Question

Students performed a procedure similar to Part II of this experiment (Analyzing Juices for Vitamin C Content) as described in the procedure section. Three 10.00mL samples of juice were titrated with DCP that had a standardized concentration of 9.98x10-4M. The three titrations took an average of 16.34mL of DCP. Calculate the mass (in mg) in 50.00mL of juice. (MM Ascorbic Acid = 176.124 g/mol)

Answer 1

`m_(AA)=14.36mgAA`

Step-by-step explanation:

Hello,

In this case, the titration is based of the equality between the moles of the titrant and the analyte, then, in terms of molarity we define it as shown below:

`M_(DCP)*V_(DCP)=M_(juice)*V_(juice)`

As the ascorbic acid is the substance reacting with the DCP, we infer that the 16.34 mL of DCP completely neutralized the ascorbic acid; in such a way, solving for the molarity of the neutralized acid, one obtains:

`M_(juice)=(M_(DCP)*V_(DCP))/(V_(juice))=(9.98x10^(-4)M*16.34mL)/(10.00mL) \\M_(juice)=1.63x10^(-3)M`

Finally, by using the molar MASS of the ascorbic acid and applying the proper units conversion, the result is shown below:

`m_(AA)=1.63x10^(-3)(molAA)/(L)*0.05000L*(176.124gAA)/(1molAA) *(1000mgAA)/(1gAA) \\m_(AA)=14.36mgAA`

Therefore, 14.36 mg of ascorbic acid are contained into 50.00mL of juice.

Best regards.

Answer 2

Students performed a procedure similar to Part II of this experiment (Analyzing Juices for Vitamin C Content) as described in the procedure section. Given that molarity of DCP is 9.98x10-4 M, it took 16.34 ml of DCP to titrate 10 mL of sample.

Amount of ascorbic acid = 0.050 L sample (0.01634 L DCP/0.01 L sample)( 9.98x10-4 mol DCP/L DCP)(1 mol Ascorbic acid/ 1mol DCP)(176.124 g/mol)(1000mg/1g)= 14.36 mg ascorbic acid