titration + balancing redux

[wanabe]

Question #1

The concentration of Mn^2+(aq) can be determined by titration with MnO_4^-(aq) in basic solution. A 22.50 mL sample of Mn^2+(aq) requires 38.28 mL of 4.026×10−2 M KMnO_4(aq) for its titration.

Mn^2+ + MnO_4^- > MnO_2(s) (not balanced)...I think the balanced eqn simply needs "2MnO_2" on the product side...My set up is this...38.28mL KMnO_4(1L/1000mL)(4.026*10^-2/1L)(1molMn^2+/1mol KMnO_4)(1/22.50mL)(1000mL)=0.0698495868 M Mn^2+...I've also tried a 1/2 mole ratio didn't work either...

 

Question #2

The titration of 4.90 mL of a saturated solution of sodium oxalate, Na_2C_2O_4, at 25 deg C requires 28.0 mL of 2.100×10−2 M KMnO_4 in acidic solution. What mass of Na_2C_2O_4 in grams would be present in 1.00L of this saturated solution?...Na_2C_2O_4+KMnO_4> Na_2MnO_4+KC_2O_4(my attempt at balancing eqn)...28.0ml KMnO_4(1L/1000mL)(2.100*10^-2/1L)(1molNa_2C_2O_4/1molKMnO_4)(133.9988g/1mol)=0.0787912944g Na_2C_2O_4

 

Question #3

A solution is to be standardized by titration KMnO_4(aq) against As_2O_3(s). A 0.1012 g sample of As_2O_3 requires 23.60 mL of the KMnO_4(aq) for its titration.

5As_2O_3 + 4MnO_4^- + 9H_2O + 12H^+ > 10H_3AsO_4 + 4Mn^2+. (how nice it's balanced)...What is the molarity of the KMnO_4}(aq)?.....I tried....0.1012As_2O_3(1mol/197.8415)(4mol MnO_4^-/5mol As_2O_3)(1/23.60mL)(1000mL/1L)=0.01733968088M(I'm lost on this one, I'm 99%sure its not right).

 

Thanks in advance.

[cypress]

Question #1

The concentration of Mn^2+(aq) can be determined by titration with MnO_4^-(aq) in basic solution. A 22.50 mL sample of Mn^2+(aq) requires 38.28 mL of 4.026×10−2 M KMnO_4(aq) for its titration.

Mn^2+ + MnO_4^- > MnO_2(s) (not balanced)...I think the balanced eqn simply needs "2MnO_2" on the product side...My set up is this...38.28mL KMnO_4(1L/1000mL)(4.026*10^-2/1L)(1molMn^2+/1mol KMnO_4)(1/22.50mL)(1000mL)=0.0698495868 M Mn^2+...I've also tried a 1/2 mole ratio didn't work either...

 

No, neither of those ratios will work because those ratios don't lead to balanced redox reactions. You have to have balanced equation both from a charge perspective and molecular perspective or you are not going to get this right. The one to one ratio leaves you one electron short.... Use that as a hint to try to figure out what other ion in the solution likely participates. Use also the description of the given solution. Once you have that figured, next use the electron shortage to figure what product removal of an electron from that ion is going to produce. Then try to balance the equation.

 

 

We will work the others next.

[wanabe]

Do I have it balanced now (The most bottom formula)? ok never mind... I'm lost as to how to balance this. That was my attempt, I know there are supposed to be H^+ ions on one side and H_2O on another but it's just going to mess up the eqn as far as I can tell...

[cypress]

It was a fair attempt, but H+ is not going to give you a source of electrons and it is electrons that you are short. Note also that the problem statement mentions that the solution is basic. So if you work it with that adjustment you might get it. You're on the right basic track.

 

Mn⁺² + MnO4^-1 +1e -> 2MnO2

 

but you need a source of one electron.... so you need to keep going....

 

Just line up the additional equation that will give you an electron and see what else you need to now balance.

[wanabe]

2H⁺ +Mn²⁺ +2MnO₄^-+e^- > 3MnO₂ + 2OH^-

 

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Is there any way we could meet on the chat?

Edited October 10, 2010 by wanabe

[cypress]

We could try chat. Ill stick around for a while.

 

You're just guessing now... These work best if you reason through them. From the description the solution is basic so you won't have any H+.

 

OH= alone is not going oxidize and generate the needed electron for the reduction reaction....

 

Let's start from the beginning.

 

All of these problems are redox reactions which can be written as two part equations. The first step is to identify the compound or ion that is the oxidizer and which is the reducer. The oxidizer is, like oxygen, going to accept electrons and the reducer is going to give up electrons.

 

Oxidizers therefore with be metal ions with large positive charges or ions with elements that easily accept additional electrons.

 

With this you can find the oxidizer and the reducing agent.

[wanabe]

I'm reduced to guessing... How can the one product be two things? That's my trouble with this.

[cypress]

Ah but it can because it is the product of both the oxidation reaction and the reduction reaction... that is what is throwing you off

 

If we write them out separate it may make it more clear

 

So the oxidizer is Mn⁷⁺O4{-} and it is reduced to Mn⁺⁴O2, right?

Edited October 10, 2010 by cypress

[wanabe]

Yes, but also it's oxidized?... chat?

[cypress]

and the reducing agent is Mn⁺²{++} and it is oxidized to Mn⁺⁴O2 ..

 

Im on chat but Im not sure I know how to use it...

[wanabe]

um somehow yes... is this about molecules or the elements(ions) that make them up can i simply always balance the elements(ions) and be correct?

 

 

 

after much strife:

the balanced eqn is 3Mn{2+}+ 4OH{-}+2MnO4{-} >5MnO2+2H2O

 

which means that the above mole ratio in the unit line eqn is 3:2 so carrying out the calculations one gets "0.10274 M"

 

yay!

Edited October 10, 2010 by wanabe

[cypress]

Nice! Don't forget about significant figures. Only four...