# Chemistry

Hi DrBob,
Sorry about that, I wasn't very clear!

A. Standardization of Na2S2O3
-Mass of KIO3 in 100mL = 0.09600g
-Calculate Molarity of KIO3= ?
-4.486 mol/0.1L= 0.004486 M

B. Titration/ Volume of Na2S2O3
1: 38.44 mL
2: 37.23 mL
3: 38.11 mL
Average Volume: 37.93mL

-Calculate molarity of Na2S2O3=?
Not sure what to use as the mass here? Help

C. V of Bleach= 1.020 mL
V of Na2S2O3= 2.010 mL
Calculate Mass of NaOCl/100mL bleach=? Help with this.
-We are told here from the concentrations and volume of the added sodium thiosulfate solution used to titrate, calculate the numbers of moles of the oxidizing agent present, assuming it to be sodium hypochlorite, NaOCl. Use this to calculate the number of grams of sodium hypochlorite that was present in the 0.5mL of bleach.

The equation for I2 is: KIO3 reacts w/ excess KI in the following reaction:
IO3 + 5I + 6H--> 3I2 + 3H20
2S2O3 + I2 --> 2I + S406

Sodium Hypochlorite reacts with sodium thiosulfate in the following reaction:
NaClO + 2H + 2I--> I2 + Cl + H20+ Na
2S203 + I2 --> 2I + S406

Thanks.

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1. A. Standardization of Na2S2O3
-Mass of KIO3 in 100mL = 0.09600g
-Calculate Molarity of KIO3= ?
-4.486 mol/0.1L= 0.004486 M
This looks ok to me except for the -4.486. I don't know why the - sign is there unless it's just a typo.

B. Titration/ Volume of Na2S2O3
1: 38.44 mL
2: 37.23 mL
3: 38.11 mL
Average Volume: 37.93mL

-Calculate molarity of Na2S2O3=?
Not sure what to use as the mass here? Help
You have 100 mL of 0.004486M KIO3. Multiply those to find mmoles of KIO3 (mL x M = mmols). Then convert moles KIO3 to moles S2O3^-2. So moles KIO3 x 6 = mmoles thiosulfate. (How do I do that. 1 KIO3 = 3I2 and 3I2 = 6S2O3^-2; therefore, 1/6 mol KIO3 = 0.4486 mmols x (6S2O3/1KIO3) = 0.4486+6 = moles S2O3). M thiosulfate = mmoles/mL

C. V of Bleach= 1.020 mL
V of Na2S2O3= 2.010 mL
Calculate Mass of NaOCl/100mL bleach=? Help with this.
-We are told here from the concentrations and volume of the added sodium thiosulfate solution used to titrate, calculate the numbers of moles of the oxidizing agent present, assuming it to be sodium hypochlorite, NaOCl. Use this to calculate the number of grams of sodium hypochlorite that was present in the 0.5mL of bleach.

mL S2O3 x M S2O3 = mmoles S2O3.
Convert mmols S2O3 to mmoles OCl using the coefficients in the equations used (as I did in the KIO3 and S2O3 above.)
Then mmoles OCl x molar mass NaOCl = g NaOCl and that will be the grams in the volume you took. Convert that to g/100 mL.

The equation for I2 is: KIO3 reacts w/ excess KI in the following reaction:
IO3 + 5I + 6H--> 3I2 + 3H20
2S2O3 + I2 --> 2I + S406

Sodium Hypochlorite reacts with sodium thiosulfate in the following reaction:
NaClO + 2H + 2I--> I2 + Cl + H20+ Na
2S203 + I2 --> 2I + S406

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2. Too late at night and I made some typos.

You have 100 mL of 0.004486M KIO3. Multiply those to find mmoles of KIO3 (mL x M = mmols). Then convert moles KIO3 to moles S2O3^-2. So moles KIO3 x 6 = mmoles thiosulfate. I've redone the explanation because I typed a + sign instead o a x sign AND I notice a 1/6 which, although correct, it's better to do it another way.
(How do I do that. 1 KIO3 = 3I2 and 3I2 = 6S2O3^-2; therefore, 1/6 mol KIO3 = 0.4486 mmols x (6S2O3/1KIO3) = 0.4486+6 = moles S2O3). M thiosulfate = mmoles/mL
Here is that section done over.
How did I did that. 1 KIO3 = 3I2 and 3I2 = 6 S2O3; therefore, convert moles KIO3 to moles S2O3 the usual way using the coefficients in the balanced equation. 0.4486 mmoles KIO3 x (6 mols S2O3/1 mole KIO3) = 0.4486 mmoles x 6 = ??mmoles S2O3. THEN, ??mmoles S2O3/mL S2O3 = M S2O3.

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