Question

Consider a different analyte for this exercise. Citric acid is found in many fruits and fruit juices. Sodium hydroxide (NaOH) is the titrant and citric acid (H₃C₆H₅O₇) the analyte according to the following balanced chemical equation.

Complete the following table for this titration.

Data Table P2: Titration of citric acid in orange juice with sodium hydroxide.

concentration of OH−	0.0702 M
volume orange juice	8.65 mL
mass orange juice	8.79 g
volume of OH− solution	27.58 mL
mmol of OH−	..... mmol
mmol of H3C6H5O7	.....mmol
mass of H3C6H5O7	......g
mass % of H3C6H5O7 in orange juice	....... %
molarity of H3C6H5O7 in orange juice	......M

Answer 1

The reaction may be represented as C3H50 (Cool) ₂ + 3 NaOH Naz C34 (coolz / +3H₂O . mmolot = volume & molarity = 27.58 ml x 0. 0702 mol = 11.936116 mma on- V ... Nach react with Imol of citric and : 1.93616 - 1936116 mmol citric acid 0.6454 mma mass = mmol x molar mass = 0.6454 mmd x 192.124 glybu - 123.99145mg 1000 mg = 1g mass and = [0.124g) H₃ CH₃O7 mass + Mange juice = massaid xloo mass Juice - 0.124 g x loo ma .799

masst = 1.4106% es warge juce molarity = mokes Volumes = 0.6454 mmol 8.65 mL [modarity = 0.074613 M ] Thus the data was completed based on the given values.