tirations ...help

[fashafosho]

One of the study questions deals with titration of two compounds....(sorry in advance if this has already been asked before)


This is from the Kaplan Study Tests for Physics (test 3)


We are given-

2 g of KI (MW= 166) is added to 50 mL of 0.01 M solution of KIO3.
The solution is titrated with 33 mL of unstandardized sodium thiosulfate solution (MW= 158). What is the molarity of thiosulfate solution?

Also given:
Iodide standardizes the thiosulfate solution. Aqueous solution of iodine is prepared by adding an excess of potassium iodide to a known volume of an acidic solution of potassium iodate:
IO3- + 5 I- + 6H+ --> 3 I2 + 3 H2O (Reaction 1)

The iodine solution resulting from Reaction 1 is titrated with the thiosulfate solution that is to be standardized. The two half-reactions that occur are:

2 S2O3(2-) --> S4O6(2-) + 2e E = -0.09

I2 + 2e ---> 2 I- E= 0.54


I know the basic formula here is molsX = molsY @ equivalence which becomes: (molarity)(volume) of X = (molarity)(volume)of Y

and so

(molarity)(volume) Iodine / volume = molarity of thiosulfate solution

but the answer choice incorporates the number of mols from the other reactions into the answers...I don't know how that works.

The answer is:
(0.01)(0.05)(6)/ (0.0032)= molarity of thiosulfate solution

the 6 comes from 3x2 (3 Iodines, each produce 2 thiosulfates= total of 6 thiosulfates). But, why are we including the 6 here?


Any help is appreciated-thanks in advance.

--------------(The reason why I had originally omitted all those details earlier was that the passage itself is very convoluted and I just wanted to know why the answer choice includes "6" in the calculation)---------

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[RSAgator]

wtf us w?

 

[Vanguard23]

What is W?
Also, the question tells us nothing except that A was titrated with B. What past that? Did it reach an equivalence or half-equivalence point? Is that what they were looking for?

 

[BloodySurgeon]

Please clarify if A, X, Y, and W are acids/bases/salts... or what not. Also if this is a buffer problem, which one is the weak acid and which is the conjugate base or the likes.

 

[fashafosho]

Sorry for the confusion, I have edited the original post.

 

[Kaustikos]

One of the study questions deals with titration of two compounds....(sorry in advance if this has already been asked before)


This is from the Kaplan Study Tests for Physics (test 3)


We are given-

2 g of KI (MW= 166) is added to 50 mL of 0.01 M solution of KIO3.
The solution is titrated with 33 mL of unstandardized sodium thiosulfate solution (MW= 158). What is the molarity of thiosulfate solution?

Also given:
Iodide standardizes the thiosulfate solution. Aqueous solution of iodine is prepared by adding an excess of potassium iodide to a known volume of an acidic solution of potassium iodate:
IO3- + 5 I- + 6H+ --> 3 I2 + 3 H2O (Reaction 1)

The iodine solution resulting from Reaction 1 is titrated with the thiosulfate solution that is to be standardized. The two half-reactions that occur are:

2 S2O3(2-) --> S4O6(2-) + 2e E = -0.09

I2 + 2e ---> 2 I- E= 0.54


I know the basic formula here is molsX = molsY @ equivalence which becomes: (molarity)(volume) of X = (molarity)(volume)of Y

and so

(molarity)(volume) Iodine / volume = molarity of thiosulfate solution

but the answer choice incorporates the number of mols from the other reactions into the answers...I don't know how that works.

The answer is:
(0.01)(0.05)(6)/ (0.0032)= molarity of thiosulfate solution

the 6 comes from 3x2 (3 Iodines, each produce 2 thiosulfates= total of 6 thiosulfates). But, why are we including the 6 here?


Any help is appreciated-thanks in advance.

--------------(The reason why I had originally omitted all those details earlier was that the passage itself is very convoluted and I just wanted to know why the answer choice includes "6" in the calculation)---------

Wow. That last part almost ****s with you if you think that the E has anything to do with it.

Anyways, this is a limiting reagant kind of problem if you look at it. You have add two original stocks to make a product that you want to titrate with the thiosulfate (the I2). So, you take this:
IO3- + 5 I- + 6H+ --> 3 I2 + 3 H2O

and figure out which is limiting in the production of I2 (it's the original IO3-) which only has 0.0005 moles, whereas you add 0.012 moles of I-. So, you're basically only reacting 0.0005 moles of IO3- with 0.012x5 moles I- (this is neglible, since you don't care how much of that you have for the product).
0.0005 moles makes 3 moles of I2. So, you're making essentially 0.0015 moles of I2 in this reaction.

I think that takes care of what you're titrating. You have a 0.0015 mole in 0.050 L solution of I2 and you're titrating it with thiosulfate, and you KNOW it's 0.0.33 liter of that. Use this:

2 S2O3(2-) --> S4O6(2-)
I2 + ---> 2 I-

And you know that you make 2 moles of thiosulfate for every 1 mole of I2.

0.0015 x 2 = 0.003 moles of thiosulfate used. Divide that by your solution to get the molarity (1 I believe).

The bolded numbers are where you get your 6 from.