Question

If concentration of OH- ions in the given reversible reaction:
Fe(OH)3 Fe3+ + 3OH-

is decreased up to 1/4th times. Then find how it will result in change of concentration of Fe3+?


Correct answer needed with proper explanation! No spammers allowed

100 points!

Answer 1

Given:

• Concentration of OH- is decreased up to 1/4 time

━━━━━━━━━━━━━━

Need to find:

• Change in concentration of Fe3+

━━━━━━━━━━━━━━

Answer:

》We know, Fe(OH)3 is a solid.

thus [Fe(OH)3] =1

━━━━━━━━━━━━━━

Stoichiometric coeff of:

• Fe =1

• OH= 3

━━━━━━━━━━━━━━

let concentration be Fe initially

be : Fe1 

and finally be Fe2

now,

$Kc = \dfrac{{concentration \: of \: product}^{coeff} }{{concentration \: of \: product}^{coeff} }$

$\implies \: Kc1 = {[Fe1]\: ({OH})^{3} }$

━━━━━━━━━━━━━━

and

$Kc2 = {[Fe2] \times ({\dfrac{1}{4}[ OH]})^{3} }$

now

━━━━━━━━━━━━━━

Kc of a reversible reaction can't be different for same reaction at same temperature,

thus

$Kc1 = Kc2 \\ \implies \: [Fe1] {[OH]}^{3} =[ Fe2] \dfrac{ {[OH]}^{3} }{64}$

$\red{\bold{\boxed{\large{ \implies \: [Fe2] = 64[Fe1]}}}}$

━━━━━━━━━━━━━━

Thus The new concentration of Fe will be 64times of initial concentration.

Answer 2

Answer:

Given:

Concentration of OH- is decreased up to 1/4 time

━━━━━━━━━━━━━━

Need to find:

Change in concentration of Fe3+

━━━━━━━━━━━━━━

Answer:

》We know, Fe(OH)3 is a solid.

thus [Fe(OH)3] =1

━━━━━━━━━━━━━━

Stoichiometric coeff of:

Fe =1

OH= 3

━━━━━━━━━━━━━━

let concentration be Fe initially

be : Fe1

and finally be Fe2

now,

Kc = \dfrac{{concentration \: of \: product}^{coeff} }{{concentration \: of \: product}^{coeff} }Kc=

concentrationofproduct

coeff

concentrationofproduct

coeff

\implies \: Kc1 = {[Fe1]\: ({OH})^{3} }⟹Kc1=[Fe1](OH)

3

━━━━━━━━━━━━━━

and

Kc2 = {[Fe2] \times ({\dfrac{1}{4}[ OH]})^{3} }Kc2=[Fe2]×(

4

1

[OH])

3

now

━━━━━━━━━━━━━━

Kc of a reversible reaction can't be different for same reaction at same temperature,

thus

\begin{gathered}Kc1 = Kc2 \\ \implies \: [Fe1] {[OH]}^{3} =[ Fe2] \dfrac{ {[OH]}^{3} }{64}\end{gathered}

Kc1=Kc2

⟹[Fe1][OH]

3

=[Fe2]

64

[OH]

3

\red{\bold{\boxed{\large{ \implies \: [Fe2] = 64[Fe1]}}}}

⟹[Fe2]=64[Fe1]

━━━━━━━━━━━━━━

Thus The new concentration of Fe will be 64times of initial concentration.