Question

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

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

Answer 1

The correct option is B $64$ times

let us consider the problem.

The concentration of solids taken to be unity.

$Fe(OH{)}_3\left(s\right)F{e}^{3+}\left(aq\right)+3O{H}^{-}\left(aq\right)$

Therefore,

$K=\left[F{e}^{3+}\right]{\left[O{H}^{-}\right]}^3$

Hence, if $\left[O{H}^{-}\right]$ ions concentration is decreased by $\frac{1}{4}$ times. then the equilibrium concentration of $F{e}^{3+}$ will increase by $64$ times.