Question

The solubility product of \[Mg{(OH)_2}{\text{ }}\;is{\text{ }}\;5 \times {10^{ - 19}}\] , pH of saturated solution of
 \[Mg{(OH)_2}\] will be:

Answer 1

Hint: The Solubility of the substance is defined as a property of solute to get dissolved in a solvent to form a solution. The solubility of ionic compounds in water varies really to the great extent. Many compounds present in nature are highly soluble which may even absorb moisture from the atmosphere whereas some of them are highly insoluble.

Complete step by step answer:
Let us check out the general equation written in the aqueous medium
\[a{A_{(s)}} \rightleftharpoons c{C_{(aq)}} + d{D_{(aq)}}\]
${K_{sp}}$Let find out a way for getting
To get the value of the ${K_{sp}}$we take the concentrations of the products formed or their molarities
and get them multiplied with each other along with the product raised to the coefficients present in front of them that can be shown as follows:
\[{K_{sp}} = {[C]^c}{[D]^d}\]
${K_{sp}}$results in the maximum extent that a solid that can get dissolved in solution under normal
 conditions.
You might be wondering about what are the factors that are responsible for affecting the value of
 ${K_{sp}}$ ?
Therefore, the factors responsible for affecting the value of the solubility product constant are:

- The common-ion effect where the existence of a common ion lowers down its value.
- The diverse-ion effect where the uncommon ions of the solutes increase its value.
- The existence of ion-pairs.

Now let’s proceed towards the solution
So, our first step would be to write the chemical equation involved in this question which is given as follows
\[Mg{(OH)_2} \rightleftharpoons M{g^{_{2 + }}} + 2O{H^ - }\]
If the molar solubility of the magnesium hydroxide is S, then it can be easily seen from the stoichiometry of the compound that the equation becomes
\[\begin{gathered}
  Mg{(OH)_2} \rightleftharpoons M{g^{_{2 + }}} + 2O{H^ - } \\
  {\text{ S 2S}} \\
\end{gathered} \]
\[\begin{gathered}
  {K_{sp}} = [M{g^{2 + }}]{[O{H^ - }]^2} \\
  {\text{ }} = (S){(2S)^2} \\
  {\text{ }} = 4{S^3} \\
  4{S^3} = 5 \times {10^{ - 19}} \\
\end{gathered} \]
On calculating the value of S we get
\[S = 5 \times {10^{ - 7}}\]
\[\begin{gathered}
  Mg{(OH)_2} \rightleftharpoons M{g^{_{2 + }}} + 2O{H^ - } \\
  {\text{ S 2S}} \\
\end{gathered} \]
In the above equation we can see that the concentration of the hydroxide ion is two times the value of S
Therefore,
\[[O{H^ - }] = 2S = 2 \times 5 \times {10^{ - 7}} = {10^{ - 6}}\]
We now recall the formula present in the Equilibrium chapter given in the NCERT that
\[pOH = - log([O{H^ - }])\]
Substituting the value of the concentration of the hydroxide ion we get
\[pOH = - log([O{H^ - }]) = 6\]
Besides we also know that
\[pH + pOH = 14\]

\[\begin{gathered}
  pH = 14 - pOH \\
  {\text{ = 14}} - 6 \\
  {\text{ = 8}} \\
\end{gathered} \]
Therefore, the value of the pH comes out to be 8.

Note: Don’t confuse between solubility and solubility product constants. The solubility of a given substance in a solvent is the total quantity of the solute that can be dissolved in the solvent given at the equilibrium condition. Whereas, the solubility product constant is found to be an equilibrium constant that provides us the brief knowledge about the equilibrium between the solid solute and its constituent ions that are dissociated across the solution.