Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: tickno on October 05, 2010, 09:13:27 PM
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Hey guys, i tried to do the follow answer and am about to give up...anyone got a idea?
A 239-mL sample of solution contains 2.73 mmol of Ca2+ ions. How many mmol of solid Na2SO4 must be added in order to cause precipitation of 99.9% of the calcium as CaSO4? Assume that the addition of solid Na2SO4 does not change the volume of the solution.
Ksp of CaSO4 = 6.1 x 10-5.
Do not include units in your answer.
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KspCaSO4=[ca2+]*[SO4 2-]
Can you calculate the concentration of Ca2+ when 99.9% calcium in the solution became CaSO4.
if you can ,all the SO42- in the solution comes from Na2SO4 you add.
Finally,don't forget the SO42- in CaSO4.
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KspCaSO4=[ca2+]*[SO4 2-]
Can you calculate the concentration of Ca2+ when 99.9% calcium in the solution became CaSO4.
if you can ,all the SO42- in the solution comes from Na2SO4 you add.
Finally,don't forget the SO42- in CaSO4.
Hey,
I dont understand completley what you are asking. So like i understand the concentration of the of the CA2+ and the SO4 will be the same. (i think)
im just not sure how to go about this question... im so confused.
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Wrong.The concentration of the of the CA2+ and the SO4 will NOT be the same.
Actually,there's a equilibrium:
Ca2+ + SO42- ::equil:: CaSO4
Think,how does the equilibrium change when you add SO42-?
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Can you calculate what should be the final concentration of Ca2+ once 99.9% was removed from the solution? Can you use calculated concentration to find concentration of SO42-?
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Hey guys.
after stressing for a very long time and trying for ages, this is what i have done, can someone tell me if it is correct or not.
Ca2+ + SO42- <=> CaSO4
Initial 0.0114 0
Change - x + x
Equilibrium 1.14*10^-5
6.1x10^-5 = [Ca2+][SO42-]
= (1.14x10^-5) * [SO42-]
[SO42-] = 0.05350877193
c = n /V
n = 12.79mmol
Is that how it should be done guys???
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You almost got it,but you still forgot the SO4 in CaSO4.
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Ca2+ + SO42- <=> CaSO4
Initial 0.0114 0
Change - x + x
Equilibrium 1.14*10^-5
6.1x10^-5 = [Ca2+][SO42-]
= (1.14x10^-5) * [SO42-]
Note that your equation used for calculation of [SO42-] doesn't use x. You know why? Because it is not ICE table question. It is simple solving Ksp for [SO42-] when [Ca2+] is known.
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ooosh, i dont quite understand what you mean.
Could you please sort of guide me by what you mean?
Sorry, i just need to learn this question as there will be a similar one on my exam on friday.
also borek i dont really understand also. What do you mean its just a simple Ksp.
would anyone be able to tell me the answer so i can work my way to it, or like tell me step by step how to do it.
Sorry guys, i just cant get my head around it.
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First of all - do you know what Ksp is?
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OK,I take this risk. But i really do not want to get a warning. :'(
Did you remember i told you better use the equilibrium :
Ca2+ + SO42- ::equil:: CaSO4
The concentration of Ca and SO4 ionized from CaSO4 is always the same,(remember this).
when you add SO4 the equilibrium will go right,the concentration of Ca and SO4 will both decrease,but the [SO42-] is different from [Ca2+](based on the Le Chatelier principle).
Let's do it! Step by step.
first,you can calculate the final [Ca2+]=2.73mol/0.239L * 0.1%
Then you can calculate the [SO42-] by using the Ksp
KspCaSO4=[Ca2+][SO42-]=6.1x10^-5
substituting the final [Ca2+] to Ksp equilibrium you can get the [SO42-] in the solution,
but don't forget the SO42- in CaSO4 ,which is also from Na2SO4 you add.
[SO42-] in the solution=KspCaSO4/final [Ca2+]
Total nSO42- =[SO42-] in the solution*V + nSO42- in CaSO4
can you follow the above to get the right answer?
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Hi there.
i understand everything up until the very last line.
How do I found out the nSO42- in the CaSO4, from the NaSO4, that i will add.
everything up until there makes perfect sense.
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Hum,how do you find out the nSO42- in the CaSO4?
Let me ask you a question, how many SO42- will be combined with 99.9% of 2.73 mmol of Ca2+ ions.