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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)`.

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To solve the problem, we will follow these steps: ### Step 1: Determine the concentration of sulfide ions (S²⁻) Given that the concentration of ammonium sulfide \((NH_4)_2S\) is \(0.021 \, M\), we know that it dissociates into two ammonium ions \((NH_4^+)\) and one sulfide ion \((S^{2-})\). Therefore, the concentration of sulfide ions is also \(0.021 \, M\). ### Step 2: Write the solubility product expression for ZnS The solubility product constant \(K_{sp}\) for the dissociation of zinc sulfide \((ZnS)\) can be expressed as: \[ ...
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