moles en = moles HCl; thus

M_en V_en = M_HCl V_HCl

V_HCl = 0.05 * 0.1 /0.2 = 25 mL

A second equivalence point occurs after adding 50.0 mL of HCl. The sample’s initial pH is

determined by the concentration of en. Although en is a dibasic weak base, we can calculate

its pH using only the Kb value for the first base dissociation reaction

Kb1 = [OH-][enH+]/[en] = x.x/0.05-x = 8.47×10^-5 = x²/0.05

x= 2.06× 10-3 M = [OH-]

pOH = 2.69 ; pH = 14-pOH=11.31

Adding HCl to a solution of en creates an enH+/en buffer, the pH of which is estimated

using the Henderson-Hasselbalch equation. After adding 12.5 mL of HCl, for example, we

find that

[en] = M_en V_en - M_HCl V_HCl / V_en + V_HCl

= 0.05 * 0.1 - 0.0125*0.2 / 0.05+0.0125

= 0.04 M

[enH+] = M_HCl V_HCl / V_en + V_HCl = 0.0125*0.2 / 0.05+0.0125 = 0.04 M

pOH = pK_b1 + log[enH+]/[en] = 4.07 + log 0.04/0.04 = 4.07

pH = 9.93

Apply same method for addition of 20and 24 mL of HCl.

[enH+] at first equivalence point = 0.025 * 0.2/ 0.05+0.025 = 0.067 M

pH at the first equivalence point (after addition of 25 mL HCl)

= 0.5(pKw + pKb1 + logC)

=0.5(14+4.07+log0.067)

=8.45

Adding additional HCl, up to the second equivalence point volume, converts enH+ to enH2+

producing an enH2+/enH+ buffer. After adding 26.0 mL of HCl, for example, we find that

[enH+] = 0.067 M * 0.075 - 0.2*0.001/0.075 + 0.001 = 0.063 M

[enH2+] = 0.2*0.001/0.075 + 0.001 = 0.00263 M

pOH = pKb2 + log [enH2+]/[enH+] = 7.01 + log 0.00263/0.063 = 7.05

After adding 37.5 mL of HCl ,

[enH+] = 0.067 M * 0.075 - 0.2*0.0125/0.075 + 0.0125 = 0.0285 M

[enH2+] = 0.2*0.0125/0.075 + 0.0125 = 0.0285 M

pOH = pKb2 + log [enH2+]/[enH+] = 7.01

Apply same method for addition of 45.00, 49.00 mL HCl.

pH = 6.99

[enH2+] at second equivalence point (after addition of 50 mL HCl)

= 0.025*0.2/0.025+0.075 = 0.05 M

pOH at second equivalence point,

= 0.5(pKw + pKb2 + logC)

= 0.5(14+7.01+log0.05)

=9.85

pH = 4.14

For volumes greater than the second equivalence point volume, the pH is determined by the

concentration of excess HCl.

After addition of 51 mL HCl

[H+] = 0.001*0.2/ 0.075+ 0.025 +0.001 = 0.00198 M

pH = 2.7

Apply same method for addition of 60.00 mL HCl.