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For the following reaction, equilibrium ...

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

Text Solution

Verified by Experts

`Fe^(2+)(aq) + S^(2-)(aq) rarr FeS(s) K_(c) = 1.6 xx 10^(17)`
Since Kc >> 1 `x approx 0.03`
`:. [S^(2-)] = 0.1 - 0.03 = 0.07 M`
`:. [Fe^(2+)][S^(2-)] = (1/1.6xx10^(17))`
`[Fe^(2+)] = 1/1.6xx0.07 xx 10^(-17) = 8.928 xx 10^(-17) = y xx 10^(-17)`
y=8.93 (Round off )
or y = 8.92 (After Truncation)
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