Optimizing Solubility: Manipulating [OH-] for Precipitation Control

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To lower the solubility of Mg(OH)2 to 1.5 x 10^-5 mol/L, the appropriate concentration of [OH-] must be calculated using the solubility product constant (Ksp) of 1.5 x 10^-11. The Ksp equation for Mg(OH)2 is expressed as Ksp = [Mg2+][OH-]^2. The discussion emphasizes that instead of using the ICE table method, one should directly determine the [OH-] concentration that results in a [Mg2+] of 1.5 x 10^-5 M. Participants suggest reviewing textbook materials or additional resources for clarity on the calculations. Understanding the relationship between [OH-] and Mg2+ is crucial for effective precipitation control.
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Homework Statement



What [OH-] would lower the solubility of Mg(OH)2
to 1.5 x 10-5 mol/L?
Ksp is 1.5 x 10-11 for Mg(OH)2.


Homework Equations



Ksp=products/reactants

The Attempt at a Solution



ICE box method

Mg OH (2) Mg(OH)2
initial: ? and I don't know what to do from here or even if this is the correct method to use.
 
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You don't need ICE table here, just check at what OH- concentration concentration of Mg2+ is 1.5x10-5M.
 
1.5 x 10^-5 = [Mg(OH)2]/[Mg][OH]

correct?
 
Write formula for Ksp of Mg(OH)2.
 
Borek said:
Write formula for Ksp of Mg(OH)2.

Ksp=products/reactants
=[Mg(OH)2]/[Mg][OH]^2

from the equation Mg+2OH = Mg(OH)2
 
No, you need to reread the book. Or check some other sources.
 
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