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o-na [289]
4 years ago
11

Calculate the solubility (in grams per 1.00×102mL of solution) of magnesium hydroxide in a solution buffered at pH = 12

Chemistry
1 answer:
GaryK [48]4 years ago
8 0
The Ksp of Mg(OH)2 in water is 1.8 x 10-<span>11. This means that in pure water, Mg(OH)2 has a solubility of:
</span>∛[(1.8 x 10-11) / 4] = 1.65 x 10-4 mol/L
<span>which is equal to
</span>1.65 x 10-4<span> mol x (58.32) / 10 x 100 mL =  9.62 x 10-4g / 1x102 mL

If the pH is 12, the hydroxide concentration in the solvent is
10^-(14-12) = 0.01 mol/L
The solubility is solve using the formula
</span>1.8 x 10-11 = x (2(0.01 + x))^2
x = 4.5x10-8 mol/L
which is equal to
4.5x10-8 mol x (58.32) / 10 x 100 mL =  2.62 x 10-7g / 1x102 mL
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\qquad\qquad\huge\underline{{\sf Answer}}

Here we go ~

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