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Molar Mass of Citric acid (H₃C₆H₅O₇) = 192.12 g/mol
H₃C₆H₅O₇(aq) + 3 NaOH(aq) → Na₃C₆H₅O₇(aq) + 3 H₂O(l)

Vida & Mindy titrated a 20.00 mL aliquot of grapefruit juice to its endpoint and it used 12.55 mL of a 0.115 M NaOH solution.

How many grams of citric acid are present in their aliquot?

What is the milligrams of citric acid per mL of juice?

Answer 1

In acid -base titrations we use the relation as

V1M1/n1 = V2M2/n2 where

V1 = volume of acid

M1 = molarity of acid

n1 = number of moles of acid in the balanced equation

V2 = volume of base

M2 = molarity of base

n2= number of moles of base in the balanced equation

Thus 20.00mL x M1 /1 = 12.55 x 0.115 /3

thus molarity of acid = 0.02405 M

It means 0.02405 moles of citric acid is present in 1L = 1000mL of solution

In 20 mL of solution the moles of citric acid present = 20mL x 0.02405 /1000mL

Mass of citric acid in 20 mL aliquot = moles in 20 mL x molar mass

= [20 x0.02405 / 1000 ] x192.12

= 0.0924 g in 20 mL

mass of citric acid present in 1 mL of juice = 0.0924 g/20 ml

= 0.004621 g/ml

= 4.621 mg /mL