Question

If concentration of OH- ions in the given reversible reaction:
Fe(OH)3<=> Fe3+ + 3OH-
is decreased up to 1/4th times. Then can you find how it will result in change of concentration of Fe3+?

Correct answer needed with proper!explanation

Answer 1

Given:

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

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

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

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

Answer 2

Nikhil hi!

বঙলাতে type করছি তুই translate করে নিস ।

Well তুই যেটা বলেছিলি সেটা যদি 5মাস আগে শুনতাম তাহলে ভাল হত। :D

But এখন শুনব।will try my best যাতে সব ঠিক হয়ে যাই।

ভালো থাকলে 13th july আসব সেদিন ভাই কে বলিস mssg করতে।

Ans টা report করে দিস :)