 # The solubility of $Fe{{(OH)}_{2}}$ is greater than $Fe{{(OH)}_{3}}$ in water. If true, write $1$ , else write $0$ . Verified
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Hint: Before talking about the answer, we should know what solubility is. Solubility is defined as the ability of a compound to get soluble. Solubility is expressed in terms of molarity. You should also know about solubility products, that is, ${{K}_{sp}}$ .

Formula used:For the following reaction: $[AB]\rightleftharpoons a[{{A}^{+}}]+b[{{B}^{-}}]$
Solubility product, ${{K}_{sp}}={{[{{A}^{+}}]}^{a}}{{[{{B}^{-}}]}^{b}}$
where, ${{K}_{sp}}$ is solubility product, ${{A}^{+}}$ is the cation, ${{B}^{-}}$ is the anion and $a,b$ are concentration of $A$ and $B$ .

Firstly, we will calculate the solubility product of $Fe{{(OH)}_{2}}$
The reaction can be written as follows:
$Fe{{(OH)}_{2}}\rightleftharpoons F{{e}^{2+}}+2O{{H}^{-}}$
Here, the solubility of $F{{e}^{2+}}$ is $'s'$ and solubility of $O{{H}^{-}}$ ions is $'2s'$
Therefore, ${{K}_{sp}}=(s){{(2s)}^{2}}$
${{K}_{sp}}=4{{s}^{3}}$
The ${{K}_{sp}}$ value of $Fe{{(OH)}_{2}}$ is $2\times {{10}^{-15}}$
$\Rightarrow 2\times {{10}^{-15}}=4{{s}^{3}} \\ \Rightarrow s={{(\dfrac{2\times {{10}^{-15}}}{4})}^{{}^{1}/{}_{3}}}=7.94\times {{10}^{-6}} \\$
Similarly, for $Fe{{(OH)}_{3}}$, the reaction will be written as:
$Fe{{(OH)}_{3}}\rightleftharpoons F{{e}^{3+}}+3O{{H}^{-}}$
and the solubility of $F{{e}^{3+}}$ is $'s'$ and solubility of $O{{H}^{-}}$ ions is $'3s'$
Therefore, ${{K}_{sp}}=(s){{(3s)}^{3}}$
${{K}_{sp}}=27{{s}^{4}}$
${{K}_{sp}}$ value of $Fe{{(OH)}_{3}}$ is $6\times {{10}^{-38}}$
$\Rightarrow 6\times {{10}^{-38}}=27{{s}^{4}} \\ \Rightarrow s={{(\dfrac{6\times {{10}^{-38}}}{27})}^{{}^{1}/{}_{4}}}=2.17\times {{10}^{-10}} \\$
Now as we can see, the solubility of $Fe{{(OH)}_{2}}$ is greater than $Fe{{(OH)}_{3}}$
So, the correct answer is $1$ .
Here, ${{K}_{sp}}$ is a solubility product which is defined as the equilibrium constant for solid to dissolve in an aqueous solution. It tells us about the level at which the solute is dissolved in the solution. It also tells us about the slightly ionic compound.

${{K}_{sp}}$ of $Fe{{(OH)}_{2}}$ is more than the ${{K}_{sp}}$ of $Fe{{(OH)}_{3}}$ and hence solubility of $Fe{{(OH)}_{2}}$ is greater than that of $Fe{{(OH)}_{3}}$ .