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Zn salt is mixed with (NH(4))(2)S of 0.0...

`Zn` salt is mixed with `(NH_(4))_(2)S` of `0.021M`. What amount of `Zn^(2+)` will remain uprecipitated in `12mL` of the solution? `K_(sp)` of `ZnS = 4.51 xx 10^(-24)`.

Text Solution

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`[(NH_(4))_(2)S] = 0.021M`
`:. [S^(2-)] = 0.021M`
`:.` At equilibrium, `[Zn^(2+)] [S^(2-)] = K_(sp) of ZnS`
`:. [Zn^(2+)] = (4.51 xx 10^(-24))/(0.021) = 2.15 xx 10^(-22)M`
`:. [Zn^(2+)]` left in solution `=2.15 xx 10^(-22) xx 65g//litre`
`= (2.15 xx 10^(-22)xx65xx12)/(1000) g//12 mL`
` = 1.677 xx 10^(-22) g(12 mL)^(-1)`
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