Spectator ions are haunting me!! Please help

[scilearner]

Hello guys ,

 

I'm sitting for the chemistry exam this year and I have started studying for it. The spectator ions have being bothering me a lot recently. In the books I'm doing they neglect them sometimes and write ionic equations and do all sorts of stuff with it. I know specatator ions aren't involved in the reaction but I get confused with them sometimes.I'll post some multiple choice problems here to make it clear where my difficulties lie. The answers are also provided. So all I really want is to understand how to get the answer.

 

Problem 1

 

A standard solution of pottassium permenganate (KMn04) has a concentration of 0.0240 M. It is titrated against a solution of iron (II) sulfate. The equation fo the reaction is. Sorry I don't know how to put numbers underneath.

 

5Fe2+ + Mn04- + 8H+ = 5Fe3+ + Mn2+ + 4H20

 

15.60 ml of the KMn04 solution reacts exactly with 20 ml of the FeSO4 solution. The concentration of the FeSO4 solution, in M, is

 

The answer in the back of the book is 0.0936

 

My problem- K and S04 are not even in the equation of the question. I think it is because they are spectator ions. Then the only concentration I can find is the concentration of Fe2+. How is this the same as the concentration of FeS04

 

Problem 2

 

50 ml of 0.020 M solution of Ba(OH)2 is added to 50 ml of a 0.060 M solution of HNO3. The hydrogen ion concentration in the resulatant solution, in mole per litre, is

 

Answer in the back of the book: 0.01 M

 

My problem: Normally I work out the excess reactant in a question like this and see the moles in excess. But when I work out if Ba(0H)2 or HNO3 is in excess. I don't work out H+ ions. Can I get the moles of H+ ions in excess because it is equal to moles of HNO3 in excess. Does that mean if the reactant is H2NO3 ( I just made this up) instead of HNO3. I times by 2 to get the H+ ions in the resulatnat solution. Another problem I have is acid and base produce H20. Why don't we add the H+ ions in the H20 to the solution

 

Problem 3

 

20.0 ml of 0.10 M HCL reacts with 20 ml of 0.30 M KOH solution. The concentration of pottassium ions in the resultant solution, in mole per litre ,is

 

The answer in the back of the book- 0.15. They are saying that K+are spectator ions, hence n(K+) does not change. However the volume changes to 40 ml hence the concentration is halved.

 

My problem: Normally I find the excess and limiting reactant in a question like this. Then I look at the moles in excess and divide that by the total volume. Why is that moles of spectator ions do not change.

 

I don't expect you to understand everything I said. I tried my best to explain. If you can help me with these questions or pass any tips when working with spectator ion questions I'll very much appreaciate it . Thanks!!

[CaptainPanic]

Regarding problem 1:

 

You are right when you say that [ce] K+ [/ce] and [ce] SO4^2^- [/ce] are not in the reaction equation because they do nothing. You could put them on both sides of the = sign, but why do that when it is ok to leave them out? Those salts are also in water, but you don't put water on both sides of the equation either. Only the water that is produced, or that reacts away is relevant

 

So you have 15.60 ml of [ce] KMnO4 [/ce] at a concentration of 0.0240 M (that is 0.0240 mol/l). This means you have 0.0003744 mol [ce] KMnO4 [/ce].

 

The reaction: [ce] 5Fe^2^+ + MnO4^- + 8H^+ = 5Fe^3^+ + Mn^2^+ + 4H2O [/ce]

 

So you know that for every mole of [ce] KMnO4 [/ce] you need 5 moles of [ce] FeSO4 [/ce]. Or, because [ce] K+ [/ce] and [ce] SO4^2^- [/ce] are irrelevant, we can say: for every mole of [ce] MnO4^- [/ce], you need 5 moles of [ce] Fe^2^+ [/ce].

 

So, you need 5 x 0.0003744 = 0.001872 moles of [ce] Fe^2^+ [/ce].

 

So, now you know how many [ce] Fe^2^+ [/ce] you need. You also know that for each ion of iron (each [ce] Fe^2^+ [/ce] ) you also need one [ce] SO4^2^- [/ce]. So, you need 0.001872 moles of [ce] FeSO4 [/ce]!

 

The concentration of the solution is unknown, but we know that 20 ml of it is enough. So, the 20 ml contain the 0.001872 moles. This means that the concentration is 0.0936 mol/l.

 

About "moles"

I have seen this before: people don't know what a "mole" actually is. It's just a number! A mole is 6.022^23. It's like a "dozen". Just a strange word for a number. Only "mole" is a very large number, because we count molecules, or ions, or atoms with it, and they're very small.

 

 

You see: the concentration of [ce] Fe^2^+ [/ce] is the same as the concentration of [ce] FeSO4 [/ce] because we count the ions ([ce] Fe^2^+ [/ce]) or in the case of [ce] FeSO4 [/ce] , we count salts [ce] FeSO4 [/ce] .

 

If you drop 12 eggs in a bath, and you also drop 12 melons in then the concentration of eggs is equal to the concentration of melons, if you count them per piece. (Obviously, the concentration in grams / liter is not identical... and this is also true for [ce] Fe^2^+ [/ce] and [ce] FeSO4 [/ce]. The concentration in mol/l is the same, the concentration in g/l is different!)

 

That was a long explanation, perhaps too much. But I had a bit of time to waste.

[scilearner]

Thanks a lot for taking your time to answer my question:-). Your explanation makes sense and now I have no problems with getting the answer to question 1. I'm still bit confused why the left the spectator ions. Can you help me further in that. Thank you once again:-)

[Fuzzwood]

Well what do you think would be easier to read: an equation polluted with ions which do nothing at all or an equation which isnt polluted with ions which do nothing at all?

 

Its really the same as with mathematics where you can cross out certain stuff against each other. In both cases, a cleaned up equation is much easier to read and you see what is actually going on a lot quicker.