To illustrate the calculation, assume that 21.95 mL of a thiosulfate solution is required to reduce the iodine liberated by 20.85 mL of 0.0200M KIO3 and that 18.63 mL of the same thiosulfate solution reacts with the iodine replaced by 20.00 mL of an unknown solution of sodium hypochlorite.
(This is in the sample calculation given in my lab manual, for reference.)
And the question I'm given in my lab manuel is:
See the question in your lab manual. Suppose that you determine that the molarity of your sodium thiosulfate solution is 0.100 M. Suppose that 10.02 mL of the same thiosulfate solution reacts with the iodine replaced by 20.00 mL of an unknown solution of sodium hypochlorite. Calculate the weight, in grams, of available Cl in a liter of the unknown solution. Enter your answer as a decimal number with 3 decimal places.
This is what I did.
Since M thio = 0.100,
2 x M OCl x 0.0200L = 0.100 M x .01002L
M OCl = .02505
0.02505 mol NaOCl x 1 mole Cl x 35.45 g Cl x 1000 mg = 888.0225 mg Cl/L =888.023 ppm
But it says the answer is incorrect. So what would I be doing wrong here?
I would look at the following. I think your value of 0.02505 M (mols/L) for NaOCl is correct.
I believe available chlorine is calculated as Cl2. Therefore, I would multiply by 70.90 instead of 35.45 and you have changed that to mg which I would not do since the problem asks for grams and not mg. Let me know how this turns out and/or shed any new information about exactly what is meant by available chlorine.
Turns out all the values were right such as .02505 M and 35.45 g Cl etc, I just didn't need to multiply by 1000 and leave it in grams as you said.
Many thanks. The literature is not clear on how to report "available chlorine"; i.e., as Cl, Cl2, or as some suggest NaOCl.
To troubleshoot the problem and find out what went wrong, let's go through the calculations step by step.
Given:
- Volume of thiosulfate solution used: 10.02 mL
- Volume of unknown solution (sodium hypochlorite) reacted: 20.00 mL
- Molarity of thiosulfate solution (Na2S2O3): 0.100 M
Let's calculate the moles of thiosulfate solution used:
Moles of Na2S2O3 = Molarity x Volume (in L)
Moles of Na2S2O3 = 0.100 mol/L x 0.01002 L = 0.001002 mol
Next, according to the balanced equation, for every 2 moles of Na2S2O3 reacting, 1 mole of Cl is used. Therefore, the moles of Cl can be calculated as follows:
Moles of Cl = 0.001002 mol / 2 = 0.000501 mol
To find the weight of available Cl in a liter of the unknown solution, we need to use the volume and molarity of the unknown solution reacted with the thiosulfate.
Given:
- Volume of unknown solution reacted: 20.00 mL
- Molarity of unknown solution (sodium hypochlorite): unknown, let's call it M
Using the stoichiometry of the balanced equation, we can set up the following equation:
Moles of Cl in unknown solution = Moles of thiosulfate used
M x Volume of unknown solution (in L) = 0.000501 mol
Solving for M:
M = 0.000501 mol / 0.02000 L = 0.02505 M
Now, the question asks for the weight of available Cl in a liter of the unknown solution, which is in grams. To convert from molarity (mol/L) to grams per liter (g/L), we need to use the molar mass of Cl, which is 35.45 g/mol.
Weight of Cl in a liter of the unknown solution = Molarity x Molar mass of Cl
Weight of Cl = 0.02505 mol/L x 35.45 g/mol = 0.88663825 g/L
Rounding to three decimal places, the answer is 0.887 g/L.
Therefore, the weight of available Cl in a liter of the unknown solution is 0.887 grams.
It's important to double-check your calculations and make sure you've followed the correct steps, including converting between units and using the correct equation stoichiometry.