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`        If the concentration of OH- ions in the reaction Fe(OH)3 (s) Fe+3 (aq.) + 3OH-(aq.) is decreased by 1/4 times, then equilibrium concen tration of Fe+3 will increase by`
3 years ago

Vikas TU
11769 Points
```							Fe(OH)3 (s) =============> Fe+3 (aq.) + 3OH-(aq.) The eqm/. constant would always attain/constant at eqm.Hence,K = (OH-)^3*(Fe3+)/(Fe(OH)3)Now here,Fe^3+ is inversely proportional to the cube of the conc. of OH-.Hence,Fe3+ = k oh-^3Fe3+’ = k(oh-/3)^3Fe3+’ = Fe3+/64hence it increased by 1/64th of previous Fe3+ concentration.
```
3 years ago
605 Points
```							Dear student,Please find the solution to your problem. Fe(OH)3 (s) → Fe+3 (aq.) + 3OH-(aq.)The eqm. constant would remain constantHence,K = [OH-]3[Fe3+]/[Fe(OH)3]Now here,Fe3+ is inversely proportional to the cube of the conc. of OH-.Hence,Fe3+ = c [OH-]–3Fe3+’ = c ([OH-]/4)–3Fe3+’ = 64c [OH-]–3Fe3+’ = 64 Fe3+hence it increased by 64th of previous Fe3+ concentration.  Thanks and regards,Kushagra
```
6 days ago
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