Question

If the concentration of $O{{H}^{-}}$ ions in the reaction:
$Fe{{\left( OH \right)}_{3}}\left( s \right)\rightleftharpoons F{{e}^{+3}}\left( aq \right)+3O{{H}^{-}}\left( aq \right)$ $Fe{{\left( OH \right)}_{3}}\left( s \right)\rightleftharpoons F{{e}^{+3}}\left( aq \right)+3O{{H}^{-}}\left( aq \right)$ is decreased by $\dfrac{1}{4}$ times, then equilibrium concentration of $F{{e}^{+3}}$ will increase by:
A) $64$ times
B) $4$ times
C) $8$ times
D) $16$ times

Answer 1

When there is no net change in the amounts of product and reactant in a reversible chemical reaction then it is known as equilibrium. In this reaction the products formed reactivate the original reactant.

Complete answer:
In this question, it is given that the concentration of hydroxide ion $\left( O{{H}^{-}} \right)$ is decreased by $\dfrac{1}{4}$ times.
Now let us see the reaction,
$Fe{{\left( OH \right)}_{3}}\left( s \right)\rightleftharpoons F{{e}^{+3}}\left( aq \right)+3O{{H}^{-}}\left( aq \right)$
The formula to calculate equilibrium constant is:
$k=\dfrac{\left[ P \right]}{\left[ R \right]}$
Where, $k$ is the equilibrium constant
$P$ is the concentration of product
$R$ is the concentration of reactant
Now let us see the reaction,
$Fe{{\left( OH \right)}_{3}}\left( s \right)\rightleftharpoons F{{e}^{+3}}\left( aq \right)+3O{{H}^{-}}\left( aq \right)$
$k=\dfrac{\left[ F{{e}^{+3}} \right]{{\left[ O{{H}^{-}} \right]}^{3}}}{\left[ Fe{{\left( OH \right)}_{3}} \right]}$
The hydroxide ion is decreased by $\dfrac{1}{4}$ times, therefore we will substitute the value in the concentration of hydroxide ion.
According to the le chatelier’s principle, here the concentration of hydroxide ion is decreasing therefore to maintain the equilibrium the reaction will move forward by increasing the concentration of $F{{e}^{+3}}$ by $x$ times.
$k=\dfrac{\left[ xF{{e}^{+3}} \right]{{\left[ \dfrac{1}{4}O{{H}^{-}} \right]}^{3}}}{\left[ Fe{{\left( OH \right)}_{3}} \right]}$
As $Fe{{\left( OH \right)}_{3}}$ is in solid form hence it will not be used in this formula.
$k=\dfrac{\left[ xF{{e}^{+3}} \right]{{\left[ O{{H}^{-}} \right]}^{3}}}{64}$
The concentration of $F{{e}^{+3}}$ is inversely proportional to concentration of $O{{H}^{-}}$
Therefore, $x=64$ times
Hence the concentration of $F{{e}^{+3}}$ is increased by 64 times.

Hence the correct answer is (A).

Additional information: Le chatelier’s principle states that if there is any change in the system equilibrium then the system changes into new equilibrium. This change can be changed in pressure, concentration temperature or volume.
If the concentration of the reactant is increased then the reaction will move towards right and if the concentration of the product is increased then the reaction will move towards left.

Note: In this question, we concluded that by decreasing the concentration of product will move the reaction forward. Here, the hydroxide ion is decreased and hence the ferric ion is multiplied by x to maintain the equilibrium.