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July 30, 2016
Posted by **Saira** on Monday, October 27, 2008 at 6:12pm.

This was the lab experiment:

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A. Standardization of the Sodium Thiosulfate Solution

Make up a standard solution of potassium iodate by accurately “weighing by difference” about 0.1 g of KIO3 and placing it in your 100 mL volumetric flask. Fill the flask to mark with distilled water. Potassium iodate reacts with excess KI in acid solution according to the following reaction:

IO3- + 5I- + 6H^+ ----->3I2 + 3H2O

RECALL:

2S2O3^2- + I2 --> 2I- + S4O6^ 2-

To perform then standardized, first prepare the buret with the given solution of Na2S2O3. Then dispense 25 mL of KIO3 solution by pipett into a 250 mL Erlenmeyer flask, add 25 mL of 2% (v/v) KI solution and then add 10 mL of 1M HCL, and mix swirling. A deep brown color should appear, indicating the presence of iodine. Titrate immediately with the Na2S2O3 solution in the burett until the brown fades to a pale yellow and then add 5 mL of starch indicator solution and continue the titration until the deep blue color of the starch indicator disappears. Record the volume of the sodium thiosulafte solution used in the titration. Determine the stoichiometric ratio between the iodate and thiosulafte and use this to calculate the accurate concentration if the Na2S2O3 solution.

A.Standardization of Na2S2O3 ( Sodium thiosulfate)

Mass of KIO3 in 100 mL = 0.1255g

Show calculation for molarity for the potassium iodate:

Moles (n)= 0.1255g/214.00= 0.00058645 mols

100mL/1000 = 0.1 L

Molarity= 0.00058645/0.1= 0.00586448

Therefore the Concentration of KIO3 is 5.86*10^-3

Titration:

Average Volume of Na2S2O3 = 27.15 mL

Calculate the concentration of Na2S2O3

_______________M.

KIO3 moles: M x L

= 0.000586448(0.025L)

= .000146612 moles

.000146612 mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) mols

= .000887972 mols S203^2-

Then M S2O3^-2 = .000887972 mols/0.02715 L.

M = 0.032400

Therefore,the concentration of Na2S2O3 is 0.032406 M.

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B.Titration of a Sample of Household bleaches:

The oxidizing agent in a household bleach is determined by placing 100 mL of distilled water in a 250 mL Erlenmeyer flask and adding 10 mL of the 10% (w/w) KI solution, swirling the contents of the flask to mix the solutions. Add an accurately measured 0.5 mL of your assigned bleach to the flask, and swirl. Then add 10 mL of 2 M H2S04 and proceed with the titration, using the standardized sodium thiosulfate solution. Add the starch indicator when the solution is pale yellow and continue the titration until the deep blue colour of the starch indicator disappear. Record the volume of thiosulfate solution used in the titration.

From the concentration and volume of the added sodium thiosulfate solution used to titrate the different brands of bleach, calculate the numbers of moles of the oxidizing agent presence, assuming it to be sodium hypochlorite, NaOCl. Use this to calculate the number of grams of sodium hypochlorite that was present in the 0.5 mL of bleach. Express the final result as the mass of sodium hypochlorite per 100 mL of bleach.

The sodium thiosulfate reacts with the sodium hypochlorite according to the following stoichiometric redox equations:

NaClO (aq) + 2H^+ + 2I- ---> I2 + Cl- +H2O + Na^+

Recall:

2S203^ 2- + I2 ---> 2I- + S4O6 ^ 2-

B.Titration of household bleach: Javex

Calculated Concentration of Na2S2O3= 0.032400 M.

Reaction # 1:

ClO^- + 2I^- ==> I2 + Cl^-

Reaction # 2:

2S2O3^2-(aq)+I2----> 2I-(aq)+S4O6^2-(aq)

Trial: 1

Volume of bleach ( 0.5mL)

Volume of Na2S2O3 ( 53 mL)

= mass of NaOCl/100 mL bleach

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moles of S203^2- =(MOLARITY)(VOLUME)

=(0.03240 mol/L)(0.053L S2O3^2-)

= 0.001717 moles S2O3^2-

moles of I2= (1/2)(0.001717)

= .0008586 moles of I2

moles of ClO- = (1/1)(0.0008586)

= 0.0008586 moles of ClO-

(Molar mass)(moles of ClO-)= mass

(74.44)(0.0008586)= 0.06391

Change mass/0.5mL to mass/100mL

Therefore: (0.06391)(100/0.5)

= 12.78

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Trail 2:

Volume of bleach ( 0.5mL)

Volume of Na2S2O3 ( 55.4 mL)

= mass of NaOCl/100 mL bleach

=_______________?

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moles of S203^2- =(MOLARITY)(VOLUME)

=(0.03240 mol/L)(0.0554L S2O3^2-)

= 0.001795 moles S2O3^2-

moles of I2= (1/2)(0.001795)

= .0008975 moles of I2

moles of ClO- = (1/1)(.0008975)

= .0008975 moles of ClO-

(Molar mass)(moles of ClO-)= mass

(74.44)(.0008975)= 0.06681

Change mass/0.5mL to mass/100mL

Therefore: (0.06681)(100/0.5)

=13.36

Average mass of NaOCl/100 mL bleach

12.78+ 13.36 = 13.07

Therefore, the average mass of NaOCl/ 100ml = 13.07

- Chemistry: please double check... -
**DrBob222**, Monday, October 27, 2008 at 8:33pmA.Standardization of Na2S2O3 ( Sodium thiosulfate)

Mass of KIO3 in 100 mL = 0.1255g

Show calculation for molarity for the potassium iodate:

Moles (n)= 0.1255g/214.00= 0.00058645 mols

100mL/1000 = 0.1 L

Molarity= 0.00058645/0.1= 0.00586448

Therefore the Concentration of KIO3 is 5.86*10^-3**No, the concentration of KIO3 is 5.8645*10^-3 M which I would round to 5.864*10^-3. You carried all the other numbers until here and you should show at least one more place. Your final answer of 0.03240 M is ok; you just need the intermediate steps to follow through the correct number of significant figures.**

KIO3 moles: M x L

= 0.000586448(0.025L)**I think this should be 0.00586448 but the final answer is ok.**

= .000146612 moles

.000146612 mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) mols

= .000887972 mols S203^2-**Shouldn't this be 0.00087972? I think you have an extra 8 in there. Same in the next step.**

Then M S2O3^-2 = .000887972 mols/0.02715 L.

M = 0.032400**You should drop the final 0 of 0.032400 to 0.03240.**

Therefore,the concentration of Na2S2O3 is 0.032406 M.**It is unclear to me how you went from 0.032400 to 0.032406. I think there was a second titration in here and it may be that you averaged that in but you don't show it anywhere.**

oles of S203^2- =(MOLARITY)(VOLUME)

=(0.03240 mol/L)(0.053L S2O3^2-)

= 0.001717 moles S2O3^2-

moles of I2= (1/2)(0.001717)

= .0008586 moles of I2

moles of ClO- = (1/1)(0.0008586)

= 0.0008586 moles of ClO-

(Molar mass)(moles of ClO-)= mass

(74.44)(0.0008586)= 0.06391

Change mass/0.5mL to mass/100mL

Therefore: (0.06391)(100/0.5)

= 12.78**The final answer looks ok except as I mentioned earlier (in your first posts) that 12.8% seems high to me. I think I pointed out that most of the commercial bleaches I see on the market run about 6%. I know this asks for g/100 mL BUT that IS mass/volume percent.**- Chemistry: please double check... -
**Saira**, Tuesday, October 28, 2008 at 4:19pmThank-You so much DrBob.. you really helped a lot..and i had made some typoz up there so thats why some things didnt make sense...and i fixed the problems you told me to before handing it in..Thank- You Once again

- Chemistry: please double check... -
- Chemistry: please double check... -
**kimberly**, Thursday, May 31, 2012 at 11:29pmcan you explain what you're doing here???

KIO3 moles: M x L

= 0.000586448(0.025L)

= .000146612 moles

.000146612 mols KIO3 x (3 mols I2/1 mol KIO3) x (2 mols S2O3^-2/1 mol I2) mols

= .000887972 mols S203^2-

Then M S2O3^-2 = .000887972 mols/0.02715 L.

M = 0.032400

Therefore,the concentration of Na2S2O3 is 0.032406 M.