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

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Discussion Overview

The discussion revolves around calculating the solubility of Al(OH)3 in a solution of 0.0010 M Al(NO3)3, focusing on the implications of the common ion effect on the solubility product constant (Ksp). The scope includes homework-related problem-solving and mathematical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant states the Ksp of Al(OH)3 is 1.0 x 10-33 and expresses confusion about how the 0.0010 M Al(NO3)3 will dissociate and affect the Ksp.
  • Another participant points out a mistake in the initial calculation and prompts for clarification on the Ksp formula for Al(OH)3.
  • A subsequent reply questions the placement of the coefficient in the Ksp expression and provides the correct formula: Ksp = [Al3+][OH-]3.
  • Further, a participant suggests that if the concentration of dissolved Al(OH)3 is x, the concentrations of Al3+ and OH- need to be determined, considering the initial concentration of Al3+ already present.
  • Another participant emphasizes that the dissolution of Al(OH)3 adds to the existing ion concentrations and suggests that approximations regarding negligible differences will be addressed later.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the problem, as there are multiple interpretations of how to apply the Ksp and the effects of the common ion. Disagreements arise regarding the correct formulation and calculation steps.

Contextual Notes

Participants express uncertainty about the correct application of the Ksp formula and the implications of the common ion effect on solubility. There are unresolved mathematical steps and assumptions regarding the concentrations of ions in solution.

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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).

Homework Equations





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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