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

Options

A. not changed
B. also decreased by `1//4` time
C. increase by 4 times
D. increase by 64 times

Explanation

`K_(C)=([Fe^(3+)][OH^(-)]^(3))/([Fe(OH)_(3)(s)])=[Fe^(3+)][OH^(-)]^(3)]` [`because` Concentration of solids is constant) When concentration of `OH^(-)` ions decreased by `1.4` times, then
`K_(C)=[Fe^(3+)]xx(([OH^(-)])/(4))^(3)=(1)/(64)[Fe^(3+)][OH^(-)]^(3)`
To maintain `K_(C)` as constant, comcentration of `Fe^(3+)` will increase by 64 times.