(B) When equal volumes are mixed,the total volume doubles,so the concentration of each species is halved. $[Ca^{2+}] = \frac{0.02}{2} = 0.01 \ M = 10^

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No,precipitation will not occur.

(B) When equal volumes are mixed,the total volume doubles,so the concentration of each species is halved. $[Ca^{2+}] = \frac{0.02}{2} = 0.01 \ M = 10^{-2} \ M$. $[SO_4^{2-}] = \frac{0.00004}{2} = 0.00002 \ M = 2 \times 10^{-5} \ M$. The ionic product $Q_{sp} = [Ca^{2+}][SO_4^{2-}] = (10^{-2}) \times (2 \times 10^{-5}) = 2 \times 10^{-7}$. Since $Q_{sp} (2 \times 10^{-7}) < K_{sp} (2.4 \times 10^{-5})$,the solution is unsaturated and no precipitation of $CaSO_4$ will occur.

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

Medium

Equal volumes of $0.02 \ M$ $CaCl_2$ and $0.00004 \ M$ $Na_2SO_4$ solutions are mixed. Will a precipitation of $CaSO_4$ occur? $(K_{sp} = 2.4 \times 10^{-5})$

A
Yes,precipitation will occur.
B
No,precipitation will not occur.
C
The solution will be saturated.
D
Data is insufficient.

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

The solubility product of $Ag_2CO_3$ and $FeCO_3$ in water at $25 \, ^\circ C$ are $4 \times 10^{-12}$ and $2.5 \times 10^{-11}$ respectively. If $S_1$ and $S_2$ are their solubilities in water respectively when dissolved separately,then $S_1/S_2$ is

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$A$ sparingly soluble salt having general formula $A_{x}B_{y}$ and molar solubility $S$ is in equilibrium with its saturated solution. Derive a relationship between solubility and the solubility product for such salt.

Medium

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Calculate the solubility product of sparingly soluble salt $BA$ at $27^{\circ} C$ if its solubility is $1.8 \times 10^{-5} \ mol \ dm^{-3}$ at same temperature.

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If the solubility of $MX_2$ in water is $0.0002 \ mol/L$,then its $K_{sp}$ is:

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In an aqueous solution containing $AgCl_{(s)}$,$AgBr_{(s)}$,and $AgI_{(s)}$,find the ratio $[Cl^{-}] : [Br^{-}] : [I^{-}]$ at equilibrium if $K_{sp}(AgCl) = 10^{-10}$,$K_{sp}(AgBr) = 10^{-13}$,and $K_{sp}(AgI) = 10^{-16}$.

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Equal volumes of $0.02 \ M$ $CaCl_2$ and $0.00004 \ M$ $Na_2SO_4$ solutions are mixed. Will a precipitation of $CaSO_4$ occur? $(K_{sp} = 2.4 \times 10^{-5})$

Similar Questions

The solubility product of $Ag_2CO_3$ and $FeCO_3$ in water at $25 \, ^\circ C$ are $4 \times 10^{-12}$ and $2.5 \times 10^{-11}$ respectively. If $S_1$ and $S_2$ are their solubilities in water respectively when dissolved separately,then $S_1/S_2$ is

$A$ sparingly soluble salt having general formula $A_{x}B_{y}$ and molar solubility $S$ is in equilibrium with its saturated solution. Derive a relationship between solubility and the solubility product for such salt.

Calculate the solubility product of sparingly soluble salt $BA$ at $27^{\circ} C$ if its solubility is $1.8 \times 10^{-5} \ mol \ dm^{-3}$ at same temperature.

If the solubility of $MX_2$ in water is $0.0002 \ mol/L$,then its $K_{sp}$ is:

In an aqueous solution containing $AgCl_{(s)}$,$AgBr_{(s)}$,and $AgI_{(s)}$,find the ratio $[Cl^{-}] : [Br^{-}] : [I^{-}]$ at equilibrium if $K_{sp}(AgCl) = 10^{-10}$,$K_{sp}(AgBr) = 10^{-13}$,and $K_{sp}(AgI) = 10^{-16}$.

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