March 28, 2024, 08:26:04 AM
Forum Rules: Read This Before Posting


Topic: Spectator ions are haunting me!! Please help  (Read 15537 times)

0 Members and 1 Guest are viewing this topic.

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Spectator ions are haunting me!! Please help
« on: April 02, 2008, 02:49:34 AM »
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!!



Offline Rabn

  • Chemist
  • Full Member
  • *
  • Posts: 284
  • Mole Snacks: +28/-13
Re: Spectator ions are haunting me!! Please help
« Reply #1 on: April 02, 2008, 03:04:10 AM »
Question 1)  What are the oxidation states of Iron and Manganese on both sides of the equation?

Q 2)  Are both of the reactants a strong acid/base?

***Some general problem solving rules for chemistry

1) Always write a balanced reaction
2) Always convert mass to moles
3) Always write down the dimensions of the answer, i.e. moles/liter, atm etc... this helps you determine what kind of equation you may need
4) Always use dimensional analysis, it is imperative to get in the habit of writing down the units.  Don't make assumptions when doing math, it makes an A** out of you and you***

Q 3) definitely double check your math...did you do it in your head?

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Re: Spectator ions are haunting me!! Please help
« Reply #2 on: April 02, 2008, 03:09:07 AM »
Question 1)  What are the oxidation states of Iron and Manganese on both sides of the equation?

Q 2)  Are both of the reactants a strong acid/base?

***Some general problem solving rules for chemistry

1) Always write a balanced reaction
2) Always convert mass to moles
3) Always write down the dimensions of the answer, i.e. moles/liter, atm etc... this helps you determine what kind of equation you may need
4) Always use dimensional analysis, it is imperative to get in the habit of writing down the units.  Don't make assumptions when doing math, it makes an A** out of you and you***

Q 3) definitely double check your math...did you do it in your head?


Thankyou for taking your time to help me wth my question :). The states are aqueous. I don't think it matters if they are strong or weak for the 2nd one.  Question 3 is right. The answer is in Litres. Its in the back of the book.

Offline Astrokel

  • Full Member
  • ****
  • Posts: 989
  • Mole Snacks: +65/-10
  • Gender: Male
Re: Spectator ions are haunting me!! Please help
« Reply #3 on: April 02, 2008, 07:36:39 AM »
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

Fe2+ + SO4 2- = FeSO4
so 1 mole of Fe2+ reacts with 1 mole of SO42- to form 1 mole of FeSO4

So when u find concentration of Fe2+, you use moles of Fe2+ divide by the volume which is 20 x 10-3 dm3.

Since the stiochiometry ratio of Fe2+ : FeSO4 is 1:1, so [Fe2+] = [FeSO4]


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

If you write out the equation correctly, you would notice 2 moles of HNO3 reacts with 1 mole of Ba(OH)2

After calculating the moles of the reagents, you would probably notice x moles of HNO3 is in excess

With the same concept as qn 1, 1 mole of H+ reacts with 1 mole of NO3- to form 1 mole of HNO3.

You can continue from here...

Yes if u use a diprotic acid such as H2SO4, you will have to times 2 to get moles of H+ ions since

2H+ + SO42-  --> H2SO4

H20 doesn't exist as ions so its impossible to calculate the moles of H+ in H20


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!!



There are 2 cases in this qn, K+ in KOH(in excess) and K+ in KCl.

However, remember you cannot find concentration for each case and sum the concentrations up.

I don't understand your book explanation, sorry.


Kelvin
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Re: Spectator ions are haunting me!! Please help
« Reply #4 on: April 02, 2008, 07:57:03 AM »
Thanks a lot Kelvin  :) :) :) :) :) your answers really helped. I totally get Q1 now. (The only slight problem I have is why did they give the equation like that instead of the normal equation). I get question 2 as well now. So with the excess HNO3 why does it break up and be H+ and NO3- in the solution and why does the H20 stay the same.

Thanks once again for your help. Really appreciate it ;)

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Re: Spectator ions are haunting me!! Please help
« Reply #5 on: April 02, 2008, 08:01:31 AM »
OH just realized you could give points to people in this forum ;D. Points for Kelvin and Rabn

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27637
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Spectator ions are haunting me!! Please help
« Reply #6 on: April 02, 2008, 08:11:29 AM »
I totally get Q1 now. (The only slight problem I have is why did they give the equation like that instead of the normal equation).

Because that's the way it goes. You could use sodium permanganate, lithium permanganate, calcium permanganate (if soluble), and perchloric acid instead of sulfuric. This gives us already 8 separate reaction equations if I am counting well. They gave one - and it describes what is happening in each separate case. Spectators are just spectators, often they only confuse.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Re: Spectator ions are haunting me!! Please help
« Reply #7 on: April 02, 2008, 08:18:02 AM »
Thank you Borek for the help :). Point for you as well. Do you have any tips on how to recognize specatator ions. Yeah I have seen the common explanation that spectator ions don't do anything in the reaction but don't they affect the mole ratios.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27637
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Spectator ions are haunting me!! Please help
« Reply #8 on: April 02, 2008, 08:26:55 AM »
don't they affect the mole ratios

Try to find an example where they will :) They may change numbers, but ratios will be the same.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Re: Spectator ions are haunting me!! Please help
« Reply #9 on: April 02, 2008, 08:30:11 AM »
don't they affect the mole ratios

Try to find an example where they will :) They may change numbers, but ratios will be the same.

Yeah :D you are right I haven't see a one like that. Anothe small question. They say that spectator ions remain unchanged through out the reaction. So if they change numbers how can they be spectator ions.

Offline Astrokel

  • Full Member
  • ****
  • Posts: 989
  • Mole Snacks: +65/-10
  • Gender: Male
Re: Spectator ions are haunting me!! Please help
« Reply #10 on: April 02, 2008, 08:46:44 AM »
Do you have any tips on how to recognize specatator ions.

1) Let's take qn 3 as an example

First write out the equation

HCl (aq) + KOH (aq) ---> KCl (aq) + H20 (l)

2) Break the components of substances in (aq) to its ions form

H+(aq) + Cl-(aq) + K+(aq) + OH- (aq) --> K+(aq) + Cl-(aq) + H20(l)

3) Cancel out the common term in each side, where the common terms are all spectator ions. In this case, the spectator ions are K+ and Cl-

What is left behind shows you what really happens during the reaction

H+(aq) + OH- (aq) ---> H20(l) (Hence, sometimes neutralisation is written in this form, because these are the ions that really participate or involve in the reaction)

well you will be able to recognize spectator ions when you do more qns !

Good Luck

Kelvin ;D
No matters what results are waiting for us, it's nothing but the DESTINY!!!!!!!!!!!!

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27637
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Spectator ions are haunting me!! Please help
« Reply #11 on: April 02, 2008, 09:28:14 AM »
Let's see. You precipitate barium sulfate from the BaCl2 solution. You add sulfuric acid:

BaCl2 + H2SO4 -> BaSO4 + 2HCl

Now the same, but with aluminum sulfate instead of sulfuric acid:

3BaCl2 + Al2(SO4)3 -> 3BaSO4 + 2AlCl3

Net ionic is in both case identical:

Ba2+ + SO42- -> BaSO4

Barium always react 1:1 with SO42-. Stoichiometric coefficients are different in full equations (1:1 in the first, 3:1 in the second), but ratio of ions that react is always the same. Now I see I wasn't too precise when writing my previous post.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Re: Spectator ions are haunting me!! Please help
« Reply #12 on: April 02, 2008, 07:13:45 PM »
Thanks a lot Astrokel and Borek for your help once again :). Good point Borek I really understood your explanation. Astrokel your explanation was very good too and I have few small questions to ask from it


2) Break the components of substances in (aq) to its ions form

H+(aq) + Cl-(aq) + K+(aq) + OH- (aq) --> K+(aq) + Cl-(aq) + H20(l)

Why is the reason we break the aqueous product to ions and not the solid, liquid or gas. Is it because aqeuous means dissolved in water and ions are dissolved in water. So does that mean the product KCL is actually in K+ and Cl- form. Also you have to balance the equation at this point right.


What is left behind shows you what really happens during the reaction


But don't the spectator ions make KCL why isn't that important. Why do we consider making H20 as what really happens during the reaction

Last one ;) when you get a question like that and let's say I found the Fe+ ion moles and they ask for the moles of Fe2SO4 (just made this non existing compound up) then do I times by half to get the moles of Fe2SO4

Thanks once again for your help guys. I'm in the brink of getting this :)

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27637
  • Mole Snacks: +1799/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Spectator ions are haunting me!! Please help
« Reply #13 on: April 03, 2008, 02:54:51 AM »
Why is the reason we break the aqueous product to ions and not the solid, liquid or gas. Is it because aqeuous means dissolved in water and ions are dissolved in water. So does that mean the product KCL is actually in K+ and Cl- form.

Exactly. Almost all ionic salts, strong bases and strong acids are fully dissociated in water.

Quote
Last one ;) when you get a question like that and let's say I found the Fe+ ion moles and they ask for the moles of Fe2SO4 (just made this non existing compound up) then do I times by half to get the moles of Fe2SO4

Yes. You may divide by two as well ;)
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline sameeralord

  • Regular Member
  • ***
  • Posts: 90
  • Mole Snacks: +12/-4
Re: Spectator ions are haunting me!! Please help
« Reply #14 on: April 03, 2008, 03:18:47 AM »
Thank you Borek once again   :). I think my problem is almost solved now. So in a reaction  if you get an aqeuous product like KCL. Does that mean actually the product is in K+ and Cl- form and they have written KCL instead of K+ and Cl- to make it easier.

Sponsored Links