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

`Fe(OH)_3(s) hArr Fe^(3+)(aq) +3OH^(-)(aq)`
`K=[Fe^(3+)][OH^-]^3`
Activity of solid is unity .
Conc. Of `OH^-` ion in the reaction is decreased by 1/4 times, then equilibrium concentration of `Fe^(3+)` will be increased by `(4)^3` =64 times in order to keep the value of K to be constant .