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AK28

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I am getting confused on this concept. Everywhere says that Common Ions decrease solubility. Yet in this problem,

Ksp of PbCl2 is 1.6x10-5. What is the solubility of PbCl2 in .01 M KCl?

The solubility is .16 M. So wouldn't that mean that common ions make the solubility increase?

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Not sure about the question you're referring to, perhaps the numbers are off. But common ions do decrease solubility as you already have some of the product present which affects equilibrium
 
Not sure about the question you're referring to, perhaps the numbers are off. But common ions do decrease solubility as you already have some of the product present which affects equilibrium

This a problem from Dat Destroyer- so the numbers are right! If you work out the problem, you will get .16 also which is why I'm confused.
 
This a problem from Dat Destroyer- so the numbers are right! If you work out the problem, you will get .16 also which is why I'm confused.

That value is S. Have you solved for S from the original Ksp and compared?
 
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Yes, number 198 in DAT destroyer 2016

Okay! I think I have the answer. The Qsp value which is the 1.6 *10^-2 is greater than the Ksp value of 1.6 * 10^-5 thus the equation is out of the equilibrium. To return it has to shift left, because there is precipitate, Q>K. So the solubility decreases!

Almost forgot all about this.
 
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Okay! I think I have the answer. The Qsp value which is the 1.6 *10^-2 is greater than the Ksp value of 1.6 * 10^-5 thus the equation is out of the equilibrium. To return it has to shift left, because there is precipitate, Q>K. So the solubility decreases!

Almost forgot all about this.
Hmm okay I think it makes sense now!
So because there is more Cl, because of Le Chatelier it will shift left? and then more product decreases K ahhh got it
 
Hmm okay I think it makes sense now!
So because there is more Cl, because of Le Chatelier it will shift left? and then more product decreases K ahhh got it

Yes, because of the extra cl there is more precipitate formed. This in order to get back to equilibrium, it must shift left.

Make sure to watch chadsvideos on this section as he explains it better than me.
 
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i understand how the answer makes sense if you compare Ksp and Qsp with and without the common ion..but do you know what the solubility of PbCl2 without any common is? i got 0.016, which is lower than 0.16.

this chemistry book online also got the solubility with no common ion to be 0.016 as well:

Okay! I think I have the answer. The Qsp value which is the 1.6 *10^-2 is greater than the Ksp value of 1.6 * 10^-5 thus the equation is out of the equilibrium. To return it has to shift left, because there is precipitate, Q>K. So the solubility decreases!

Almost forgot all about this.
 

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i understand how the answer makes sense if you compare Ksp and Qsp with and without the common ion..but do you know what the solubility of PbCl2 without any common is? i got 0.016, which is lower than 0.16.

this chemistry book online also got the solubility with no common ion to be 0.016 as well:

1.6 times ten to the negative two is .016
 
1.6 times ten to the negative two is .016

yes, 0.016 is the molar solubility of PbCl2 in pure water..i'm confused about why the molar solubility of PbCl2 in 0.1 KCl is 0.16, which is a higher number than the solubility of PbCl2 in pure water.

Chad did an example in one of his videos with AgCl in pure water verse AgCl in 0.1 HCl..the x value (the solubility) ended up lower in the solution with the common ion.
 
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