Question

If the concentration of $O{H}^{-}$ions in the reaction:
$Fe(OH{)}_3(s)\rightleftharpoons F{e}^{3+}(aq)+3O{H}^{-}$(aq.) is decreased by $1/4$ times, then equilibrium concentration of $F{e}^{3+}$ will increase by:

• A. 8 times
• B. 16 times
• C. 64 times
• D. 4 times

Answer 1

Answer: (C)

If the concentration of $O{H}^{-}$ions in the reaction:
$Fe(OH{)}_3(s)\rightleftharpoons F{e}^{3+}(aq)+3O{H}^{-}$(aq.)
is decreased by $1/4$ times, then equilibrium concentration of $F{e}^{3+}$ will increase by $64$ times.
$K={\left[F{e}^{3+}\right]}_i{\left[O{H}^{-}\right]}_i^3\dots .(1)$
$K={\left[F{e}^{3+}\right]}_f{\left[O{H}^{-}\right]}_f^3\dots (2)$
${\left[F{e}^{3+}\right]}_i$ and ${\left[O{H}^{-}\right]}_i^3$ are the inital concentrations.
${\left[F{e}^{3+}\right]}_f$ and ${\left[O{H}^{-}\right]}_f^3$ are the final concentrations.
From (1) and (2)
${\left[F{e}^{3+}\right]}_i{\left[O{H}^{-}\right]}_i^3={\left[F{e}^{3+}\right]}_f{\left[O{H}^{-}\right]}_f^3\dots ..(3)$
The concentration of $O{H}^{-}$ions is decreased by $1/4$ times
${\left[O{H}^{-}\right]}_f=\frac{{\left[O{H}^{-}\right]}_i}{4}\dots \text{ (4) }$
Substitute equation (4) in equation (3)
${\left[F{e}^{3+}\right]}_i{\left[O{H}^{-}\right]}_i^3={\left[F{e}^{3+}\right]}_f{\left(\frac{{\left[O{H}^{-}\right]}_i}{4}\right)}^3$
${\left[F{e}^{3+}\right]}_i={\left[F{e}^{3+}\right]}_f{\left(\frac{1}{4}\right)}^3$
$4^3\times {\left[F{e}^{3+}\right]}_i={\left[F{e}^{3+}\right]}_f$
$64\times {\left[F{e}^{3+}\right]}_i={\left[F{e}^{3+}\right]}_f$