The citric acid in a lemon juice sample was neutralized by titration with NaOH solution.If 5.00 mL of lemon juice required 47.8 mL of 0.121 M NaOH for neutralization,what was the molarity of the citric acid in the lemon juice? The reaction is 3 NaOH + H₃C₆H₅O₇ $\rightarrow$3 H₂O + Na₃C₆H₅O₇.

Question

Quizplus · Accepted Answer

From the balanced chemical equation provided, we can see that 3 moles of NaOH reacts with 1 mole of citric acid (HS(C6H7O7)).

Therefore, the number of moles of NaOH used in the titration is:

0.121 M * 47.8 mL = 0.00577 moles NaOH

Since 3 moles of NaOH reacts with 1 mole of citric acid, the number of moles of citric acid in the lemon juice sample is:

0.00577 moles NaOH / 3 = 0.00192 moles citric acid

The volume of lemon juice used in the titration is 5.00 mL, or 0.00500 L.

Therefore, the molarity of citric acid in the lemon juice sample is:

0.00192 moles / 0.00500 L = 0.384 M

However, since citric acid has three acidic protons, it is a triprotic acid, which means that each molecule of citric acid can donate three protons. This means that the molarity of the citric acid solution is actually three times higher:

0.384 M * 3 = 1.15 M

Rounded to two significant figures, the answer is 0.39 M or D.

The Citric Acid in a Lemon Juice Sample Was Neutralized →\rightarrow→

The Citric Acid in a Lemon Juice Sample Was Neutralized $\rightarrow$