For the following reaction, equilibrium constant `K_(c)` at 298 K is `1.6xx10^(17)`
`Fe_((aq))^(2+)+S_((aq))^(2-) hArr FeS(s)`
When equal volume of 0.06 `M Fe^(2+)` and 0.2 `Ms^(-2)` solution are mixed, then equilibrium concentration of `Fe^(2+)` is found to be `Yxx10^(-17) M`. Y is

Answer

8.93

Explanation

For the reaction `Fe^(2+) (aq) +S^(2-)(aq) to Fe(S), K_(c)` is given by `k_(C)=([FeS])/([Fe^(2+)][S^(2-)])=(1)/([Fe^(2+)] [S^(2-)])`
Before mixing, `[Fe^(2+)] =0.06 M and [S^(2-)]=0.2M`, and after mixing equal volumes of the solutions, the concentration become `[Fe^(2+)]=0.03 M and [S^(2-)]=0.1M`
`Fe^(2+) (aq)+S^(2-) (aq) to FeS(aq)`
Now, we have Initial `"0.03 0.1"`
At eq. `"0.03 -x 0.1-x"`
Final `"Y 0.07"`
As `K_(c)` is very high `(1.6 xx 10^(17))` we make the approximation that 0.03-x=0 `rArr x=0.03`.
Now, `K_(c)=1/([Fe^(2+)] [S^(2-)]) rArr 1.6 xx 10^(17)=1/([Fe^(2+)] xx (0.07))`
Therefore, `[Fe^(2+)] =(1)/((1.6 xx 10^(17)) xx (0.07))=8.93 xx 10^(-17)`