Question

A student followed the procedure in this experiment to determine the number of grams of C_3H_5O(COOH)_3 per 1 mL of an apple juice sample. The titration of 20.0 mL of the undiluted juice required 12.84 mL of 9.580 times 10^-2M NaOH solution. Calculate the number of moles of NaOH required for the titration. Calculate the number of moles of C_3H_5O(COOH)_3 titrated. Calculate the mass of C_3H_5O(COOH)_3 present in the juice sample. Calculate the mass of C_3H_5O(COOH)_3 present in 1 mL of apple juice.

Answer 1

Given,

The volume of undiluted juice = 20.0 mL x ( 1L /1000 mL) = 0.02 L

The concentration of NaOH = 9.580 x 10^-2 M

The volume of NaOH solution required for the titration = 12.84 mL x ( 1L /1000 mL) = 0.01284 L

1) Calculating the number of moles of NaOH required for the titration,

We know, the formula,

Molarity = Number of moles / L of solution

9.580 x 10^-2 M = Number of moles / 0.01284 L

Number of moles = 1.23 x 10^-3 mol of NaOH

2) Calculating the number of moles of C₃H₅O(COOH)₃ titrated,

The reaction between acid and NaOH is,

C₃H₅O(COOH)₃(aq) + 3NaOH(aq) $\rightarrow$ C₃H₅O(COONa)₃(aq) + 3H₂O(l)

Using the calculated number of moles of NaOH and the mole ratio from the balanced chemical reaction, calculating the number of moles of C₃H₅O(COOH)₃,

= 0.00123 mol of NaOH x ( 1 mol of C₃H₅O(COOH)₃ / 3 mol of NaOH)

= 4.10 x 10^-4 mol of C₃H₅O(COOH)₃ are titrated.

3) Calculating the mass of C₃H₅O(COOH)₃ present in the juice sample.

Converting calculated moles of C₃H₅O(COOH)₃ to grams by using the molar mass of C₃H₅O(COOH)_3.

We know, the molar mass of C₃H₅O(COOH)₃ = 192.124 g/mol

Thus,

= 4.1 x 10^-4 mol of C₃H₅O(COOH)₃ x ( 192.124 g / 1 mol)

= 0.0788 g of C₃H₅O(COOH)₃ in the juice sample.

4) calculating the mass of C₃H₅O(COOH)₃ present in 1 mL of apple juice.

= 1 mL apple juice x (0.07878 g of C₃H₅O(COOH)₃ / 20.0 mL apple juice)

= 0.00394 g of C₃H₅O(COOH)₃ in 1 mL of apple juice.