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) rarrFe^(3+) +3OH^(-)`
`{:("Conc.", x, x ,3x):}`
`K_C = [Fe^(3+)] [OH^(-)]^(3)`
`= x (3x)^(3) " "...(i)`
If concentration is decreased by 1/4 then,
`Fe(OH)_(3) rarrFe^(3+) +3OH^(-)`
`{:(x.,x.,3x.):}`
`K_C = [Fe^(3+)][OH^(-0)]^3`
`=x.(3x.)^3 " "...(ii)`
`3x. =(3x)/4 " " ` (as concentration of `OH^(-)` decreased by 1/4)
`x=x.((3x)/4)^2`
Than equating (i) and (ii)
`x(3x)^3 = x.((3x)/4)^3`
`64x =x.. `