Consider a 26.0 mL aliquot of a 0.1180 M aqueous solution of the weak base (B),

Question

Consider a 26.0 mL aliquot of a 0.1180 M aqueous solution of the weak base (B),

ethyl 3-aminobenzoate, characterized by a base-dissociation constant, Kb = 4.96E8.

This solution is titrated with 0.0772 M HCl. Calculate or determine the following

(work must be shown for credit; to simplify grading, please use the small x

approximation it is valid here): 30 points:

(a) The initial pH (i.e. the pH of the analyte solution): .

(b) The volume (in mL) of titrant needed to reach the endpoint: .

(c) The pH at the equivalence point: .

(d) The pH after the addition of 23.5 mL of titrant: .

(e) The pH after the addition of 42.0 mL of titrant: .

(f) The pH after the addition of a large excess of titrant:

Explanation / Answer

(a) The initial pH (i.e. the pH of the analyte solution): .

here the concentration of base is 0.1180 M

the base will dissociate on its own:
....B + H2O <----> BH+ + OH-
I. . 0.1180 M. . . . . . . 0. . . . . 0
E. 0.1180 M -x. . . . . .x. . . . . .x

kb = [OH-][BH+]/[B]

here, Kb = 4.96*10^-8

4.96*10^-8= x^2 / 0.1180 -x

0.1180 –x is approximately = 0.1180
x = [OH-] = 7.65E-5 M

[H+] = kw/[OH-] = 1E-14/7.65E-5 = 1.30E-10
pH = 9.88

(b) The volume (in mL) of titrant needed to reach the endpoint:

26.0 mL aliquot of a 0.1180 M aqueous solution of the weak base means number of moles = 0.1180 mol / 1000 ml *26.0 ml

=3.068 *10^-3 mol of base

to neutrilize base need same number of acid at equivalent point then volume of acid will be:

1000 ml HCl = 0.0772 mol HCl

1000 ml / 0.0772 mol * 3.068 *10^-3 mol = 39.74 ml of HCl

(c) The pH at the equivalence point:

At the equivalence point, we added the same amount of HCl and Base Therefore the moles of HCl will completely react with the moles of base to completely produce the conjugate acid form.

At the equivalence point, the moles of base = 3.068 *10^-3, so HCl will have the same. Therefore you will produce 3.068 *10^-3 base chloride. That will dissociate because it is a conjugate acid of a weak base.

Since we're no longer taking ratios, find the new concentration by dividing by the total volume:

3.068 *10^-3 BHCl / total volume; = 0.047 M BHCl

....BHCl ----> B + H+
I...0.0136. . . 0. . .0
E.0.047-x. .x. . .x

ka = [H+][B] / [BHCl]

The dissociation produces H+ ions, so you need the Ka rather than the Kb.

ka = kw/kb = 1E-14/4.96E-8= 2.02E-7

2.02E-7 = x^2 / 0.047-x

x = [H+] = 9.742E-5

pH = 4.01

(d) The pH after the addition of 23.5 mL of titrant:

Adding HCl completely reacts with the base to form its conjugate acid pair:
B + HCl ----> BHCl

0.1180 M * 26.0mL = 3.068 mmol base
0.0772 M * 23.5 mL = 1.81 mmol HCl

3.068- 1.81 = 1.258 mmol base leftover
1.81 mmol base chloride produced

Now use the Henderson-Hasselbalch equation:

pOH = pKb + log[base/ conjugated cation]

Kb = 4.96E8. then Kb = 7.30

pOH = 7.30+ log(1.258/1.81)

pOH = 7.30+ log(1.258/1.81)

pOH =7.30-0.158

= 7.142
pH = 14-7.142 = 6.858

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