Polyprotic acids Focus Question: What are the major species in solution during t

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Polyprotic acids Focus Question: What are the major species in solution during the titration of citric acid with a strong base? Citric acid is a weak organic acid that occurs naturally in citrus fruits and is mainly used as a flavoring and preservative in food and beverages, especially soft drinks. Citric acid is a triprotic acid. Although its formula C_6H_8O_7 indicates the presence of 8 hydrogens, only the hydrogens attached to the three carboxylic groups (COOH) are acidic. In this part of the lab, you will calculate 8 data points of a titration of 100 mL of 0.50 M citric acid and 0.50 M NaOH and then draw the graph. 1. Acid dissociation: Start by writing the equations for the three steps of dissociation of citric acid. Make sure you give each species the correct charge. 2. Initial pH: Calculate the initial pH of the 0.50 M citric acid solution. The initial pH is determined by K_a1 and the initial concentration of the acid. 3. Midpoints: Determine the pH of the 3 midpoints of the titration. At each midpoint, the pH = pK_a of the species you are titrating. 4. Equivalence points: Calculate the pH of the 3 equivalence points of the titration. At the first and the second equivalence point, the pH is the average of the pK_a values above and below. At the third equivalence point (the end point), the pH is determined by the K_b of the conjugate base of the weakest acid and its concentration. 5. Past the equivalence point: Calculate the pH after 400 mL of NaOH were added. At this point the pH depends only on the concentration of excess NaOH. 6. Titration curve: The titration curve is a pH vs. volume of 0.50 M NaOH graph. Plot the 8 points and connect them to determine the titration curve of citric acid. The curve should be relatively fiat around each of the midpoints (buffering region) and should sharply increase around the equivalence points. For each of the 8 points write the major species (other than H_2O and Na^+) next to the point in the graph.

Explanation / Answer

Dissociation equations for citric acid,

let the triprotic acid be represented as H3A, then,

First dissociation : H3A + H2O <==> H3O+ + H2A-

Second dissociation : H2A- + H2O <==> H3O+ + HA^2-

third dissociation : HA^2- + H2O <==> H3O+ + A^3-

pH calculation

a) initial

H3A + H2O <==> H3O+ + H2A-

let x amount has dissociated

Ka1 = [H3O+][H2A-]/[H3A]

7.5 x 10^-4 = x^2/0.5

x = [H3O+] = 0.02 M

pH = -log[H3O+] = 1.71

Major species : H3A

b) pH at Ist half-equivalence point

[H3A] left = [H2A-] formed

pH = pKa1 = -log[Ka1] = 3.12

major species : [H3A] and [H2A-]

c) pH at Ist equivalence point

= 1/2(pKa1 + pKa2)

= 1/2(3.12 + 4.77)

= 3.945

major species : H2A-

d) pH at IInd half-equivalence point

[H2A-] left = [HA^2-] formed

pH = pKa2 = -log[Ka2] = 4.77

major species : [H2A-] and [HA^2-]

e) pH at IInd equivalence point

= 1/2(pKa2 + pKa3)

= 1/2(4.77 + 6.40)

= 5.585

major species : HA^2-

f) pH at IIIrd half-equivalence point

[HA^2-] left = [A^3-] formed

pH = pKa3 = -log[Ka3] = 6.40

major species : [HA^2-] and [A^3-]

g) pH at IIIrd equivalence point

[A^3-] formed = 0.5 M x 100 ml/400 ml = 0.125 M

A^3- + H2O <==> HA^2- + OH-

let x amount has hydrolyzed

Kb1 = Kw/Ka3 = [HA^2-][OH-]/[A^3-]

1 x 10^-14/4 x 10^-7 = x^2/0.125

x = [OH-] = 5.59 x 10^-5 M

pOH = -log[OH-] = 4.25

pH = 14 - pOH = 9.75

major species : [A^3-]

h) pH after 400 ml of NaOH was added

This is past equivalence point

excess [OH-] = 0.5 m x 100 ml/500 ml = 0.1 M

pOH = -log[OH-] = 1

pH = 14 - pOH = 13

major species : [OH-]

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