# Homework Help: Electrochemistry Lab

1. Feb 7, 2010

### chemnerd666

Hello all this is my problem:
1. The problem statement, all variables and given/known data
I have to determine the concentration of a sodium thiosulfate solution used to titrate a potassium iodate solution. The only variable that is known is the concentration of the potassium iodate solution, 0.01 mol/L. There is 20 mL of 0.01 M potassium iodate mixed with 25 mL of water, 2 mL of 3M H2SO4 and 3 mL of 1M KI. This is the analyte. The titration required 12.35 mL of the sodium thiosulfate solution to reach endpoint.

2. Relevant equations
here is the reaction equation for the titration (note: K is omitted because it plays no role in the reaction):IO3-(aq) + 6 S2O32-(aq) + 6 H+(aq)→ I-(aq) + 3 S4O62-(aq) +3 H2O(l)

3. The attempt at a solution
In order to determine the molarity of sodium thiosulfate, the molarity of the 50 mL potassium thiosulfate must be determined, then the amount of moles can be determined from that and using stoichiometric ratios, the moles of sodium thiosulfate can be determined and so forth... I have tried to use the equation C1V1=C2V2 but i get a weird answer:
C2= (0.01M)(1L)/(0.05L) = 0.2 M.
There is no way this can be correct as why would the concentration in a diluted solution be greater than the initial concentration?

2. Feb 7, 2010

3. Feb 7, 2010

### chemnerd666

interesting, but it has to be possible to calculate it somehow using standard equations, thanks anyways.

4. Feb 7, 2010

### Staff: Mentor

There is no such thing as "standard equations". You have to write correct reaction equation first, then use stoichiometry to solve the problem. That's the only correct approach.

No idea what you are trying to calculate with C1V1=C2V2. Where did you get 1L from?

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5. Feb 7, 2010

### chemnerd666

C1 is the initial concentration of the solution of potassium iodate, 0.01 mol/L, V1 is the initial volume of the solution, I assumed it was 1L, 20 mL of potassium iodate was diluted with 25 mL distilled water and 2 mL of 3M H2SO4 and 3mL of 1M KI were added. This would make the volume of the diluted solution 50 mL. C2 is the final concentration of the potassium iodate solution and V2 is the final volume, which I thought would be 50 mL. Now that I look at it, this equation may not really apply to this situation, I am just unsure about how to calculate the moles of potassium iodate.

6. Feb 7, 2010

### Staff: Mentor

Doesn't matter what it was mixed with - this is your amount of iodate used for titration. Now, get the correct reaction equation, calculate correct amount of iodine produced, caculate amount of thiosulphate and so on.

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methods

7. Feb 7, 2010

### chemnerd666

But can I use 0.01 mol/L to find the amount of moles in 20 mL of the potassium iodate?

8. Feb 7, 2010

9. Feb 7, 2010

### chemnerd666

Thank you for your help :)