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

Answer

A

Explanation

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