Question

The solubility product of $Fe{{(OH)}_{3}}$ is $1\text{ x 1}{{\text{0}}^{-36}}$ . What is the minimum concentration of $O{{H}^{-}}$ ions required to precipitate $Fe{{(OH)}_{3}}$ from 0.001 M solution of $FeC{{l}_{3}}$ ?

Answer 1

Solubility product of ferric oxyhydroxide is the product of the concentration of iron and hydroxyl ions. The concentration of the hydroxyl ions can be calculated by dividing the solubility product of the compound to the concentration of the solution.

Complete step by step answer:
The solubility product of an electrolyte at a specified temperature may be defined as the product of the molar concentration of its ions in a saturated solution, each concentration raised to the power equal to the number of ions produced on the dissociation of one molecule of the electrolyte.
So, when the $Fe{{(OH)}_{3}}$ dissociates into ions, it form and
The equilibrium equation will be:
$Fe{{(OH)}_{3}}\rightleftharpoons F{{e}^{3+}}+3O{{H}^{-}}$
Given the solubility product of $Fe{{(OH)}_{3}}$ is $1\text{ x 1}{{\text{0}}^{-36}}$
So, by the definition we can write,
${{K}_{sp}}\text{ of }Fe{{(OH)}_{3}}=\text{ }1\text{ x 1}{{\text{0}}^{-36}}$
$therefore,\text{ }\\!\\![\\!\\!\text{ }F{{e}^{3+}}]{{[O{{H}^{-}}]}^{3}}=\text{ }1\text{ x 1}{{\text{0}}^{-36}}$
For finding the concentration required to precipitate $Fe{{(OH)}_{3}}$ from 0.001 M solution of$FeC{{l}_{3}}$ ,
We have to divide the solubility product with the given concentration of the solution, we get:
${{[O{{H}^{-}}]}^{3}}=\dfrac{{{K}_{sp}}}{0.001}$
${{[O{{H}^{-}}]}^{3}}=\dfrac{1\text{ x 1}{{\text{0}}^{-36}}}{0.001}$
${{[O{{H}^{-}}]}^{3}}={{10}^{-33}}$
Now the concentration of hydroxyl ions is three times hence we have to cube root the value, to get the minimum value of hydroxyl ions.
$[O{{H}^{-}}]={{({{10}^{-33}})}^{\dfrac{1}{3}}}$
$[O{{H}^{-}}]={{10}^{-11}}M$
Hence, the minimum concentration of $O{{H}^{-}}$ ions required will be ${{10}^{-11}}M$ .
Additional information: The solubility product of a sparingly soluble salt at a particular temperature can be calculated from a knowledge of the solubility of the salt in water at a particular temperature. Some examples are:

& AgCl\to A{{g}^{+}}+C{{l}^{-}} \\\ & {{K}_{sp}}\text{ of }AgCl=[A{{g}^{+}}][C{{l}^{-}}] \\\ & Mg{{(OH)}_{2}}\to M{{g}^{2+}}+2O{{H}^{-}} \\\ & {{K}_{sp}}\text{ of }Mg{{(OH)}_{2}}=[M{{g}^{2+}}]{{[O{{H}^{-}}]}^{2}} \\\ \end{aligned}$$ **Note:** Both the ionic product and solubility product represent the product of the concentration of the ions in the solution. But ionic products are applicable to all types of solution and solubility products are only applicable to those solutions which occur in dynamic equilibrium.