How Does pH Affect the Solubility of Mg(OH)2?

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The discussion centers on calculating the solubility of Mg(OH)2 in a buffered solution at pH 8.80, with the expected solubility being 0.45 M or 26 g/L. The participant calculates the hydroxide ion concentration as [OH-] = 6.3 x 10^-6 and sets up the equilibrium expression for Mg(OH)2 dissociation. They derive a concentration of x = 1.65 x 10^-4 M but are unsure of the discrepancy with the expected answer. A key point raised is that in a buffered solution, the correct approach is to consider [OH-] solely as 6.3 x 10^-6, without adding 2x. The discussion highlights the importance of accurately accounting for hydroxide concentration in solubility calculations.
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Homework Statement


What is the solubility of Mg(OH)2 in a solution buffered at pH 8.80.

Homework Equations


The Attempt at a Solution


The answer is .45 M or 26 g/L but I can't get that for some reason.

I can find the concentration of hydroxide [OH-] = 6.3 x 10-6
Mg(OH)2 -> Mg+2 + 2OH-
...0+x,...6.3x10-6 + 2x

1.8x10-11 = x(6.3x10-6 + 2x)^2

x = 1.65 x 10-4 M (neglecting the 6.3 x10-6)
I've also tried solving the cubic with similar results.
Does anyone where I went wrong?
 
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Solution is buffered. [OH-] is not 6.3x10-6 + 2x but JUST 6.3x10-6.
 
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