(D) The precipitation of $ZnCl_2$ occurs when the ionic product $Q_{sp} > K_{sp}$.$Q_{sp} = [Zn^{2+}][Cl^{-}]^2$.For option $A$: $Q_{sp} = (10^{-8})(1

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$[Zn^{2+}] = 10^{-5} \ M$; $[Cl^{-}] = 10^{-3} \ M$

(D) The precipitation of $ZnCl_2$ occurs when the ionic product $Q_{sp} > K_{sp}$.$Q_{sp} = [Zn^{2+}][Cl^{-}]^2$.For option $A$: $Q_{sp} = (10^{-8})(10^{-8})^2 = 10^{-24} For option $B$: $Q_{sp} = (10^{-5})(10^{-4})^2 = 10^{-13} For option $C$: $Q_{sp} = (10^{-6})(10^{-5})^2 = 10^{-16} For option $D$: $Q_{sp} = (10^{-5})(10^{-3})^2 = 10^{-11} > 1.2 \times 10^{-12}$.Since $Q_{sp} > K_{sp}$ only in option $D$,precipitation will occur.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

Medium

Which of the following sets of concentrations will cause the precipitation of $ZnCl_2$ $(K_{sp} = 1.2 \times 10^{-12} \ M^3)$?

A
$[Zn^{2+}] = 10^{-8} \ M$; $[Cl^{-}] = 10^{-8} \ M$
B
$[Zn^{2+}] = 10^{-5} \ M$; $[Cl^{-}] = 10^{-4} \ M$
C
$[Zn^{2+}] = 10^{-6} \ M$; $[Cl^{-}] = 10^{-5} \ M$
D
$[Zn^{2+}] = 10^{-5} \ M$; $[Cl^{-}] = 10^{-3} \ M$

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Solubility product

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Which of the following sets of concentrations will cause the precipitation of $ZnCl_2$ $(K_{sp} = 1.2 \times 10^{-12} \ M^3)$?

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