I had to perform an experiment for rate of reaction.

The point of the experiment was to show how changes in reactant concentration, temperature, and catalyst presence can affect the rate of a reaction. The reaction that we studied was
H2O2 + 2I^- + 2H^+ -> I2 + 2H2O

We had to set up 6 flasks each with and certain amount of KI, starch, DI water, buffer, acid, Na2S2O3, and H2O2 all in mL. Then we had to measure the time it took for a reaction to happen which was indictated when the solution began to turn blue. For Na2S2O3 each flask had 2.50mL in it which is a total of 15.

I have to calculate the concentration of Na2S2O3 in the reaction solutions based on the volume used in each reaction flask and the total volume calculated.

I hope I explained this better this time.

I did this same experiment a couple weeks ago. We are measuring the amount of time it takes for the thiosulfate to run out.

So to calculate the concentration of the thiosulfate ion. You can multiply the milliliters of thiosulfate in whatever trail you are calculating it for by the concentration of the thiosulfate divided by the total volume in milliliters of all the KI, dionizedwater, buffer etc. Which mine was 150ml but yours is probably different. That should give you your concentration. You could use the same process to find concentration of hydrogen peroxide and the others if needed in your data calc.
Does that help any?

Hey I'm pretty sure I'm in the same lab as Hannah and had the same question.

In class we were told that we could calculate concentrations using the formula:

Concentration 1 x Vol. 1 = Concentration 2 x Vol. 2

Is that essentially what you said in the answer?

And to Hannah, given that the volume 1 was the same for all six flasks (2.50mL) and the total volume was also the same for all six flasks (75mL), I'm thinking the concentration of thiosulfate will be the same for all reactions…

Yes, exactly! Since we are diluting we can use M1V1=M2V2.

So Concentration of thiosulfate X Amt.(volume) of added thiosulfate = (75ml)total volume X the Concentration of the new concentration of thiosulfate in the flask(what you are trying to find)

The thiosulfate concentration wouldn't be the sane in all rxns I'd you added different milliliters to each flask. If you added same any. Of milliliters of thiosulfate in each trail you did then yes, it would be the same concentration for all.

I just wrote mine in terms of the concentration of the thiosulfate. Pretty much I isolated the variable I was trying to find. :) same thing though!

Yes I think that helps me. Im not very good at this stuff so let me make sure I understand this.

The concentration of the thiosulfate used was 0.10M and for each flask we used 2.50mL. So I have to multiply these two (2.50mL)(0.10M) and then divide by the total amount combined for all the other solutions used, is that correct?

Yes.

To calculate the concentration of Na2S2O3 in the reaction solutions, you need to determine the amount of Na2S2O3 in moles and divide it by the total volume of the reaction solution.

Step 1: Convert the volume of Na2S2O3 used in each flask to moles.

Since each flask contains 2.50 mL of Na2S2O3, and there are 6 flasks, the total volume of Na2S2O3 used is 2.50 mL/flask x 6 flasks = 15 mL.

Step 2: Convert the volume of Na2S2O3 to moles.

To do this, you need to know the molar concentration (also known as molarity) of Na2S2O3, expressed in moles per liter (mol/L).

Let's assume the molar concentration of Na2S2O3 in your experiment was 0.1 mol/L.

Therefore, the amount of Na2S2O3 in moles can be calculated using the formula:

moles = concentration (mol/L) x volume (L)

moles = 0.1 mol/L x 0.015 L (since 15 mL is equal to 0.015 L)

moles = 0.0015 mol of Na2S2O3

Step 3: Calculate the concentration of Na2S2O3 in the reaction solutions.

To determine the concentration, divide the moles of Na2S2O3 by the total volume of the reaction solution.

Let's assume the total volume of the reaction solution is 250 mL (0.250 L).

concentration = moles/volume

concentration = 0.0015 mol/0.250 L

concentration = 0.006 mol/L or 6 mmol/L (millimoles per liter)

Therefore, the concentration of Na2S2O3 in the reaction solutions is 6 mmol/L.