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)` ions 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)(aq)`
If `[OH^(-)]` is decreased to `y/4`, to keep K constant, `x.(y/4)^(3)=xy^(3)orx.=64x`