If the concentration of OH^- ions in the reaction Fe(OH)_(3(s)) hArr Fe_((aq))^(3+) + 3OH_((aq))^(-)is decreased by 1/4 times,then equilibrium concentration of Fe^(3+) will increase by:

If the concentration of OH^- ions in the reaction Fe(OH)_(3(s)) hArr F

1.	If the concentration of OH^- ions in the reaction Fe(OH)_(3(s)) hArr Fe_((aq))^(3+) + 3OH_((aq))^(-)is decreased by 1/4 times,then equilibrium concentration of Fe^(3+) will increase by:
Answer» 8 times 16 times 64 times 4 times Solution :For this reaction `K_(eq)` is GIVEN by `K=([Fe^(3+)][OH^-]^3)/([Fe(OH)_3])` `=[Fe^(3+)][OH^-]^3` [ `because` [solid]=1] If `[OH^+]` is decreased by `1/4` times then for reaction equilibrium constant to remain constant, we have to increase the CONCENTRATION of `[Fe^(3+)]` by a factor of `4^3` i.e. 4 x 4 x 4 =64 .