Question

Calculate pH at which ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ being to precipitate form a solution containing {\text{0}}{\text{.1M}}$$$${\text{M}}{{\text{g}}^{{\text{2 + }}}} ions.
${K_{sp}}$ of${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}{\text{ = 1}} \times {\text{1}}{{\text{0}}^{{\text{ - 11}}}}$.

Answer 1

We need to know that the ${K_{sp}}$ is known as solubility product of the compound. The solubility product of the compound is defined as the product of the molar concentration of the product in form of constituent ion formed in the decomposition reaction. In these each constituent ion raised to the power of its stoichiometric co-efficient in the equilibrium balanced reaction. For the decomposition of the reaction form cation and anion. The positive charge of the ion is called as cation. The negative charge of the ion is called as anion. The sum of the formed ion is equal to the zero that mean neutral.
Formula used:
We must have to know that the stability product of a compound is equal to the product of the molar concentration of respective ion is multiply by number of formed ions in the decomposition reaction.
For example, the equilibrium reaction
${X_m}{Y_n} \rightleftarrows {X^{n + }} + n{Y^{m - }}$
Here, two ions are formed in the decomposition reaction of ${X_m}{Y_n}$ are ${X^{n + }}$ and ${Y^m}$ . The n number Ym- and m number of Xn+ are formed.
Hence, the solubility product of the above decomposition reaction is,
Ksp = [Xn+]m [Ym-]n
The pH of the solution is
${\text{pH = 14 - pOH}}$
{\text{pH = 14 - }}\left\\{ {{\text{ - log}}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]} \right\\}
${\text{pH = 14 + log}}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]$

Complete answer:
Solubility product of the ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ is
Ksp = ${\text{[M}}{{\text{g}}^{{\text{ + 2}}}}]{[{\text{O}}{{\text{H}}^{\text{ - }}}{\text{]}}^{_{\text{2}}}}$
The Ksp of ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ is ${\text{1}} \times {\text{1}}{{\text{0}}^{{\text{ - 11}}}}$
The concentration of ${\text{M}}{{\text{g}}^{{\text{2 + }}}}$ is ${\text{0}}{\text{.1M M}}{{\text{g}}^{{\text{2 + }}}}$.
We changed the formula for find the concentration of ${\text{O}}{{\text{H}}^{\text{ - }}}$ is
${\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]^{\text{2}}}{\text{ = }}\dfrac{{{{\text{K}}_{{\text{sp}}}}}}{{\left[ {{\text{M}}{{\text{g}}^{{\text{ + 2}}}}} \right]}}$
${\text{ = }}\dfrac{{{\text{1}} \times {\text{1}}{{\text{0}}^{{\text{ - 11}}}}}}{{{\text{ 0}}{\text{.1}}}}$
${\text{ = 1}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}$
The concentration of ${\text{O}}{{\text{H}}^{\text{ - }}}$ is formed in the reaction is$= 1 \times {10^{ - 10}}{\text{M}}$.
The pH of the solution is
${\text{pH = 14 - pOH}}$
{\text{pH = 14 - }}\left\\{ {{\text{ - log}}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]} \right\\}
${\text{pH = 14 + log}}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]$
Calculate pH at which ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ being to precipitate form a solution is calculated,
${\text{pH = 14 + log}}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]$
${\text{pH = 14 + log}}1 \times {10^{ - 10}}$
$\Rightarrow {\text{pH = 4}}$
According to the calculation, we conclude pH at which ${\text{Mg}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ being to precipitate form a solution containing ${\text{0}}{\text{.1M M}}{{\text{g}}^{{\text{2 + }}}}$ ions is $4$.

Note:
We must know that the solubility product of the compound is used to find the molar concentration of ions. If we find the one molar concentration of ion, it is easy to calculate the molar concentration of other ions in the equilibrium reaction. The solubility product of the compound is divided by the molar concentration of ion to find the molar concentration of other ions. The solubility product of the compound is used to find the nature of the reaction and precipitation condition. If the solubility product of the compound is greater than ionic product means, precipitation will occur and the solution in the form of supersaturation. If the solubility product of the compound is lesser than ionic product means, no precipitation and solution in form of unsaturation. If the solubility product of the compound is equal to the ionic product, the reaction in the state of equilibrium and solution in form of saturation.