Question

In the system : $Fe(OH{)}_{3\left(s\right)}\rightleftharpoons F{e}_{\left(aq\right)}^{3+}+3O{H}_{\left(aq\right)}^{-}$, decreasing the conc. of $O{H}^{-}$ ions $\frac{1}{3}$ times will cause the equilibrium conc of $F{e}^{3+}$ to increase __________ times.

• A. $3$
• B. $9$
• C. $18$
• D. $27$

Answer 1

Answer: (D)

$27$

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

The equilibrium constant,

$K=\frac{\left[F{e}^{3+}\right][O{H}^{-}{]}^3}{\left[Fe\right(OH{)}_3]}$

If $\left[O{H}^{-}\right]$ is increased $3$ times then,

$\Rightarrow K^1=\frac{\left[F{e}^{3+}\right]\left(3{)}^3\right[O{H}^{-}{]}^3}{\left[Fe\right(OH{)}_3]}$

$\Rightarrow K^1=27K$

It causes $\left[F{e}^{3+}\right]$ to increase $27$ times.