Answer to Question #153959 in Organic Chemistry for fdf

Question

Procedure

1. Use a pipet bulb and a 5-mL transfer pipet to measure EXACTLY 5.00 mL of a commercial bleach solution into a 100-mL volumetric flask.
2. Dilute to the mark with distilled water, stopper, and mix well.
3. Weigh out approximately 2 g solid KI. This is a large excess over that which is needed.
4. Pipet EXACTLY 25.0 mL of the dilute bleach into an Erlenmeyer flask.
5. Add the solid KI and about 25 mL of distilled water. Swirl to dissolve the KI.
6. Working in a fume hood, add approximately 2 mL of 3 M HCl while stirring the solution. The solution should be dark yellow to red-brown from the presence of the I3- complex ions.
7. Fill the buret with 0.100 M sodium thiosulfate solution (this may have already been done for you). Record the initial buret reading and the solution molarity in the Data table.
8. Titrate with the standard 0.100 M sodium thiosulfate solution until the iodine color fades to light yellow.
9. Add one dropperful of starch solution. The blue color of the starch-iodine complex should appear.
10. Continue the titration until one drop of Na2S2O3 solution causes the blue color to disappear. Record the final buret reading in the Data Table.

Question--> How would each of the following laboratory mistake affect the calculated value of the percent NaClO in the commercial bleach (too high, too low, no change)? Explain. a. In step 6, some of the iodine that formed sublimed from the solution. 

Answer 1

Here volume = 100 ml =0.100 L

Molarity = number of moles / volume in L

1. $7.08\times10$ ^-4 moles Vitamin C / 0.100 L

$= 7.08\times10$ ^-3

$M1V1 = M2V2 Here = M1 = 7.08*10^-3M V1= 5.0 ML V2 =100 ml M2 = ? M1V1 = M2V2 7.08*10^-3M * 5.0 ml = M2*100 ml M2 = 7.08*10^-3M * 5.0 ml /100 ml M2= 3.54*10^-4 M 3.54*10^-4 M is this diluted concentratio$