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The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined - HSC - SSCE Chemistry - Question 32 - 2022 - Paper 1
Question 32
The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined. Step 1: A solution of NaOH(aq) was standardised by titrating it... show full transcript
Worked Solution & Example Answer:The concentration of citric acid, a triprotic acid, in a carbonated soft drink was to be determined - HSC - SSCE Chemistry - Question 32 - 2022 - Paper 1
Step 1
Titration 1
Answer
To determine the moles of KHP:
Amount of KHP = 4.989 g / 204.22 g mol⁻¹ = 0.02442593677 moles.
The concentration of KHP in mol/L:
Concentration of KHP = 0.02442593677 mol / 0.1000 L = 0.2442593677 mol L⁻¹.
For the NaOH:
Volume of NaOH = 27.40 mL = 0.02740 L.
Calculating moles of NaOH:
Moles NaOH = Concentration KHP × Volume KHP = 0.02442593677 mol L⁻¹ × 0.02500 L = 0.00061076349 moles.
Concentration of NaOH in the titration:
Concentration NaOH = 0.00061076349 moles / 0.02740 L = 0.2228972333 mol L⁻¹.
Step 2
Titration 2
Answer
Volume NaOH = 13.10 mL = 0.01310 L.
From Titration 1:
Concentration NaOH = 0.2228972333 mol L⁻¹.
Moles NaOH = Concentration NaOH × Volume NaOH = 0.2228972333 mol L⁻¹ × 0.01310 L = 0.00291995375 moles.
Volume of citric acid used = 25.00 mL = 0.02500 L.
The ratio of citric acid to NaOH is 1:3 (for triprotic acid).
Amount of citric acid = moles NaOH / 3 = 0.00097331791 moles (in 25 mL).
Amount of citric acid in 250 mL = 0.00097331791 moles × (250.00 / 25.00) = 0.0097331791 moles.
Concentration of citric acid:
Concentration c = 0.0097331791 mol / 0.07500 L = 0.129775213 mol L⁻¹ = 0.1298 mol L⁻¹.