Citric acid , (C6H8O7molar mass = 192.12 G/MOL) is a triprotic acid extracted from citrus fruits and pineapple

Question

How many grams of citric acid are contained in a 27.0ML sample that requires 37.6ML of 1.05 m for NAOH neutralization to occur?



Give the formula of the calcium salt of this compound.

Answer 1

First calculate the millimoles of base present:

37.6 ml * 1.05 M = 39.48 mmoles OH-

For the base to neutralize the acid, we need the same amount of H+, but since this is a triprotic acid, it has 3 moles of H+ per mole of Citric Acid (CA).

39.48 mmol H+ (1 mmol CA/3 mmol H+) = 13.16 mmol CA

13.16 mmol (1 mol/1000 mmol) = 0.01316 mol CA

0.01316 mol * 192.12 g/mol = 2.53 g Citric Acid

The volume of the citric acid is immaterial since you want the grams of the acid rather than the molarity of the solution.

Answer 2

Nowadays, you would call it, citric acid, calcium salt.

For these problems, you MUST, MUST do your calculations in moles and then convert to mass. So......
moles of NaOH to neutralize= 0.0376 * 1.05 M= 0.04 appx. This will neutralize 0.0133 moles of the acid. So compute the corresponding mass from 0.0133 moles * 192 g/mole.

Answer 3

haha... is this for Dr. McWilliam's online assignment? lol just joking....