help?! Name: Date Score:125 Experiment 11: Determining the Amount of Vitamin C i

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help?!

Name: Date Score:125 Experiment 11: Determining the Amount of Vitamin C in the Siena Drink Mix Pre-Lab Assignment Please answer the following question in your lab notebook after the Hazard Table When standardizi 10.00 mL of a 5 mg/mL solution of ascorbic acid. What is the molarity of ng an iodine solution, 30.37 mL of the solution was required to completely oxidize molarity of the titranto oxidize Post-Lab Assignment Enter your raw data in the first three columns in the table below. Then, perform a sample calculation that determines the molarity of your iodine titrant from the ascorbic acid standardization data. Next,. perform a Q-test to determine if you can remove any of your data points as an outlier. Finally. 1. determine the average molarity of your titrant (with any outiers removed) and place it in the box. Ascorbic Acid Standardization Data Molarity Trial 2 molesof GveQ t Sample Molanty Calculation: 8.886x lo ma 8 2-M zees,.torhaus0.29SX /"A O-test Average Molarity O 287 xlo

Explanation / Answer

For the first one , the reaction is:

C6H8O6 + I2 + H2O C6H6O6 + 2I- + 2H+

10 ml of 5 mg / ml ascorbic acid, to get the mg just multiply these valus

50 mg or 0.05 grams of AA , molar mass of AA is 176 g/gmol

moles of AA are = 0.05/ 176 = 0.000284 moles of AA

since is a 1:1 reaction, moles of AA must be equal to moles of Iodine I2 so

Molarity of iodine is 0.000284 / 0.03037 = 0.00935 M

From the calculations you made I have repeated them and they are correct, for the Q test im using the Dixons Q test

Lets put the molarities in order

2.35x 10-9 , 2.93x 10-9 , 3.32x 10-9, I suspect the first point is an outlier

Calculate Q with

Q = gap / range, gap is the difference between the value we suspect and the closest value to it and the range the difference between the max and min value

gap = 2.93x 10-9 - 2.35x 10-9 = 5.81 x 10-10

range = 9.71x 10-10

Q =  5.81 x 10-10 / 9.71x 10-10 = 0.559

Now we have to look at the Q table for critical values of Q, im getting a value of 0.97 for a 95% confidence level (value gotten online) so

0.595 < 0.97

Q < Q table so this point is not an outlier

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