Help With Titration Calculation?

[匿名]

2020-08-26 12:26 pm

A student measured the mass of an empty micropicnometer to be 2.000g. A student places 0.500mL of citric acid solution into a micropicnometer. The student reweighed the micropicnometer and found the mass of the citric acid solution and micropicnometer to be 2.525g.

A student titrates 0.101mL of citric acid solution using the standardized 0.1M solution (use the exact concentration calculated from the previous titration problem). The sample required 0.819mL of NaOH solution to reach the end point.

Given the following chemical equation:
NaOH + H3C6H5O7 → 3Na+ + 3H2O + C6H5O73-

Calculate the mass of citric acid in the 0.101mL citric acid solution.
Calculate the total mass of the 0.101mL citric acid solution.

回答 (1)

戇戇居士

2020-08-26 3:40 pm

Refer to: https://en.wikipedia.org/wiki/Citric_acid
Molar mass of H₃C₆H₅O₇ = 192.123 g/mol = 192.123 mg/mmol

3NaOH + H₃C₆H₅O₇ → 3Na⁺ + 3H₂O + C₆H₅O₇⁻
Mole ratio NaOH : H₃C₆H₅O₇ = 3 : 1

(0.1011 mmol NaOH / 1 mL NaOH solution) × (0.819 mL NaOH) × (1 mmol H₃C₆H₅O₇ / 3 mmol NaOH) × (192.123 mg H₃C₆H₅O₇ / 1 mmol H₃C₆H₅O₇)
= 5.30 mg H₃C₆H₅O₇
= 0.00530 g H₃C₆H₅O₇

Mass of 0.500 mL of H₃C₆H₅O₇ solution = (2.525 - 2.000) g = 0.525 g
Total mass of 0.101 mL H₃C₆H₅O₇ solution = 0.525 × (0.101/0.500) g = 0.106 g

The mass of citric acid in the 0.101mL citric acid solution = 0.00530 g
The total mass of the 0.101mL citric acid solution = 0.106 g

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