Question
If the concentration of $$O{H^ - }$$ $$ions$$ in the reaction, $$Fe{\left( {OH} \right)_3}\left( s \right) \rightleftharpoons $$ $$F{e^{3 + }}\left( {aq} \right) + 3O{H^ - }\left( {aq} \right)$$ is decreased by $$\frac{1}{4}$$ times, then equilibrium concentration of $$F{e^{3 + }}$$ will increase by
A.
8 times
B.
16 times
C.
64 times
D.
4 times
Answer :
64 times
Solution :
$$Fe{\left( {OH} \right)_3}\left( s \right) \rightleftharpoons $$ $$F{e^{3 + }}\left( {aq} \right) + 3O{H^ - }\left( {aq} \right)$$
$$K = \frac{{\left[ {F{e^{3 + }}} \right]{{\left[ {O{H^ - }} \right]}^3}}}{{\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}\,\,\,...{\text{(i)}}$$
To maintain equilibrium constant, let the concentration of $${F{e^{3 + }}}$$ is increased $$x$$ times, on decreasing the concentration of $${O{H^ - }}$$ by $$\frac{1}{4}$$ times
$$\eqalign{
& K = \frac{{\left[ {xF{e^{3 + }}} \right]{{\left[ {\frac{1}{4} \times O{H^ - }} \right]}^3}}}{{\left[ {Fe{{\left( {OH} \right)}_3}} \right]}}\,\,\,...{\text{(ii)}} \cr
& {\text{By dividing eq}}{\text{. (ii) by (i) we get}} \cr
& \frac{1}{{64}} \times x = 1\,\, \Rightarrow \,\,x = 64\,{\text{times}} \cr} $$