Destroyer 2008 Ochem #82

[jdpaul14]

Advertisement - Members don't see this ad

Q: The Ksp of PbCl2 is 1.6 x 10^-5 @ 25 degrees Celsius. What is the solubility of PbCl2 in 0.01M KCl?

A: 0.16M


This is a common ion effect problem, and I do not understand how they got:

Ksp = x(0.01)^2

.....shouldn't it be Ksp = x(2x + 0.01)^2?

The solution says that since Ksp is really small you can discount the 2x, why is that? And what if the Ksp is very large, then what do I do?

Hope this wasn't too confusing, hopefully someone will be able to answer my question.

Thanks a lot

 

[nze82]

Q: The Ksp of PbCl2 is 1.6 x 10^-5 @ 25 degrees Celsius. What is the solubility of PbCl2 in 0.01M KCl?

A: 0.16M


This is a common ion effect problem, and I do not understand how they got:

Ksp = x(0.01)^2

.....shouldn't it be Ksp = x(2x + 0.01)^2?

The solution says that since Ksp is really small you can discount the 2x, why is that? And what if the Ksp is very large, then what do I do?

Hope this wasn't too confusing, hopefully someone will be able to answer my question.

Thanks a lot

In order to see why we can ignore 2x, lets proceed as follows:

Step#1: Let's first ignore the common ion effect to see how small x really is:

PbCl2 <--> Pb2+ + 2Cl-
Ksp = [Pb2+][Cl-]^2 = (x)(2x)^2 = 4x^3 = 1.6E-5
x = 0.02

Do the above calculation for some larger Ksp, and you'll see that x ends up being a larger number, and therefore we wouldn't be able to ignore it in our calculations.

Step#2: Now let's consider the common ion effect:

PbCl2 <--> Pb2+ + 2Cl-
For this reaction [Cl-] = (2x)^2 = (2x0.02)^2 = 0.0016M ~ 0

KCl <--> K+ + Cl-
For this reaction [Cl-] = 0.01

Notice the insignificant contribution of the first reaction to the total Cl- concentration compared to that of the second reaction.

Now let's rewrite the Ksp, while considering the common ion effect:
Ksp = [Pb2+][Cl-]^2
Ksp = (x)(x+0.01)^2
As we showed above, the first reaction's contribution to the total Cl- concentration is insignificant, so (x+0.01) ~ 0.01. Therefore:
Ksp = (x)(0.01)^2
(x) = Ksp / (0.01)^2 = 1.6E-5/(0.01)^2 = 0.16M

 

[5HTburb]

http://video.google.com/videoplay?docid=-7609889215419912625

Check that out and if you understand the basics of solubility product constant, the common ion effect starts around the 4 minute mark. I wouldn't say it is the greatest, but its always better to hear and visualize it rather than trying to figure it out by staring in a book... Hope this helps

 

[jdpaul14]

Ok, I understand, but what is considered a small Ksp? Is it like anything smaller than 10^-5?

Thanks again,

 

[bbwreckerz04]

Ok, I understand, but what is considered a small Ksp? Is it like anything smaller than 10^-5?

Thanks again,

Same as Ka less than 5%.

 

[jdpaul14]

Less that 5% of what? Sorry if I sound lost, but what would that correspond to in scientific notation?

 

[lonhosilver]

The rule of thumb... that I have heard is anything less than 10^-2 is considered small...but for the purpose of the Real DAT and the practice tests....you can always cancel out anything variable type that would make you do the quad. formula....they dont want you to do that

 

[nze82]

Ok, I understand, but what is considered a small Ksp? Is it like anything smaller than 10^-5?

Thanks again,

In almost all questions of this nature, you're Ksp is small enough so that you can ignore x in your calculations.