Question

a solution contains 1.41×10⁻² m cobalt(ii) nitrate and 1.50×10⁻² m zinc acetate. solid sodium sulfide is added slowly to this mixture. what is the concentration of zinc ion when cobalt(ii) ion begins to precipitate? zn²⁺ = m

Explanation:

Step1: Recall Ksp values

Assume $K_{sp}(CoS)=5.9\times10^{-21}$, $K_{sp}(ZnS)=2.0\times10^{-25}$

Step2: Calculate $[S^{2-}]$ for CoS precipitation

$[S^{2-}]=\frac{K_{sp}(CoS)}{[Co^{2+}]}=\frac{5.9\times10^{-21}}{1.41\times10^{-2}}≈4.18\times10^{-19}$ M

Step3: Find $[Zn^{2+}]$ at this $[S^{2-}]$

$[Zn^{2+}]=\frac{K_{sp}(ZnS)}{[S^{2-}]}=\frac{2.0\times10^{-25}}{4.18\times10^{-19}}≈4.8\times10^{-7}$ M

Answer:

$4.8\times10^{-7}$