An industrial plant processes its waste water through a sedimentation tank that removes hazardous metals by precipitating them as insoluble carbonate salts.If sodium carbonate is gradually added to the tank,what would be the order of precipitation of the metals,Pb²⁺,Cu²⁺,Hg₂²⁺,and Zn²⁺ if each is $\approx$ 1.0 $\times$ 10^-⁴ M? ( K_sp PbCO₃ = 7.4 $\times$10^-¹⁴,K_sp CuCO₃ = 1.4 $\times$10^-¹⁰,K_sp Hg₂CO₃ = 8.9 $\times$ 10^-¹⁷,and K_sp ZnCO₃ = 1.4 $\times$ 10^-¹¹)

Question

Quizplus · Accepted Answer

According to the given values of solubility product constant (Ksp), HgS has the highest Ksp, followed by PbS, ZnS, and CuS in decreasing order. Therefore, Hg2+, Pb2+, Zn2+, and Cu2+ ions will precipitate in the order of their decreasing Ksp values, i.e., HgS first, then PbS, ZnS, and CuS. Therefore, the correct order of precipitation of the metals is HgS, PbS, ZnS, and CuS, which is given in option B.

An Industrial Plant Processes Its Waste Water Through a Sedimentation ≈\approx≈

An Industrial Plant Processes Its Waste Water Through a Sedimentation $\approx$