Q88 · Practice questionComplex unfamiliar7 marks
QUESTION 88 (7 marks)
A pure monoprotic carboxylic acid contains 40.0% C, 6.67% H and 53.3% O by mass. A 0.360 g sample is made up to 250.0 mL; a 25.00 mL aliquot requires 12.00 mL of 0.05000 mol L⁻¹ NaOH. The undiluted 250.0 mL acid solution has pH 3.19. Use C = 12.0, H = 1.0, O = 16.0.
a) Determine the empirical formula. [2 marks]
b) Use the titration to determine the molar mass and identify the acid. [3 marks]
c) Calculate its Ka from the concentration and measured pH. [2 marks]
WORKED SOLUTION
7 marksPractice marking scheme
Empirical formula CH₂O; molar mass 60.0 g mol⁻¹; ethanoic acid; .
Mole ratio is . Aliquot acid moles=0.0006000, total=0.006000, so . Molecular formula is C₂H₄O₂, and the monoprotic carboxylic acid is ethanoic acid. ; ; .
- Converts mass composition to mole ratio. [1 marks]
- Determines CH₂O. [1 marks]
- Calculates total acid moles using aliquot factor. [1 marks]
- Calculates molar mass 60.0 g mol⁻¹. [1 marks]
- Identifies C₂H₄O₂/ethanoic acid. [1 marks]
- Calculates concentration and [H⁺]. [1 marks]
- Calculates Ka using equilibrium acid concentration. [1 marks]
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