For the reaction:
`2Fe^(3+)(aq)+(Hg_(2))^(2+)(aq) hArr 2Fe^(2+)(aq)`
`K_(c)=9.14xx10^(-6)` at `25^(@)C`. If the initial concentration of the ions are `Fe^(3+)=0.5 M, (Hg_(2))^(2+)=0.5 M, Fe^(2+)=0.03 M` and `Hg^(2+)=0.03 M,` what will be the concentration of ions at equilibrium.

Answer

B, C, D

Explanation

`{:(2Fe^(3+)(aq),+,(Hg_(2))^(2+)(aq),hArr,2Fe^(2+)(aq),+,2Hg^(2+)(aq)),(0.5,,-.5,,0.03,,0.03),(,,,,,,("conc before reaction")),((0.5-a),,(0.5-a//2),,(0.03+a),,(0.03+a)),(,,,,,,"Concentration after reaction at equilibrium"):}`
`K_(c)=9.14xx10^(-6)=([Fe^(2+)]^(2)[Hg^(+2)]^(2))/([Fe^(3+)]^(2)[Hg_(2)^(+2)])`
`:. 9.14xx10^(-6)=((0.03+a)^(2)(0.03+a)^(2))/((0.5-a)^(2)(0.5-a//2))`
`:. a=0.0027`
`:. [Fe^(3+)]=0.5-0.0027=0.4973 M`
`[Hg_(2)^(2+)]=0.5-0.0027/2=0.4987 M`
`[Fe^(2+)]=0.03+0.0027=0.0327 M`
`[Hg^(2+)]=0.03+0.0027=0.0327 M`