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Topic: Please solve it completely,a tough question of Ionic equilibrium  (Read 11091 times)

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Offline rohitarura

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After solid SrCO3  was equilibrated with a pH 8.60 buffer,the solution was found to have [Sr2+]=2.2x10-4. What is the solubility product constant for SrCO3??

Many of u may be having its solution,but don't just copy it,please explain the steps also...

Thanks in advance

Offline plankk

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #1 on: April 24, 2009, 12:29:54 PM »
At first write the equation for Kso. The value of what you have to find?

The next step is to write a reaction which shows from where you have OH- - find the constant of that equilibrium. You also have to notice a dependence between [OH] and all forms of carbonic acid. Thanks to that you can calculate the value of something which you need.  ;)

Good luck!

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #2 on: April 24, 2009, 01:41:00 PM »
The next step is to write a reaction which shows from where you have OH- - find the constant of that equilibrium. You also have to notice a dependence between [OH] and all forms of carbonic acid. Thanks to that you can calculate the value of something which you need.  ;)

Why OH? Using H+ is faster and more obvious.
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Offline plankk

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #3 on: April 24, 2009, 01:53:59 PM »
It depends for who. ;)

Offline rohitarura

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #4 on: April 24, 2009, 04:20:27 PM »
Can u please show the working??

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #5 on: April 24, 2009, 04:30:58 PM »
Please read forum rules.

Try to follow plankk advice.
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Offline rohitarura

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #6 on: April 25, 2009, 02:50:57 PM »
Sir,the way plankk has told i had done the very same way then i also got the answer but at one step I couldn't understand why did i made that assumption.If you'll ask then I will post my solution but I wanted a more proper and better method,so i posted the question..
No offense meant,
Thank you.

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #7 on: April 25, 2009, 03:41:17 PM »
Show your work and elaborate on the assumption that you are not sure about, we will try to discuss it in detalis.
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Offline rohitarura

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #8 on: April 26, 2009, 03:26:12 PM »
SrCO3 ::equil:: Sr2+ + CO32-

[Sr2+]=s(say)=2.2x10-4                                (1)
[CO32-]=KSP/s                                 (2)

CO32- + H2::equil:: HCO3- + OH-

Its Kb =10-3/4.7
[CO32-]=KSP/s                     
[OH-]=10-5/2.51                                      (3)
[HCO3-]=s                                       (4)

Now,Equating Kb to the concentrations, we get KSP=8.9x10-10  ,which is the answer.

Now,the problematic assumption is the equation (4).The problem is that it is assumed in step (4) that the formed amount of CO32- in first reaction i.e., s is almost decomposed to HCO3- ,and also after this much decomposition of CO32- also,the Sr2+ concentration remains s only..

So,this was something weird that i found in the solution.

Please see to it, Sir

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #9 on: April 26, 2009, 04:31:39 PM »
Its Kb =10-3/4.7
[CO32-]=KSP/s                     
[OH-]=10-5/2.51                                      (3)
[HCO3-]=s                                       (4)

No idea what you are doing here.

Hint: knowing pH you also know ratio of concentrations of CO32- and HCO3-. You also know sum of their concentrations...
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Offline rohitarura

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #10 on: April 27, 2009, 04:54:30 AM »
OK,Sir then please tell what is the sum of concentrations of CO32- and HCO-.
I think it is equal to the concentration of Sr2+,but unable to understand why so??

Offline plankk

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #11 on: April 27, 2009, 05:53:33 AM »
You're right. Why? Because the only source of ions CO32- and HCO3- is your salt - SrCO3.

Of course it is right, when we establish that ions HCO3- don't hydrolyze. We can make that simplification because the Kb2 is very small. In our solution the concentration of H2CO3 is very small, so we can exlude it.

Offline rohitarura

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #12 on: April 27, 2009, 11:20:30 AM »
Thanks,plankk.
Can u lastly,please explain the following.
The initial concentrations of each Sr2+ and CO32- is equal to square root of the KSP,which is 2.98x10-5.And this the amount whose part has hydrolysed to HCO3-.
According to you, sum of concentrations of CO32- and HCO3- must be equal to the present amount of Sr2+ i.e., 2.2x10-4.But shouldn't it be equal to the square root of the KSP,which is 2.98x10-5 ??
Also,in the Schaum's series book the reason for [Sr2+]=[CO32-]+[HCO3-] was given as "The carbonate ion which dissolves forms HCO3- in a 1:1 molar ratio or remains unreacted."  What does this statement mean??

This is the book from where i got the question-  http://books.google.co.in/books?id=OWdOf9ECB1QC&dq=3000+Solved+Problems+in+chemistry&printsec=frontcover&source=bl&ots=i3tnh73q8a&sig=_9xe1g4uGn5IU9KPpEyuzcxEdU8&hl=en&ei=6M31SducEouGkQWX0MDtCg&sa=X&oi=book_result&ct=result&resnum=3#PPP3,M1

Offline plankk

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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #13 on: April 27, 2009, 01:29:26 PM »
The equilibrium concentration of Sr2+ is constant. You couldn't it count by taking square root of Ksp because CO32- hydrolyzes and the concentartion of it changes during solution. Thanks to that process (the hydrolysis) the solubity of SrCO3 is bigger than it result from square root of Ksp (2,98x10-5 < 2,2x10-4). During the solution when CO32- hydrolyzes, the concentration of it decreases, so the Ksp isn't exeeded, so bigger amount of salt can be soluted.

Quote
"The carbonate ion which dissolves forms HCO3- in a 1:1 molar ratio or remains unreacted."  What does this statement mean??
It means only that CO32- can dissolves to HCO3- or not. And if one molecules of CO32- hydrolizes, one molecule of HCO3- comes into being (so the molar ratio 1:1). That deduction results from the reaction's equation.

I hope I clarify it to you. If not, ask.



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Re: Please solve it completely,a tough question of Ionic equilibrium
« Reply #14 on: April 27, 2009, 02:46:49 PM »
The initial concentrations of each Sr2+ and CO32- is equal to square root of the KSP

This holds ONLY when there are no side reactions - that is, you can be sure [Sr2+]=[CO32-].
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