Question

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

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

Answer 1

Answer: (C)

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