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Reply
| | From:  Abalo3 (Original Message) | Sent: 1/13/2008 3:58 AM |
Hi Steve,
I had difficulties with this lab until the teacher sent me the lab results to finish up it. I am totally out of touch and need some suggestions.Your inputs will be greatly appreciated.
Thank you.
Procedure 1:
Standard versus iodine stock solution:
16.00 mL was too much
13.00 mL was too much
12.40 mL just changed color
12.35 mL just changed color
12.35 mL just changed color
LAB:
The prepared iodine solution on the Chemicals shelf with a stated (anticipated) concentration of 0.0015M is standardized (to determine the current actual concentration) by performing the following titration:
1. Prepare a solution of pure ascorbic acid of known strength.
1b. Weight out and add exactly 0.05 g ascorbic acid to the volumetric flask.
1c. Fill the volumetric flask with water. (This is the most precise way of making a 100 mL solution.
2. Do a rough (approximate) titration of the iodine stock solution. (I do not understand when they ask about doinig a rough approximate????????????)
2a. Take a 150 Erlenmeyer flask from the Glassware shelf and place it on the workbench.
2b. Pour 10.0 mL of the ascorbic acid solution into the flask.
2c. Add 10.0 mL of distilled water to the flask to dilute the ascorbic acid solution added.
2d. Add 1.0 mL of the starch indicator to the flask.
2f. Fill the burette with 50.00 mL of the iodine stock solution (with nominal strength of 0.0015M.)
2g. Drag the flask and drop it on the lower half of the burette (the "base of the stand".)
2h. Open the Properties window and move the window to a convenient location on the computer screen. Click back on the burette, and then click the pushpin in the corner of the Properties window to lock it on the burette.
2i. Titrate the ascorbic acid sample by adding iodine until the solution in the flask turns dark blue. (As you learned in the Titration Tutorial, you should do this first rough titration by adding the iodine solution 1-2 mL at a time in order to quickly find the range in which the endpoint is reached.)
2j. Refill the burette with additional iodine solution to bring it back to the 50.00 mL mark. Be careful not to overfill the burette, since it would make a mess on a real lab bench.
3. Do two or more "fine" (or precise) titrations of other samples of the standard ascorbic acid solution. Strive to cause the starch indicator change color when adding only one more drop (the "last 0.05 mL") of iodine solution.
4. Calculate the current molarity (today's "standardized value") of the iodine solution using the average of your best titration results of the ascorbic acid solution.
Questions
1. Calculate the molarity of the freshly-prepared ascorbic acid standard (known strength) solution:
(a) Mass of ascorbic acid used:
Do I use 0.05 g of ascorbic acid use in the first flask????
(b) Moles of ascorbic acid (MW=176.1 g/mol):
(c) Total volume of standard solution prepared (mL):
(d) Ascorbic acid concentration in standard solution (mol/L):
2. For each titration, record and calculate the following. (Be sure to show your data and calculations clearly).
(a) Volume of iodine solution added (mL):
(b) Concentration of the iodine solution, using the formula M1*V1 = M2*V2 (where V1 and V2 are the volumes of the two concentrated chemical solutions added to the flask):
3. From your two or more best "fine" titrations, calculate the average iodine concentration, thus determining today's standard value for the iodine solution.
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Reply
| | | From: Abalo3 | Sent: 1/14/2008 5:04 PM |
(c) Total volume of standard solution prepared (mL) was 100 ml
d. Ascorbic acid concentration in standard solution (mol/L):
Converting 100 ml= 0.1 L
C6H8O6= (12*6) + (1*8) + (16*6)
= 176 g/mole
= 176g
Find the mole of Ascorbic first:
= 1g * (1 mole / 176g)
= 0.0057 mole
Ascorbic acid concentration
= 0.0057mole / 0.1 L
=0.057 mole / L |
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Reply
| | | From: ·Steve· | Sent: 1/14/2008 6:11 PM |
= 176 g/mole
= 176g
176 g/mol, or 176.12412 g/mol using accurate atomic weights. (We'll round off the molarity later.) Find the mole of Ascorbic first:
= 1g * (1 mole / 176g) Don't forget, you used 0.05 g, not 1 g. The molarity should be given to three significant figures at least, four would be better (more common for accurate work like this). The given information is a little vague about significant figures, unfortunately! |
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Reply
| | | From: Abalo3 | Sent: 1/14/2008 7:19 PM |
c) Total volume of standard solution prepared (mL) was 100 ml
d. Ascorbic acid concentration in standard solution (mol/L):
Converting 100 ml= 0.1 L
C6H8O6= 176.12412 g/mole
= 176.12412 g
Find the mole of Ascorbic first:
= 0.05g * (1 mole / 176g)
= 2.83891* 10^-4 mole
Ascorbic acid concentration :
=2.83891*10^-4mole / 0.1 L
=0.0028 mole/L
2.a.Volume of Iodine (I2) added
=50 ml - 12.35 ml
= 37.65 ml
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Reply
| | | From: ·Steve· | Sent: 1/15/2008 2:37 AM |
=0.0028 mole/L That's good, but I would not round the answer off so much, since this is an analytical lab. Three significant digits is 0.00284 M and four significant digits is 0.002839 M. 2.a.Volume of Iodine (I2) added
=50 ml - 12.35 ml
= 37.65 ml
I got the impression that the numbers below were the actual volumes, but if the buret starts with 50 mL at the top and these are actually the readings from the buret, then that is right, you would subtract each number from 50 mL to obtain the volume of iodine solution added. Burets are normally the other way with 0 at the top and 50 mL at the bottom. That way, the reading from the buret is the volume delivered and you don't have to always subtract the reading from 50. But the procedure says to refill the buret "back to the 50.00 mL mark," which indicates the 50 is at the top. Does the results printout specify if the numbers such as 12.40 mL are buret readings or actual volumes? This is an important point, so we want to get it right. 12.40 mL just changed color
12.35 mL just changed color
12.35 mL just changed color |
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Reply
| | | From: Abalo3 | Sent: 1/16/2008 1:57 PM |
Question????
Does the results printout specify if the numbers such as 12.40 mL are buret readings or actual volumes? This is an important point, so we want to get it right.
12.40 mL just changed color
12.35 mL just changed color
12.35 mL just changed color
Answers>>>>>>
These values are the volumes of titrant that fell from
the burette and into the titration flask to cause the
color change. |
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Reply
| | | From: Abalo3 | Sent: 1/16/2008 11:55 PM |
To recap everything from the beginning
1.a. The mass of ascorbic used is 0.05g
b. Moles of ascorbic acid
=0.05g ascorbic * (1 mol ascorbic/ 176.1 g ascorbic)
=2.84 *10^-4 moles
(c) Total volume of standard solution prepared (mL) was 100 ml
d.
d. Ascorbic acid concentration in standard solution (mol/L):
Converting 100 ml= 0.1 L
C6H8O6= 176.12412 g/mole
= 176.12412 g
Find the mole of Ascorbic first:
= 0.05g * (1 mole / 176g)
= 2.83891* 10^-4 mole
Ascorbic acid concentration :
=2.83891*10^-4mole / 0.1 L
=0.002839 M
2.a.Volume of iodine solution added (mL) is 12.35 ml
(b) Concentration of the iodine solution, using the formula M1*V1 = M2*V2 (where V1 and V2 are the volumes of the two concentrated chemical solutions added to the flask)
A little confuse here?????
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Reply
| | | From: ·Steve· | Sent: 1/17/2008 4:06 AM |
Let M1 and V1 be for the ascorbic acid solution and M2 and V2 be for the iodine solution. M1 = 0.002839 M from your calculation. V1 = 10.0 mL according to the procedure. M2 = molarity of the iodine solution to solve for. V2 = volume of iodine solution used to titrate the ascorbic acid solution = 12.35 mL. That's it! The formula M1V1 = M2V2 works because the reaction between ascorbic acid and iodine is one-to-one. Volume in mL times molarity = mmoles, so this formula is actually "mmoles of ascorbic acid = mmoles of iodine" if mL volumes are used for V1 and V2. You do not have to convert the volumes to liters here. |
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Reply
| | | From: ·Steve· | Sent: 1/17/2008 4:17 AM |
I see you already did this in Part II, good! The procedure says to take an average of the iodine molarities calculated from the best titrations. These would be with the volumes 12.40 mL, 12.35 mL, and 12.35 mL of iodine solution. All three of these gave end points that "just changed color." That will finish Part I. Molarity of the iodine solution (Part I):
M1V1= M2V2
M2= (0.002839M * 0.010L) / 0.01235 L
= 0.002299 M |
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Reply
| | | From: Abalo3 | Sent: 1/17/2008 1:47 PM |
2.
b. Concentration of the iodine solution, using the formula M1*V1 = M2*V2 (where V1 and V2 are the volumes of the two concentrated chemical solutions added to the flask)
Molarity of the iodine solution (Part I):
M1V1= M2V2
M2= (0.002839M * 10.0 mL) / 12.35 mL
= 0.002299 M
3. From your two or more best "fine" titrations, calculate the average iodine concentration, thus determining today's standard value for the iodine solution
=12.40 mL+ 12.35 mL +12.35 mL
=12.3667 mL |
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Reply
| | | From: Abalo3 | Sent: 1/17/2008 3:41 PM |
3. From your two or more best "fine" titrations, calculate the average iodine concentration, thus determining today's standard value for the iodine solution
= (0.002299 M + 2.4139 *10^-4M + 0.002425M) / 3
= 0.00165513M |
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Reply
| | | From: ·Steve· | Sent: 1/17/2008 8:22 PM |
Now use this average volume to calculate the iodine molarity. That will be the average molarity. Do it the same way you did before, using M1V1 = M2V2. M1V1= M2V2
M2= (0.002839M * 10.0 mL) / 12.35 mL <–�?Change this volume to 12.3667 mL
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Reply
| | | From: Abalo3 | Sent: 1/17/2008 9:55 PM |
I believe this the end for this part of the problem
4. From your two or more best "fine" titrations, calculate the average iodine concentration, thus determining today's standard value for the iodine solution.
M1V1 = M2V2.
M1V1= M2V2
M2= (0.002839M * 10.0 mL) / 12.3667 mL
= 0.0022957M
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