Question

Lemon juice has a pH of about 2.5. Assuming that the acidity of lemon juice is due solely to citric acid, that citric acid is a monoprotic acid, and that the density of lemon juice is 1.0 g/mL, then the citric acid concentration calculates to 0.5% by mass. Estimate the volume of 0.0100 M NaOH required to neutralize a 3.71-g sample of lemon juice. The molar mass of citric acid is 190.12 g/mol.

Answer 1

Answer:

9.76 mL

Explanation:

To solve this problem we first need to calculate the amount of citric acid contained in a 3.71 g sample of lemon juice.

• The only data of concentration we're given is a mass procent, that 0.5% of the mass of a sample is citric acid:

• 3.71 g * 0.5/100 = 0.01855 g citric acid

Now with the molar mass of citric acid we can calculate the moles:

• 0.01855 g citric acid ÷ 190.12 g/mol = 9.757x10⁻⁵ mol citric acid

Because citric acid is a monoprotic acid, one mol of acid will react with one mol of NaOH, thus we calculate the volume of 0.0100 M NaOH that has 9.757x10⁻⁵ mol:

Concentration = mol/Volume

0.0100 M = 9.757x10⁻⁵ mol / Volume

Volume = 9.76x10⁻³L = 9.76 mL

Answer 2

Answer:

The volume to neutralize the sample of lemon juice is 9,76mL

Explanation:

3,71 g sample of lemon juice are:

3,71 g ×$\frac{0,5 g citric acid}{100 g LemonJuice}$ = 0,01855 g of citric acid. In moles:

0,01855 g × $\frac{1mol}{190,12g}$ = 9,76x10⁻⁵ moles of citric acid.

Assuming this acid is a monoprotic acid, the moles of NaOH required to neutralized the citric acid are 9,76x10⁻⁵ moles

As the molar concentration of NaOH is 0,0100 mol/L. The volume to neutralize the sample of lemon juice is:

9,76x10⁻⁵ moles × $\frac{1L}{0,0100mol}$ = 9,76x10⁻³L ≡ 9,76 mL

I hope it helps!