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(s)hArrFe^(3+)(aq)+3OH^(-)(aq)`
`K=([Fe^(3+)][OH^(-)]^(3)]/([Fe(OH)_(3)])`
To maintain equilibrium constant, let the concentration of `Fe^(3+)` is increased x times, on decreasing the concentration of `OH^(-)` by `(1)/(4)` times
`K=([xFe^(3+)][OH^(-)]^(3))/([Fe(OH)_(3)])`.....(ii)
By dividing eq. (ii) by (i) we get
`(1)/(64)xxX=1rArrX=64` times.