For dilute solutions
The pH of the solution at the equivalence point is
pH = 1/2 pKw + 1/2 pKa - 1/2 pC
Where
= the ionic product of water = 1× 10^14
= the dissociation constant of acetic acid = 1.82 ×10^-5
= the concentration of the salt in moldm-3
Accordingly, for the neutralization of 0.1M acetic acid with 0.1M NaOH at the equivalence point,
pH = 7 + 2.37 - 1/2 (1.3) = 8.37
The initial pH of the 0.1M acetic acid can be calculated using its Ka value to be 2.87.
For other concentrations corresponding pH value can be calculated by,
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Procedure
Pipette out 25.00mL of the given acetic acid solution into a titration flask and add few drops of thymolphthalein indicator. Titrate this solution with 0.10M NaOH solution.
- Calculate the concentration of the acetic acid solution.
- Sketch the titration curve.
- Comment on choosing indicators for this titration.
- Derive an equation for the effective or conditional formation constant of water on the employment of a weak acid, HA, with an acid dissociation constant (Ka)
- Calculate the conditional formation constant of water for the above titration.
- Calculate the standard free energy change of this titration (Room temperature = 30 oC
- A sample of vinegar, weighing 12.0g, was titrated with a 0.500M NaOH solution. 17.50 mL being required to obtain a phenolphthalein end point and 5.5 mL being required to obtain a methyl orange end point. Explain the use of correct indicator system in this titration and calculate the percentage of acetic acid in vinegar.
Discussion
Acetic acid is a weak acid and NaOH is a strong base. When acetic acid is titrated with NaOH, with the final drop of NaOH at the end point, the medium turns alkaline. So an alkaline indicator is suitable for this titration. Thymolphthalein is an alkaline indicator with the pH range of 9.3-10.5, the pH range where this reaction has its end point.
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