Apr 11, 2010
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I keep getting different answers than the book...

1)Calculate the molar solubility of Al(OH)3 (ksp=2 x 10^-32)


I tried: ksp= [x][3x]^3

= 27x^4 = 2 x 10^-32
Then I solved for x to get 5.22 x 10^-9
but the answer should be 2 x 10^-11
What did I do wrong?

2) Same thing really... Calculate the molar solubility of Fe(OH)3 (ksp=4 x 10^-38) in water.


(answer: 4 x 10^-17)
I tried the same method as a the first problem and it wasn't right. I am missing something.

Thanks a bunch!!!

Gold3nLily
 
Last edited:
Apr 11, 2010
10
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Status
Pre-Dental
They are from my chemistry book. I though the answer in the back might have been wrong when I missed the first one, but after missing the next related one I didn't think it was probable for it to be the book's error.

Does the fact that one of the compounds dissociating in water is a really strong base have an effect here?



First one looks right the way you did it.

Where are these problems from?
 

UndergradGuy7

10+ Year Member
Jun 23, 2007
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They are from my chemistry book. I though the answer in the back might have been wrong when I missed the first one, but after missing the next related one I didn't think it was probable for it to be the book's error.

Does the fact that one of the compounds dissociating in water is a really strong base have an effect here?
No, the way you did it right. Even if you try googling similar problems, you can find problems on school websites that do it the way you did it.

The book probably used a wrong ksp value or multiplied something wrong.

Edit:
I tried doing since it is in h2o then H+ = 1.0E-7 and OH- = 1.0E-7
So (x)(1.0E-7)^3 = Ksp
and x = 2 x 10^-11

Book is right. Ignore what I wrote in other post.
 
Last edited:
Apr 11, 2010
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Oh, I see. It is a common ion problem.

That is weird because when it is Al(OH)2 instead of Al(OH)3 or Fe(OH)2 instead of Fe(OH)3 it is not a common ion problem but it is done the way I tried it before. I don't know why they are different, but I will just memorize it.

Thank you so much, I feel so much better now that this is figured out!

Gold3nLily

"Edit:

I tried doing since it is in h2o then H+ = 1.0E-7 and OH- = 1.0E-7
So (x)(1.0E-7)^3 = Ksp
and x = 2 x 10^-11"
 

evanyou

7+ Year Member
Jan 26, 2010
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Oh, I see. It is a common ion problem.

That is weird because when it is Al(OH)2 instead of Al(OH)3 or Fe(OH)2 instead of Fe(OH)3 it is not a common ion problem but it is done the way I tried it before. I don't know why they are different, but I will just memorize it.

Thank you so much, I feel so much better now that this is figured out!

Gold3nLily

"Edit:
I tried doing since it is in h2o then H+ = 1.0E-7 and OH- = 1.0E-7
So (x)(1.0E-7)^3 = Ksp
and x = 2 x 10^-11"

Can you provide Ksp of Al(OH)2 and Fe(OH)2?
I am just guessing that if those two have have relatively larger ksp, the common ion effect of [OH] 1.0e-7 can be ignored.