Question

Calculate the pH at which Mg(OH)2 begin to precipitate from solution containing 0.1M Mg2+ ions. Ksp of Mg(OH)2= 1*10^-11

Answer 1

pH should be 9.. calculations are in the pic above.

Answer 2

Given:

Concentration of $Mg^{2+}$ = 0.1M

$K_{sp}$ of Mg$(OH)_2$= $1\times 10^{-11}$

To find:

pH of Mg$(OH)_2$ = ?

Calculation:

The products formed due to dissociation of Mg$(OH)_2$ are as follows:

Mg$(OH)_2$⇄$Mg^{2+} + 2OH^-$

∴ $K_{sp}$  = $[Mg^{2+}]\times [OH^-]^2$

when the formula is rearranged it yields

[O$H^-$]= $\sqrt\frac{K_{sp}}{[Mg^{2+}]}$= $\sqrt \frac{1 \times 10^-{11}}{0.1} = \sqrt{1.0\times 10^{-10}} = 10^{-5}$

With [O$H^-$] value, pOH of the solution can be calculated as

pOH = -log[O$H^-$]

pOH = -log[$10^{-5}$]

pOH = 5

As pH = 14 - pOH

∴ pH = 14 - 5

pH = 9

Conclusion:

Mg$(OH)_2$ begin to precipitate from solution containing 0.1M Mg2+ ions at pH 9.

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