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:

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!

Answer 2

The volume required to neutralize the sample of juice is ;  9.76 mL

Given data:

pH of lemon juice = 2.5

Density of lemon juice = 1.0 g/mL

citric acid concentration = 0.5% by mass

molar conc of NaOH = 0.0100 M

First step :   Determine the mass of citric acid

3.71 * ( 0.5g citric acid  / 100 g of lemon juice )

= 3.71 * ( 0.5 / 100 )  = 0.01855 g citric acid

Next : Moles of citric acid ( NaOH ) required to neutralize the sample juice

0.01855 * ( 1 / 190.12 g ) = 9.76x10⁻⁵  moles

∴ Volume of NaOH required to neutralize the sample

   = molar conc * moles

   = 0.0100 * 9.76x10⁻⁵  = 9.76x10⁻³L

    ≈ 9,76 mL

 

Hence we can conclude that The volume required to neutralize the sample of juice is ;  9.76 mL

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