# Difficult AP Chemistry Question (Titration + Calculating Moles)

1. Nov 22, 2006

### twotaileddemon

Okay, so I was browsing through an AP chemistry book because I really want to major in either chemistry of physics and want to take advantage over thanksgiving break to get ahead on chemistry because I really love it ^^. So I was wondering if someone could check my work on this problem. Thanks ^^!!

I. 2(Mn2)+ + 4(OH-) + (O2) -> 2(MnO2) + 2(H2O)
II. (MnO2) + 2(I-) + 4(H+) -> Mn(2+) + (I2) + 2(H2O)
III. 2((S2O3)2-)) + (I2) -> ((S4O6)2-)) + 2(I-)

The amount of oxygen dissolved in water can be determined by titration. First, (MnSO4) and NaOH are added to a sample of water to convert all of the dissolved oxygen to (MnO2), as shown in equation I above. Then (H2SO4) and KI are added and the reaction represented by equation II proceeds. Finally, the (I2) that is formed is titrated with standard sodium thiosulfate (Na2S2O3), according to equation III.

1) A student found that a 50 mL sample of water required 4.86 mL of .0112 M (Na2S2O3) to reach the equivalence point. Calculate the number of moles of oxygen dissolved in this sample.

After using Hess' Law, I got
4(OH-) + (O2) + 8(H+) + 4((S2O3)2-)) -> 6(H2O) + 2((S4O6)2-)
Anyway.. I did this
4.86 mL x (11.2 mol/mL) = 54.4 mol (Na2S2O3)
Then I converted that to moles of oxygen by dividing by four (what I got from Hess' law)
So 13.6 moles (O2)
Then I divided by .05L, the amount of water we are given.
Resulting in 272.2 mol (O2/L)

My question here is, did I do the set up correctly? There were more L/mL units than I'm used to, so I got confused with how to cancel them or set them up in equations. If not, any advice would be very helpful and appreciated ^^.

2) How would the results above be affected if some (I2) were lost before the ((S2O3)2-) was added? Explain.

I said the results would yield more moles of oxygen because you would need more of everything else to compensate for the lost. Though, I'm not really sure because when I did Hess' law iodine was gone and canceled out.

3) What volume of dry (O2) measured at 25*C and 1 atm of pressure would have to be dissolved in 1 L of pure water in order to prepare a solution of the same concentration that was obtained in (1)?

Well, if I did 1 wrong, then I'm going to do this wrong.. but here's my dilemma. I am continually confused with all the liter information I keep getting.. But here I go..
The density of water is 1 g/L, and you have 1 L of water, therefore you have 1 g of water or 1/18 moles of water.
(1/18) x (1 mol 02 / 6mol H20) = 1/108 mol 02

Now.. the mol's of oxygen in part I were quite large.. and this is very small.. leading me to believe I made some huge error. So I'm requesting any guidance if anyone is willing to help. Thanks again ^^.. even just for reading this!

2. Nov 23, 2006

### siddharth

Whenever I do problems involving redox reactions, I use equivalent weights and normality. The calculations are much simpler that way.

So for your question, the millieq of O2 which reacted will be equal to the milli eq of thiosulfate consumed.

Last edited: Nov 23, 2006
3. Nov 23, 2006

### twotaileddemon

Hm.. I'm still a little confused.
The amount of oxygen has to equal the amount of sodium thiosolfate consumed?
(I'm going to be without computer access for 3 days so sorry if I respond late)

4. Nov 23, 2006

### GCT

You should use the ideal gas law for the third part.

5. Nov 24, 2006

### konartist

There is no way that it is 54.4 mol. You need to convert ML to L before you do the calculations. Therefore, .00486L x (11.2/1L) = .0544 Molarity.

6. Nov 25, 2006

### siddharth

No! The amount of oxygen will NOT be equal to the amount of thio consumed. Are you familiar with the concept of equivalent mass?

For redox reactions, the equivalent weight of an oxidizing or reducing agent will be the molecular weight divided by the net change in oxidation number.

For example, for $$MnO_4^{2-} \rightarrow Mn^{2+}$$, the equivalent weight of $$MnO_4^{2-}$$ will be (molecular weight)/5

If a certain mass of the substance is present, the number of equivalents will be (mass)/(equivalent weight).

The advantage of calculating this is that, for a series of redox reactions, the number of equivalents consumed or generated will be equal for different species. This does not mean that the amount of species consumed is same, but only the number of equivalents.

This is all probably very confusing, so I'll work out an example. Let's say we need to find what volume of a 0.2 M solution of $$MnO_4^{2-}$$ reacts with 50ml of a 0.1M solution of $$C_2O_4^{2-}$$

So the reactions are
$$MnO_4^{2-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O$$
$$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}$$

Notice that the equivalent weight of $$MnO_4^{2-}$$ in the first reaction will be (molecular weight/5), since the change in oxidation number is 5

Similarly, the equivalent weight of $$C_2O_4^{2-}$$ in the second reaction is (molecular weight/2)

You can calculate the initial mass of $$C_2O_4^{2-}$$ present, which will be (50)(0.1)(Mwt) milligrams (Do you see why?). Since all this reacts, the number of equivalents which react will be (50)(0.1)(Mwt)/(Equivalent weight), which is (50)(0.1)(2) = 10 milli eqs.

Now here's where the advantage of using equivalents come. Since the number of equivalents of reacting species is the same, the number of equivalents of $$MnO_4^{2-}$$ which is consumed is also 10.

If x ml of $$MnO_4^{2-}$$ reacts, then the mass which reacts is (x)(0.2)(mwt). The number of equivalents which react can be calculated by dividing by the equivalent weight (which in this case is mwt/5).

This gives 10 = (x)(0.2)(5), and 10 ml of the 0.2 M solution of $$MnO_4^{2-}$$ solution reacts. If you have a series of redox reactions, you can proceed easily in a similar fashion.

Now, can you apply this concept to solve your original question?

Last edited: Nov 25, 2006
7. Nov 29, 2006

### twotaileddemon

I got the answers ^^. Thanks