Molar solubility

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drillers

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I need clarification on this problem:

100ml of 0.1 M NaCl is mixed with 900ml of 0.01 M PbNO3 solution. If the ksp of PbCl2 is 1.6E-5, is the solution saturated or unsaturated?

Heres where i get confused. When we get down to PbCl2 --> Pb+ + 2Cl- ; (x)(2x)^2
Is the equation (0.009 M Pb)(0.01 M Cl)^2
OR (0.009 M Pb)(0.02 M Cl)^2

Because we have 0.01 mol Cl- in 1000ml I thought we should use 0.02^2 (since 2Cl-) but EK said 0.01^2? Anyone know which is correct and why?


Lastly, is saturation based on the ksp of PbCl2 or its molar solubility? I ask bc the molar solubility of PbCl2 is around 1.6E-2 (cube root est) while its ksp is 1.6E-5.

EK really has me confused on this bc most of the time they say saturation is based on the molar solubility, but then in the answer they say well the Q<ksp so unsat

But if the Q were 1.6E-4 that would be more than the ksp but less than the molar solubility, so wouldn't that still mean unsaturated?
 
I need clarification on this problem:

100ml of 0.1 M NaCl is mixed with 900ml of 0.01 M PbNO3 solution. If the ksp of PbCl2 is 1.6E-5, is the solution saturated or unsaturated?

Heres where i get confused. When we get down to PbCl2 --> Pb+ + 2Cl- ; (x)(2x)^2
Is the equation (0.009 M Pb)(0.01 M Cl)^2
OR (0.009 M Pb)(0.02 M Cl)^2

Because we have 0.01 mol Cl- in 1000ml I thought we should use 0.02^2 (since 2Cl-) but EK said 0.01^2? Anyone know which is correct and why?


Lastly, is saturation based on the ksp of PbCl2 or its molar solubility? I ask bc the molar solubility of PbCl2 is around 1.6E-2 (cube root est) while its ksp is 1.6E-5.

EK really has me confused on this bc most of the time they say saturation is based on the molar solubility, but then in the answer they say well the Q<ksp so unsat

But if the Q were 1.6E-4 that would be more than the ksp but less than the molar solubility, so wouldn't that still mean unsaturated?

Can't you simply use the solubility product (Ksp) and compare that to Q?
I believe that if your current situation (Q) is less than the Ksp you are unsaturated and nothing precipitates.

and use 0.02 for cl because there is 2 moles for every 1 mole of pb, I believe

also be on the look out for the common ion effectTM
 
Yeah I was thinking to use .02^2 instead of .01^2 but EK said to use .01 which through me off
 
Yeah I was thinking to use .02^2 instead of .01^2 but EK said to use .01 which through me off

ohh yeah you're right. Because that is what you have. the Q is independent of what actually forms, just a measure of what is actually in your solution.
 
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