Question

In the reaction $Fe{(OH{)}_3}_{\left(s\right)}$ $\rightleftharpoons F{e}_{aq}^{3+}+3O{H}_{aq}^{-}$, if the concentration of $O{H}^{-}$ is decreased by $\frac{1}{4}$ times then the equilibrium concentration $ofF{e}^{3+}$ increased by how many times?

Answer 1

The equilibrium constant for a reaction at the given temperature is constant; it does not depend on the concentration of reactants of products. So, it will not change on changing the concentration of reactant or product.

Now, given reaction is:-

$Fe{\left(OH\right)}_{3\left(S\right)}\rightleftharpoons {Fe}^{+}\left(aq\right)+3{OH}^{-}\left(aq\right)$

$K_C=\left[{Fe}^{+}\right]{\left[{OH}^{-}\right]}^3\to \left(I\right)$

Now, at eqm,

${\left[{OH}^{-}\right]}^{{}^{\prime }}=\frac{1}{4}\left[OH\right]\to \left(II\right)$

At eqm,

$\Rightarrow {KC}^{{}^{\prime }}={\left[{Fe}^{3+}\right]}^{{}^{\prime }}{\left({\left[{OH}^{-}\right]}^2\right)}^3\to \left(III\right)$

But, $K_C={K_C}^{{}^{\prime }}$

$\Rightarrow {\left[{Fe}^{3+}\right]}^2{\left(\frac{\left[{OH}^{-}\right]}{4}\right)}^3=\left[{Fe}^{3+}\right]{\left[{OH}^{-}\right]}^3$

$\Rightarrow {\left[{Fe}^{3+}\right]}^{{}^{\prime }}=64\left[{Fe}^{3+}\right]$