What is the solubility of Al(OH)3 in 0.0010 M Al(NO3)3?

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The solubility of Al(OH)3 in 0.0010 M Al(NO3)3 is influenced by the common ion effect due to the presence of Al3+ ions. The Ksp expression for Al(OH)3 is Ksp = [Al3+][OH-]^3, where the concentration of Al3+ is already 0.0010 M from the dissociation of Al(NO3)3. The concentration of OH- ions must be determined considering the additional Al3+ ions from the dissolution of Al(OH)3. The correct approach involves setting up the Ksp equation with the adjusted concentrations and solving for x, the solubility of Al(OH)3. The final calculated solubility is 3.3 x 10^-11 M.
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Homework Statement




The Ksp of Al(OH)3 is 1.0 x 10-33. What is the solubility of Al(OH)3 in 0.0010 M Al(NO3)3? Give your answer using scientific notation and to 2 significant figures (i.e., one decimal place).

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The Attempt at a Solution



I can solve easier questions than this but I'm confused on how the .0010 M Al(NO3)3 will dissociate into and how it will effect the Ksp.

From what I gather AL(NO3)3 will dissociate into 4 ions, AL, and 3(NO3). Since it's at .001 M the 3(NO3) has a value of .003M.

1.8E-33 = (x)(.003) = 6E-31. Which I know isn't even close. The correct answer is 3.3e-11 but I can only guess on how to get that answer.

Any help would be appreciated.
 
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dolpho said:
1.8E-33 = (x)(.003)

Not bad, with one important mistake.

Ignoring everything else, how dos the formula for Ksp of Al(OH)3 look like?
 
Borek said:
Not bad, with one important mistake.

Ignoring everything else, how dos the formula for Ksp of Al(OH)3 look like?

Ksp = ( Al^3+) ( 3OH)^3?
 
Why do you throw this 3 in every possible place? Apparently you are not sure where it should be, aren't you?

K_{sp} = [Al^{3+}][OH^-]^3

Now, if concentration of dissolved Al(OH) is x, and there were already 0.0010 M of Al3+ present, what are concentrations of both Al3+ and OH-?

Note: it may turn out we will need an additional step, but it will become obvious later.
 
Borek said:
Why do you throw this 3 in every possible place? Apparently you are not sure where it should be, aren't you?

K_{sp} = [Al^{3+}][OH^-]^3

Now, if concentration of dissolved Al(OH) is x, and there were already 0.0010 M of Al3+ present, what are concentrations of both Al3+ and OH-?

Note: it may turn out we will need an additional step, but it will become obvious later.

Al would be .001 and OH would be .001^3 ?
 
No. When the Al(OH)3 is dissolved it adds to the ions already present in the solution.

In some cases the difference in the concentration will be negligible, but let's start by getting these things right, we will think about approximations and negligibility later.
 
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