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The ${K_{sp}}$ of $Mg{\left( {OH} \right)_2}$ is $1 \times {10^{ - 12}}$. $0.01m$ $Mg{\left( {OH} \right)_2}$ will precipitate at the limiting pH equal to:

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Last updated date: 29th May 2024
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Answer
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Hint: We know that the solubility product steady${K_{sp}}$ is the harmony consistent for a strong substance dissolving in a fluid solution. It addresses the level at which a solute disintegrates in solution.
Consider the overall disintegration response beneath (in watery solutions):
\[aA\left( s \right) \rightleftharpoons cC\left( {aq} \right) + dD\left( {aq} \right)\]
To settle for the${K_{sp}}$it is important to take the molarities of the products (cC and dD ) and duplicate them. If there are any coefficients present before the products, it is important to raise the products to that coefficient power and likewise duplicate the focus by that coefficient
\[{K_{sp}} = {\left[ C \right]^c}{\left[ D \right]^d}\]

Complete answer:
Considering the connection among dissolvability and Ksp is significant while portraying the solvency of somewhat ionic mixtures. Notwithstanding, this article examines ionic mixtures that are hard to disintegrate; they are considered "marginally solvent" or "practically insoluble." Solubility products constants (${K_{sp}}$ ) are given to those solutes, and these constants can be utilized to track down the molar dissolvability of the mixtures that make the solute. This relationship additionally works with discovering the ${K_{sp}}$ of a somewhat dissolvable solute from its dissolvability.
The expression of the solubility product is given as,
\[{K_{SP}}{\text{ }} = \left[ {Mg{{\text{ }}^{2 + }}} \right]\left[ {OH{{\text{ }}^ - }} \right]{{\text{ }}^2}\]
Modify the above articulation:
\[\left[ {OH{{\text{ }}^ - }} \right] = {\text{ }}\left[ {Mg{{\text{ }}^{2 + }}} \right]{K_{sp}}\]
Substitute qualities in the above articulation and ascertain hydroxide particle focus
\[\left[ {OH{{\text{ }}^ - }} \right] = 0.011 \times {10^{ - 12}}\]
\[\left[ {OH{{\text{ }}^ - }} \right] = 1 \times {10^{ - 5}}M\]
Ascertain pOH from the hydroxide particle focus:
\[pOH = - log\left[ {O{H^ - }} \right]\]
Now we can substitute the known values we get,
\[pOH = - log\left( {1 \times 10{{\text{ }}^{ - 5}}} \right)\]
On simplification we get,
\[pOH = 5\]
Ascertain pH from pOH-
\[pH = 14 - pOH\]
Now we can substitute the known values we get,
$pH = 14 - 5$
On simplification we get,
$pH = 9$
The pH of the solution is $9$

Note:
We need to remember that for exceptionally dissolvable ionic mixtures the ionic exercises should be found rather than the fixations that are found in a somewhat solvent solution.
The dissolvability of the response is decreased by the basic particle. For a given harmony, a response with a typical particle present has a lower ${K_{sp}}$ and the response without the particle has a more prominent ${K_{sp}}$.
Having a contradicting impact on the ${K_{sp}}$ esteem contrasted with the normal particle impact, extraordinary particles increment the Ksp esteem. Phenomenal particles are particles other than those associated with harmony.
With an ionic pair (a cation and an anion), the ${K_{sp}}$ esteem determined is not exactly the test esteem because of particles engaged with matching. To come to the determined ${K_{sp}}$ esteem, more solute should be added.
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