Question

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

Answer 1

Hint: $ {K_{sp}} $ is called the solubility product because it is the product of the solubilities of the ions in moles per liter. Here the given compound is $ Mg{(OH)_2} $ which will dissociate into $ 1 $ . Hence, Solubility product expression becomes
 $ {K_{sp}} $ = $ [M{g^{2 + }}]{[O{H^ - }]^2} $
The above expression can be simplified to find the $ O{H^ - } $ concentration.
 $ [O{H^ - }] $ = $ \sqrt {\dfrac{{{K_{sp}}}}{{[M{g^{2 + }}]}}} $

Complete step by step solution:
 $ Mg{(OH)_2} $ will release $ M{g^{2 + }} $ and $ O{H^ - } $ ions according to the reaction,
 $ \rightleftharpoons $ $ Mg{(OH)_2} $ $ M{g^{2 + }} $ $ + $ $ 2O{H^ - } $
Thus, the solubility product is
 $ {K_{sp}} $ = $ [M{g^{2 + }}]{[O{H^ - }]^2} $
Where $ {K_{sp}} $ = solubility product constant
 $ [M{g^{2 + }}] $ =concentration of $ M{g^{2 + }} $ ions
 $ [O{H^ - }] $ =concentration of $ O{H^ - } $ ions
In this question we are given the concentration of $ M{g^{2 + }} $ ions ( $ 0.01 $ m) and the value of $ {K_{sp}} $ ( $ 1 \times {10^{ - 12}} $ ).
Using the simplified formula we can find the concentration of $ O{H^ - } $ ions.
  $ [O{H^ - }] $ = $ \sqrt {\dfrac{{{K_{sp}}}}{{[M{g^{2 + }}]}}} $
 $ \Rightarrow $ $ [O{H^ - }] $ = $ \sqrt {\dfrac{{1 \times {{10}^{ - 12}}}}{{0.01}}} $ m
 $ \Rightarrow $ $ [O{H^ - }] $ = $ \sqrt {{{10}^{ - 10}}} $ m
 $ \Rightarrow $ $ [O{H^ - }] $ = $ {10^{ - 5}} $ m
Now we got the concentration of $ O{H^ - } $ ions. By using the formula mentioned below, we can find the $ $ $ pOH $ when the concentration of $ O{H^ - } $ ion is known.
 $ pOH = - \log [O{H^ - }] $
 $ \Rightarrow $ $ pOH = - \log [{10^{ - 5}}] $
 $ \Rightarrow pOH = - 1 \times - 5 $
 $ \Rightarrow pOH = 5 $
Now the general equation in terms of $ pH $ and $ pOH $ is,
 $ pH + pOH = 14 $
 $ \Rightarrow pH = 14 - pOH $
As we know $ pOH = 5 $ , thus by substituting the value we get,
 $ \Rightarrow pH = 14 - 5 $
 $ \Rightarrow pH = 9 $
Therefore, $ 0.01 $ m $ Mg{(OH)_2} $ will precipitate at the limiting $ pH $ of $ 9 $ .

Additional information:
The solubility product is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol $ {K_{sp}} $ . The value of $ {K_{sp}} $ depends on temperature and is different for every salt. $ {K_{sp}} $ value generally increases with the increase in temperature due to increased solubility. Some of the factors which affect the value of $ {K_{sp}} $ are:
 $ \bullet $ Common-ion effect (the presence of a common ion lowers the value of $ {K_{sp}} $ .
 $ \bullet $ The diverse ion effect (if the ions of solute are uncommon, the value of $ {K_{sp}} $ will be high).
 $ \bullet $ Ion pair presence.

Note:
The idea here is that you need to use magnesium hydroxide’s solubility product constant to determine what concentration of $ O{H^ - } $ ions would cause the solid to precipitate out of solution.