Question

Cedrick & Astrid titrated a 20.00 mL aliquot of grapefruit juice with a 0.165 M NaOH solution to the end point. The initial buret reading was 1.72 mL and the final buret reading was 15.51 mL. They calculated that there was 0.1457 g of citric acid present in the juice sample. What is the amount mg of citric acid present per mL of juice?

Answer 1

Explanation:

Citric acid upon dissociation given three hydrogen ion that is, citric acid is a triprotic acid.

It is given that volume of juice is 20 ml and concentration of NaOH is 0.165 M.

    Initial volume = 1.72 ml,       Final volume = 15.51 ml

Hence, volume of the base will be calculated as follows.

                 $V_{base} = V_{final} - V_{initial}$

                              = (15.51 - 1.72) ml

                              = 13.79 ml

Now, moles of NaOH will be calculated as follows.

                 No. of moles = $Molarity \times Volume$

                                       = $0.165 M \times 13.79 ml$

                                       = 2.275 mmol

As,   1 mol of  citric acid = 3 mol of NaOH

Hence, moles of citric acid will be calculated as follows.

                 Mole of citric acid = $\frac{1}{3} \times 2.275 mmol$

                                                = 0.758 mmol

Since, molar mass of citric acid is 192.124 g/mol. Hence, mass of citric acid will be calculated as follows.

                      Mass = No. of moles × Molar mass

                                = 0.758 mmol × 192.124 g/mol

                                = 145.72 mg

Thus, we can conclude that there is 145.72 mg of citric acid present per mL of juice.