(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 2 Equilibrium: 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

The $K_{sp}$ value for $Gd(OH)_3$ is $2.7 \times 10^{-23}$. At what $pH$ will $Gd(OH)_3$ begin to precipitate? (Given: $\log 3 = 0.48$)

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The moles of $Ag^{+}$ which must be added to decrease the concentration of $Cl^{-}$ from $4 \times 10^{-5} \ M$ to $10^{-5} \ M$ in $100 \ mL$ solution,if $K_{sp}$ for $AgCl$ is $10^{-10} \ M^2$ at $25 \ ^oC$.

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At $310 \, K$,the solubility of $CaF_{2}$ in water is $2.34 \times 10^{-3} \, g / 100 \, mL$. The solubility product of $CaF_{2}$ is $x \times 10^{-8} \, (mol / L)^{3}$. Find the value of $x$. (Given molar mass: $CaF_{2} = 78 \, g \, mol^{-1}$)

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The solubility product of $AgCl$ is $4.0 \times 10^{-10}$ at $298 \ K$. The solubility of $AgCl$ in $0.04 \ m \ CaCl_2$ will be

Medium

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The solubility product of $PbS$ is $3.4 \times 10^{-28}$. If $[Pb^{2+}] = 1 \times 10^{-2} \ mol/L$,what concentration of $[S^{2-}]$ is required to precipitate $PbS$?

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

Similar Questions

The $K_{sp}$ value for $Gd(OH)_3$ is $2.7 \times 10^{-23}$. At what $pH$ will $Gd(OH)_3$ begin to precipitate? (Given: $\log 3 = 0.48$)

The moles of $Ag^{+}$ which must be added to decrease the concentration of $Cl^{-}$ from $4 \times 10^{-5} \ M$ to $10^{-5} \ M$ in $100 \ mL$ solution,if $K_{sp}$ for $AgCl$ is $10^{-10} \ M^2$ at $25 \ ^oC$.

At $310 \, K$,the solubility of $CaF_{2}$ in water is $2.34 \times 10^{-3} \, g / 100 \, mL$. The solubility product of $CaF_{2}$ is $x \times 10^{-8} \, (mol / L)^{3}$. Find the value of $x$. (Given molar mass: $CaF_{2} = 78 \, g \, mol^{-1}$)

The solubility product of $AgCl$ is $4.0 \times 10^{-10}$ at $298 \ K$. The solubility of $AgCl$ in $0.04 \ m \ CaCl_2$ will be

The solubility product of $PbS$ is $3.4 \times 10^{-28}$. If $[Pb^{2+}] = 1 \times 10^{-2} \ mol/L$,what concentration of $[S^{2-}]$ is required to precipitate $PbS$?

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