Question

In the system; Fe(OH)3(s) Fe3 +(aq) + 3OH-(aq), decreasing the concentration of OH- ions 1/3 times will cause the equilibrium concentration of Fe3 + to increase by:-


a) 3 times
b) 9 times
c) 18 times
d) 27 times
Kindly answer sir/mam.

Answer 1

The correct option is d.

For the reaction iron hydroxide dissociates to iron (iii) and hydroxide ion as follows:

$Fe (OH)_3 (s)  =  Fe^3^+ (aq) + 3 OH^- (aq)$

The equilibrium constant can be shown as:

$K = [Fe^3^+ (aq)] [ OH^- (aq)]^3 / [Fe (OH)_3]$

if concentration of hydroxide reduces by 3 times then$If, [ OH^- (aq)] = (1 /3 ) [ OH^- (aq)]$

$K^’ = [Fe^3^+ (aq)] [ (1 /3 )  OH^- (aq)]^3 / [Fe (OH)_3]$

$= (1/27) * [Fe^3^+ (aq)] [ OH^- (aq)]^3 / [Fe (OH)_3]$

$= (1/27) * K$

$K = 27 K^’$

Thus, equilibrium concentration increases by 27 times.

Answer 2

Answer:

d)27 times

Explanation:

Fe(OH)3(s)<---->Fe^3+ +3OH^-

At equilibrium, conc. of solid is taken as unity.

Kc=[Fe^3+][OH^1-]^3

Kc only depends on temperature.

For this reaction, temperature remains constant ,so Kc is also constant.

For second reaction,

Conc. of products=

[Fe^3+][1/3OH^1-]^3

As Kc is same, we can equate this equation to conc. if products at equilibrium.

[Fe^3+][OH^1-]^3=[Fe^3+][1/3OH^1-]^3

This gives us,

[Fe^3+][OH^-1]^3= 1/27[Fe^3+][OH^1-]^3

Cancelling out OH^1- from each side, we get,

[Fe^3+] (at equilibrium)×27= [Fe^3+](after changing conditions)

Thus, equilibrium conc. of Fe^3+ increases by 27 times.