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# chemistry A-level impossible question????? watch

1. Is this question impossible or what- with the amount of info provided??

An excess of finely divided tin and lead was shaken with a solution containing 1mol/dm^3 lead(II) ions and 0.50mol/dm^3 tin(II) ions. Calculate the equilibrium concentrations of lead(II) and tin(II) ions: given Kc as 2.2 at 25 degrees Celsius.

Sn (s) + Pb{2+} (aq) = Pb (s) + Sn{2+} (aq)

2. Why do you think it is impossible?
3. you know the concentration of lead and tin before. You know KcGiven the stoichiometry you also know that the concentration of lead and tin will be the same in the products.So find the product total concentration and square root it.
4. (Original post by Pigster)
Why do you think it is impossible?
Because they havnt provided the starting concentration or anything.

Adding Sn the equilibrium will shift on the right
at equilibrium [Sn2+]= 0.50+x and [Pb2+]= 1.0-x
After this bit its simple algebra but how did they get that^^
5. (Original post by Abbas khoja)
they havnt provided the starting concentration
What concentration haven't they provided?

Also (and I suspect this might be the problem), what do you think the expression for Kc for this equilibrium is?
6. (Original post by Pigster)
What concentration haven't they provided?

Also (and I suspect this might be the problem), what do you think the expression for Kc for this equilibrium is?
Okay so kc = Sn {2+} / Pb {2+}
Starting concentrations: Pb = 1M
Sn = 0 ( as it is product and reaction hasnt occured yet)
Equilibrium concentrations- need to find

Problem I am having
Where does Sn = 0.5M come into this
7. (Original post by Abbas khoja)
and 0.50mol/dm^3 tin(II) ions.
8. (Original post by Pigster)
I know that tin ion conc is 0.5 but is that starting conc? How can it be if tin ion is on the product side. And it cant be eq.conc because that is what the questions is asking for. If you dont want to help go away lol
9. (Original post by Pigster)
Not to mention the question.
10. (Original post by TutorsChemistry)
Not to mention the question.
Why is everyone on here so mean haha shits gone to your heads
11. (Original post by Abbas khoja)
Okay so kc = Sn {2+} / Pb {2+}
Starting concentrations: Pb = 1M
Sn = 0 ( as it is product and reaction hasnt occured yet)
Equilibrium concentrations- need to find

Problem I am having
Where does Sn = 0.5M come into this
My advise is have another think about the equilibrium constant. Why have you omitted the Pb and Sn terms, and only included the ions?
12. (Original post by Abbas khoja)
Why is everyone on here so mean haha shits gone to your heads
Examiners and markers are just as bad, I'm afraid.
13. (Original post by Abbas khoja)
Is this question impossible or what- with the amount of info provided??

An excess of finely divided tin and lead was shaken with a solution containing 1mol/dm^3 lead(II) ions and 0.50mol/dm^3 tin(II) ions. Calculate the equilibrium concentrations of lead(II) and tin(II) ions: given Kc as 2.2 at 25 degrees Celsius.

Sn (s) + Pb{2+} (aq) = Pb (s) + Sn{2+} (aq)

The question gives you the initial concentrations of lead and tin ions. Then in the reaction, tin ions are the product, but there were already some tin ions present (and lead ions present too) at the beginning.

Suppose x moles/dm^3 of lead ions react to form x moles/dm^3 of tin ions. It follows that the final concentrations will be initial - x for lead ions and initial + x for tin ions (and initial conc. for tin ions is 0.5mol/dm^3).

14. (Original post by K-Man_PhysCheM)
The question gives you the initial concentrations of lead and tin ions. Then in the reaction, tin ions are the product, but there were already some tin ions present (and lead ions present too) at the beginning.

Suppose x moles/dm^3 of lead ions react to form x moles/dm^3 of tin ions. It follows that the final concentrations will be initial - x for lead ions and initial + x for tin ions (and initial conc. for tin ions is 0.5mol/dm^3).

Thank you! I get it now 😀
I didnt realise some product can be present initially.
Appreciate it 👍
15. (Original post by Abbas khoja)
Starting concentrations: Pb = 1M
Sn = 0
(Original post by Abbas khoja)
I know that tin ion conc is 0.5 but is that starting conc?
Why did you accept that the starting conc of Pb2+ was 1M but questioned that Sn2+ was 0.5?

(Original post by Abbas khoja)
If you dont want to help go away lol
I, like many here, believe that simply giving people answers is of little help to them. It is far better for you to understand where you are having difficulties and how to solve problems if you come across something similar in the future.
16. (Original post by Abbas khoja)
I didnt realise some product can be present initially.
That'll be the answer to my question from moments ago.

When completing an ICE table, the values for the products doesn't have to be 0.
17. (Original post by TutorsChemistry)
My advise is have another think about the equilibrium constant. Why have you omitted the Pb and Sn terms, and only included the ions?
Pure substances such as solids can not have a "concentration" and do not appear in the equilibrium expression.
18. (Original post by Abbas khoja)
Is this question impossible or what- with the amount of info provided??

An excess of finely divided tin and lead was shaken with a solution containing 1mol/dm^3 lead(II) ions and 0.50mol/dm^3 tin(II) ions. Calculate the equilibrium concentrations of lead(II) and tin(II) ions: given Kc as 2.2 at 25 degrees Celsius.

Sn (s) + Pb{2+} (aq) = Pb (s) + Sn{2+} (aq)

what examboard is this from?
19. (Original post by charco)
Pure substances such as solids can not have a "concentration" and do not appear in the equilibrium expression.
While this makes sense, for A-level we've been taught that solids/solvents are not included in the equilibrium expression because they are in massive excess and so their concentration remains essentially constant. The effect of the solid/solvent in excess is thus included in the equilibrium constant.

A solid can have a concentration: number of moles of atoms per unit volume. But this would be a constant anyway, assuming the solid doesn't change its packing pattern during the reaction. Though I suppose this is usually described by density for solids and pure liquids...
20. (Original post by LLearnt)
what examboard is this from?
Its from jim clarks calculations book.

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