Hi DrBob,

Sorry about that, I wasn't very clear!

Can someone please help me complete these calculations for my lab; I am having some trouble.

A. Standardization of Na2S2O3
-Mass of KIO3 in 100mL = 0.09600g
-Calculate Molarity of KIO3= ?
Please check: 0.09600g/(214.0g/mol)=4.486x 10^-4 mol
-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.

A. Standardization of Na2S2O3

-Mass of KIO3 in 100mL = 0.09600g
-Calculate Molarity of KIO3= ?
Please check: 0.09600g/(214.0g/mol)=4.486x 10^-4 mol
-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

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.

No problem, I'm here to help. Let's go through each of your calculations step by step.

A. Standardization of Na2S2O3:
To calculate the molarity of KIO3, you need to know the mass of KIO3 and its molar mass. In this case, the mass of KIO3 is given as 0.09600g and its molar mass is 214.0 g/mol. To calculate the number of moles of KIO3, divide the mass of KIO3 by its molar mass:

Number of moles of KIO3 = 0.09600g / 214.0 g/mol = 4.486 x 10^-4 mol

Next, you need to calculate the molarity of KIO3. The volume used for this calculation is 100mL or 0.1L. Divide the number of moles of KIO3 by the volume in liters:

Molarity of KIO3 = 4.486 x 10^-4 mol / 0.1 L = 0.004486 M

B. Titration/Volume of Na2S2O3:
To calculate the molarity of Na2S2O3, you need to use the volume of Na2S2O3 used in each titration and the average volume. However, you mentioned that you're not sure what mass to use.

In the case of Na2S2O3, you don't need to know the mass. Instead, you use the stoichiometry of the reaction between Na2S2O3 and the analyte (in this case, I2) to calculate the molarity of Na2S2O3.

C. V of Bleach and V of Na2S2O3:
To calculate the mass of NaOCl in the given volume of bleach (1.020 mL), you need to perform stoichiometric calculations based on the balanced equation for the reaction between NaOCl and Na2S2O3.

First, calculate the number of moles of Na2S2O3 used in the titration. This can be done using the molarity of Na2S2O3 and its volume:

Number of moles of Na2S2O3 = Molarity of Na2S2O3 x Volume of Na2S2O3 (in liters)

Next, use the balanced equation to determine the stoichiometric ratio between Na2S2O3 and NaOCl. The equation shows that 2 moles of Na2S2O3 react with 1 mole of NaOCl. Based on this ratio, you can calculate the number of moles of NaOCl present in the titrated volume of Na2S2O3.

Finally, to calculate the mass of NaOCl in the given volume of bleach, you need to convert the number of moles to grams. To do this, multiply the number of moles of NaOCl by its molar mass.

Remember to perform the calculations using the given concentrations and volumes.

I hope this explanation helps. If you have any further questions, feel free to ask.