If the concentration of `OH^(-)` ions in the reaction
`Fe(OH)_(3)(s)hArrFe^(3+)(aq.)+3OH^(-)(aq.)`
is decreased by `1//4` times, then the equilibrium concentration of `Fe^(3+)` will increase by

Options

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

Answer

C

Explanation

`K= ([Fe^(3+)][OH^(-)]^(3))/([Fe (OH)_(3)(s)])=[Fe^(3+)] [OH^(-)]^(3)=xy^(3)(say)`
If `[OH^(-)]` is decreased to y/4, to keep K constant
` x'(y/4)^(3)= x +y ^(3) :. x' = 64 x `