Weak acids are titrated with solutions of strong bases of known concentration (standard solutions). — Physical Chemistry — Kinetics Chemistry Question
Titration of weak acids
Weak acids are titrated with solutions of strong bases of known concentration (standard solutions). The solution of weak acid (analyte) is transferred into a 250 cm3 conical flask and the solution of strong base (titrant) is delivered from a burette. The equivalence point of the titration is reached when the amount of added titrant is chemically equivalent to the amount of analyte titrated. The graph, which shows the change of pH as a function of volume of titrant added, is called titration curve. The equivalence point of a titration is the theoretical point, which cannot be determined experimentally. It can only be estimated by observing some physical change associated with the process of the titration. In acid–base titrations, the end point is detected by using acid/base indicators.
Construct the titration curve by calculating a few characteristic points and select indicator for the titration of 50.00 cm3 of acetic acid solution (CH3COOH, Ka = 1.8×105 , c = 0.1000 mol dm3) with a sodium hydroxide solution (c = 0.1000 mol dm3 ). You may consult Table 1.
Table 1: Some common acid/base indicators
Common name | Transition range, pH | Color change
Methyl orange | 3.2 – 4.4 | red–orange
Methyl red | 4.2 – 6.2 | red–yellow
Bromothymol blue | 6.0 – 7.6 | yellow–blue
Phenol red | 6.8 – 8.2 | yellow–red
Phenolphthalein | 8.0 – 9.8 | colorless–red
Thymolphthalein | 9.3 – 10.5 | colorless–blue
Model Answer
The titration reaction is
CH3COOH + OH– → CH3COO– + H2O
a) Initial pH
The pH of the solution before the titration begins, is calculated by the acid dissociation constant and the initial concentration of CH3COOH:
[CH3COO-][H+]/[CH3COOH] = Ka = 1.8×10-5
[H+] = 1.8×10-5 × 0.1000 = 1.34×10-3 and pH = 2.87
b) pH after the addition of 10.00 cm3 of the titrant
The solution contains acetic acid and sodium acetate. Therefore it is a buffer solution.
The concentration of each constituent is calculated:
c(CH3COOH) = ((0.05000 dm3 × 0.1000 mol dm-3) – (0.01000 dm3 × 0.1000 mol dm-3)) / 0.06000 dm3 = 0.0667 mol dm-3
[CH3COOH] = 0.0667 mol dm-3
[CH3COO-] = (0.01000 dm3 × 0.1000 mol dm-3) / 0.06000 dm3 = 0.01667 mol dm-3
These concentrations are then substituted into the dissociation constant expression of acetic acid for calculating the concentration of [H+]:
[H+] = 1.8×10-5 × 0.0667 / 0.01667 = 7.20×10-5
pH = 4.14
c) pH at the equivalence point
At the equivalence point, all acetic acid has been converted to sodium acetate and the pH is calculated from the hydrolysis of acetate ions:
CH3COO– + H2O <-> CH3COOH + OH–
The volume of the titrant required for the equivalence point (Vep) is calculated:
Vep = (0.05000 dm3 × 0.1000 mol dm-3) / 0.1000 mol dm-3 = 0.05000 dm3 = 50.00 cm3.
The total volume of the solution is 100.0 cm3. Therefore, at this point of the titration [CH3COOH] = [OH–] and [CH3COO–] = (0.05000 dm3 × 0.1000 mol dm-3) / 0.1000 dm3 = 0.0500 mol dm-3
[OH-]2 / 0.0500 = Kw / Ka = 1.00×10-14 / 1.8×10-5 = 5.56×10-10
[OH-] = 0.0500 × 5.56×10-10 = 5.27×10-6
pOH = 5.28 and thus pH = 14 – 5.28 = 8.72
d) pH after the addition of 50.10 cm3 of titrant
At this stage, all acetic acid has been converted to sodium acetate and the pH of the solution is calculated by the excess of sodium hydroxide, which has been added:
[OH-] = ((0.0501 dm3 × 0.1000 mol dm-3) – (0.0500 dm3 × 0.1000 mol dm-3)) / 0.1001 dm3 = 1.0×10-4
pOH = 4 and pH = 10
[VISUAL]
e) Selection of indicator
Since the pH at the equivalence point is 8.72, the appropriate acid base indicator is phenolphthalein.
Ascorbic acid (vitamin C, C6H8O6) is a weak acid and undergoes the following dissociation steps:
C6H8O6 <-> C6H7O6– + H+ Ka1 = 6.8×10–5
C6H7O6– <-> C6H6O62– + H+ Ka2 = 2.7×10–12
Hence, ascorbic acid solution can be titrated with sodium hydroxide solution according to the first acid dissociation constant.
50.00 cm3 of ascorbic acid solution with a concentration of 0.1000 mol dm–3 are titrated with a sodium hydroxide solution with a concentration of 0.2000 mol dm–3. In the following calculations you can ignore the second dissociation step.
Based on the calculation choose the correct answers from the given possibilities:
(i) The initial pH of the solution is:
a) 7.00, b) 2.58, c) 4.17, d) 1.00
(ii) The volume of the titrant required for the equivalence point is:
a) 50.00 cm3 , b) 35.00 cm3 , c) 25.00 cm3 , d) 20.00 cm3
(iii) The pH of the solution after the addition of 12.5 cm3 of the titrant is equal to:
a) 4.17, b) 2.58, c) 7.00, d) 4.58
(iv). The pH at the equivalence point is:
a) 7.00, b) 8.50, c) 8.43, d) 8.58
(v) The appropriate indicator for the titration is (refer to Table 1):
a) bromothymol blue, b) phenol red, c) phenolphthalein, d) thymolphthalein
(vi) The pH of the solution after the addition of 26.00 cm3 of the titrant is equal to:
a) 13.30, b) 11.30, c) 11.00, d) 11.42
Model Answer
(i) b, (ii) c, (iii) a, (iv) b, (v) c, (vi) d