To `500 mL` of `0.150 M AgNO_(3)` solution were added `500 mL` of `1.09 M Fe^(2+)` solution and the reaction is allowed to reach an equilibrium at `25^(@)C`
`Ag^(o+)(aq)+Fe^(2+)(aq) hArr Fe^(3+)(aq)+Ag(s)`
For `25` mL of the solution, `30 mL` of `0.0832 M KMnO_(4)` was required for oxidation. Calculate the equilibrium constant for the the reaction `25^(@)C`.

Answer

A, B, C, D

Explanation

`{:(Ag^(o+)(aq),+,Fe^(2+)(aq),hArr,Fe^(3+)(aq),+,Ag(s)),(500xx0.150,,500xx0.109,,0,,0),(=75,,=545,,0,,0),(,,,,,,("Milli moles before reaction")),((75-x),,(545-x),,x,,x),(,,,,,,("Milli moles after reaction")):}`
`:' mM=mEq ("both" Ag^(o+)//Ag "and" Fe^(2+)//Fe^(3+) "have valency factor unity")`
`:' K_(c)=([Fe^(3+)])/([Ag^(o+)][Fe^(2+)])`
`:. K_(c)=(x/1000)/(((75-x)/1000)((545-x)/1000))` ...(i)
`:'` concenration `=("Millimole")/("Total volume")`
`[Ag^(o+)]=(75-x)/1000`
`[Fe^(2+)]=(545-x)/1000`
`[Fe^(3+)]=x/1000`
Now `25 mL` of mixture require `30 mL` of `0.0832 M` or `0.0832xx5 N KMnO_(4)`.
`:' Fe^(2+)` is oxidised by `KMnO_(4)`. `:.` Milliequivalent of `Fe^(2+)` left at equilibrium in `1000 mL`
=Milliequivalent of `KMnO_(4)` for `1000 mL`
`=(30xx0.0832xx5xx1000)/(25)=499.2`
`:. 545-x=499.2`
`:. x=545-499.2=45.8`
Thus, by equation (i),
`K_(c)=(45.8/1000)/(((75-45.8)/1000)xx((545-45.8)/1000))=(45.8xx1000)/(29.2xx499.2)`
`K_(c)=3.1420`