Solubility product Questions in English

(B) The dissociation of $Al_2(SO_4)_3$ in water is represented as:
$Al_2(SO_4)_3 (s) \rightleftharpoons 2Al^{3+} (aq) + 3SO_4^{2-} (aq)$
For a general salt $A_x B_y$,the solubility product expression is $K_{sp} = [A^{y+}]^x [B^{x-}]^y$.
Applying this to $Al_2(SO_4)_3$,where $x = 2$ and $y = 3$:
$K_{sp} = [Al^{3+}]^2 [SO_4^{2-}]^3$.

(B) For a salt of type $MX_2$,the dissociation is given by: $MX_2(s) ⇌ M^{2+}(aq) + 2X^-(aq)$.
Let the solubility be $S \ M$. Then,$[M^{2+}] = S$ and $[X^-] = 2S$.
The solubility product expression is $K_{sp} = [M^{2+}][X^-]^2$.
Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 4 \times 10^{-12}$,we have $4S^3 = 4 \times 10^{-12}$.
$S^3 = 10^{-12}$,so $S = \sqrt[3]{10^{-12}} = 10^{-4} \ M$.
Since $[M^{2+}] = S$,the concentration of $M^{2+}$ ions is $1.0 \times 10^{-4} \ M$.

(B) The equilibrium constant for a chemical reaction in which a solid ionic compound dissolves to yield its ions in solution is called the solubility product $(K_{sp})$.
For example,consider the equilibrium: $CuSO_{4(s)} \rightleftharpoons Cu^{2+}_{(aq)} + S{O_{4}}^{2-}_{(aq)}$
The solubility product for the above equilibrium is given by $K_{sp} = [Cu^{2+}][SO_{4}^{2-}]$,where $[Cu^{2+}]$ and $[SO_{4}^{2-}]$ are the molar concentrations of the ions in the saturated solution.

(B) For a sparingly soluble salt $MX_2$,the dissociation equilibrium is:
$MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^{-}(aq)$
Let the solubility of the salt be $S \, \text{mol L}^{-1}$. Then,$[M^{2+}] = S$ and $[X^{-}] = 2S$.
The solubility product expression is:
$K_{sp} = [M^{2+}][X^{-}]^2 = (S)(2S)^2 = 4S^3$
Given $K_{sp} = 1.0 \times 10^{-11}$,we have:
$4S^3 = 1.0 \times 10^{-11}$
$S^3 = 0.25 \times 10^{-11} = 2.5 \times 10^{-12}$
$S = \sqrt[3]{2.5 \times 10^{-12}} \approx 1.357 \times 10^{-4} \, \text{mol L}^{-1}$
Rounding to the nearest value,$S \approx 1.36 \times 10^{-4} \, \text{mol L}^{-1}$.

(C) The ionic product for each metal sulphide is calculated as: $Q = [M^{2+}][S^{2-}] = 10^{-3} \times 10^{-16} = 10^{-19}$.
Precipitation occurs when the ionic product exceeds the solubility product $(K_{sp})$.
Comparing the $K_{sp}$ values: $MnS (10^{-15}), FeS (10^{-23}), ZnS (10^{-20}), HgS (10^{-54})$.
Since $HgS$ has the lowest $K_{sp}$ $(10^{-54})$,it will be the first to satisfy the condition $Q > K_{sp}$ as the sulphide ion concentration increases,and it will precipitate first.

(A) The dissociation of $Mg(OH)_2$ in an aqueous solution is represented as:
$Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$
If the solubility is $x$,then the concentration of ions at equilibrium will be:
$[Mg^{2+}] = x$
$[OH^-] = 2x$
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [Mg^{2+}][OH^-]^2$
Substituting the values:
$K_{sp} = (x)(2x)^2 = (x)(4x^2) = 4x^3$
Therefore,the correct option is $A$.

(C) The dissolution equilibrium for $BaSO_4$ is: $BaSO_4(s) ⇌ Ba^{2+}(aq) + SO_4^{2-}(aq)$.
The solubility product expression is: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.
Given $[Ba^{2+}] = 0.01 \ M = 10^{-2} \ M$ and $K_{sp} = 1.0 \times 10^{-9}$.
To precipitate $BaSO_4$,the ionic product must exceed $K_{sp}$,so $[SO_4^{2-}] > \frac{K_{sp}}{[Ba^{2+}]}$.
$[SO_4^{2-}] > \frac{1.0 \times 10^{-9}}{10^{-2}} = 10^{-7} \ M$.
Since $H_2SO_4$ is a strong acid that dissociates completely into $2H^+$ and $SO_4^{2-}$,the concentration of $SO_4^{2-}$ ions is equal to the concentration of $H_2SO_4$.
Therefore,the concentration of $H_2SO_4$ must be greater than $10^{-7} \ M$.

(A) The dissociation of the salt is given by: $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^{-}(aq)$
Let the solubility be $s = 1.0 \times 10^{-5} \ mol \ L^{-1}$.
The concentrations of the ions are $[A^{2+}] = s$ and $[B^{-}] = 2s$.
The solubility product constant $K_{sp}$ is defined as: $K_{sp} = [A^{2+}][B^{-}]^2$
Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$
Calculating the value: $K_{sp} = 4 \times (1.0 \times 10^{-5})^3 = 4 \times 10^{-15}$

(B) The dissociation of $CaF_2$ is given by:
$CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)$
Let the solubility be $s$ mol/$L$.
At equilibrium,$[Ca^{2+}] = s$ and $[F^{-}] = 2s$.
The solubility product constant $K_{sp}$ is defined as:
$K_{sp} = [Ca^{2+}][F^{-}]^2$
Substituting the values:
$K_{sp} = (s)(2s)^2 = (s)(4s^2) = 4s^3$

(A) The dissociation of $Ag_2CrO_4$ in water is given by the equation:
$Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$
The solubility product constant $(K_{sp})$ is defined as the product of the molar concentrations of the ions,each raised to the power of its stoichiometric coefficient.
Therefore,$K_{sp} = [Ag^{+}]^2 [CrO_4^{2-}]$.

(D) The dissociation of $CaF_2$ is given by: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.
Let the solubility be $s = 2 \times 10^{-4} \, mol/L$.
The solubility product expression is $K_{sp} = [Ca^{2+}][F^-]^2 = (s)(2s)^2 = 4s^3$.
Substituting the value of $s$: $K_{sp} = 4 \times (2 \times 10^{-4})^3$.
$K_{sp} = 4 \times (8 \times 10^{-12}) = 3.2 \times 10^{-11}$.

(D) In an ammoniacal solution,the concentration of $S^{2-}$ ions is significantly high because $NH_3$ acts as a base and reacts with $H_2S$ to produce a high concentration of $S^{2-}$ ions.
Since the ionic product for both $MS$ and $NS$ will exceed their respective solubility products $(K_{sp})$,both sulphides will precipitate.
Therefore,the correct option is $D$.

(C) For the precipitation of $AgCl$ to occur,the ionic product $(Q_{sp})$ must exceed the solubility product $(K_{sp} = 10^{-10})$.
When equal volumes are mixed,the concentration of each ion is halved.
For option $(C)$: $[Ag^+] = \frac{10^{-5}}{2} \ M$ and $[Cl^-] = \frac{10^{-4}}{2} \ M$.
$Q_{sp} = [Ag^+][Cl^-] = (\frac{10^{-5}}{2}) \times (\frac{10^{-4}}{2}) = \frac{10^{-9}}{4} = 0.25 \times 10^{-9} = 2.5 \times 10^{-10}$.
Since $2.5 \times 10^{-10} > 10^{-10}$,the ionic product exceeds $K_{sp}$,and precipitation occurs.

(D) The dissociation of silver chromate is given by: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$.
Let the solubility of $Ag_2CrO_4$ be $S = 2 \times 10^{-8} \, mol \, dm^{-3}$.
In the presence of $0.01 \, M$ $K_2CrO_4$,the concentration of $CrO_4^{2-}$ ions is approximately $0.01 \, M$ due to the common ion effect.
The concentration of $Ag^+$ ions is $[Ag^+] = 2S = 2 \times (2 \times 10^{-8}) = 4 \times 10^{-8} \, M$.
The solubility product $K_{sp}$ is calculated as: $K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$.
Substituting the values: $K_{sp} = (4 \times 10^{-8})^2 \times (0.01) = (16 \times 10^{-16}) \times 10^{-2} = 16 \times 10^{-18}$.

(A) The dissociation of $SnS_2$ in water is represented by the equation:
$SnS_2(s) \rightleftharpoons Sn^{4+}(aq) + 2S^{2-}(aq)$
The solubility product constant $(K_{sp})$ is defined as the product of the molar concentrations of the ions,each raised to the power of its stoichiometric coefficient.
Therefore,$K_{sp} = [Sn^{4+}] [S^{2-}]^2$.

(D) Calcium oxalate $(CaC_2O_4)$ is a salt of a strong base $(Ca(OH)_2)$ and a weak acid (oxalic acid,$H_2C_2O_4$).
It dissolves in strong mineral acids like $HCl$ and $HNO_3$ because the oxalate ions $(C_2O_4^{2-})$ react with $H^+$ ions to form the weak acid $H_2C_2O_4$,shifting the equilibrium to the right.
However,acetic acid $(CH_3COOH)$ is a weak acid and does not provide a sufficiently high concentration of $H^+$ ions to significantly shift the equilibrium and dissolve the precipitate.

(A) The dissociation of $BaCl_2$ is represented as: $BaCl_2(s) \rightleftharpoons Ba^{2+}(aq) + 2Cl^-(aq)$.
Let the solubility be $S \ mol/L$. Then $[Ba^{2+}] = S$ and $[Cl^-] = 2S$.
The solubility product expression is $K_{sp} = [Ba^{2+}][Cl^-]^2$.
Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 4 \times 10^{-9}$,we have $4S^3 = 4 \times 10^{-9}$.
$S^3 = 10^{-9}$.
Taking the cube root,$S = 10^{-3} \ mol/L$.

(D) The solubility of alkaline earth metal hydroxides increases down the group from $Be$ to $Ba$.
Since $Be(OH)_2$ is the least soluble among the given hydroxides,it has the lowest concentration of ions in a saturated solution.
Therefore,$Be(OH)_2$ has the lowest value of the solubility product $(K_{sp})$ at $25\,^oC$.

(D) The solubility of ionic compounds depends on the lattice energy and hydration energy.
$Ag_2S$ has a very low solubility product constant $(K_{sp})$ compared to the other silver halides.
Due to the high covalent character and extremely strong lattice energy of $Ag_2S$,it is the least soluble in water among the given options.

(A) Solubility product $(K_{sp})$ refers to the product of the molar concentrations of the ions present in a saturated solution of a sparingly soluble ionic compound,each raised to the power of its stoichiometric coefficient.
For a general salt $A_{a}B_{b(s)} \rightleftharpoons aA_{(aq)}^{b+} + bB_{(aq)}^{a-}$,
The equilibrium constant is $K_{c} = \frac{[A^{b+}]^a [B^{a-}]^b}{[A_{a}B_{b}]}$.
Since the concentration of the pure solid $[A_{a}B_{b}]$ is constant,the solubility product constant is defined as $K_{sp} = [A^{b+}]^a [B^{a-}]^b$.
Thus,it is the product of the ionic concentrations in a saturated solution.

(A) For a salt of the type $AB$,the solubility product is given by ${K_{sp}} = S^2$,where $S$ is the solubility.
Given ${K_{sp}} = 6.4 \times 10^{-5}$.
Therefore,$S = \sqrt{{K_{sp}}} = \sqrt{6.4 \times 10^{-5}} = \sqrt{64 \times 10^{-6}}$.
$S = 8 \times 10^{-3} \, mol/L$.

(B) The solubility of $BaSO_4$ in $g/L$ is given as $2.33 \times 10^{-3} \ g/L$.
First,convert the solubility into $mol/L$ $(S)$:
$S = \frac{\text{solubility in } g/L}{\text{molar mass}} = \frac{2.33 \times 10^{-3}}{233} = 1 \times 10^{-5} \ mol/L$.
Since $BaSO_4$ is a $1:1$ electrolyte,it dissociates as: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = S \times S = S^2$.
Substituting the value of $S$:
$K_{sp} = (1 \times 10^{-5})^2 = 1 \times 10^{-10}$.

(D) $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$
$NaCl(aq) \rightarrow Na^+(aq) + Cl^-(aq)$
Since $NaCl$ is a strong electrolyte,the concentration of $Cl^-$ ions from $NaCl$ is $0.2 \ M$.
The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.
Substituting the values: $1.20 \times 10^{-10} = [Ag^+] \times 0.2$.
Therefore,the solubility of $AgCl$ is $[Ag^+] = \frac{1.20 \times 10^{-10}}{0.2} = 6 \times 10^{-10} \ M$.

(C) The solubility product expression for $BaSO_4$ is $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.
Given $K_{sp} = 1.5 \times 10^{-9}$ and $[Ba^{2+}] = 0.01 \ M$.
Precipitation starts when the ionic product exceeds the solubility product.
$[SO_4^{2-}] = \frac{K_{sp}}{[Ba^{2+}]} = \frac{1.5 \times 10^{-9}}{0.01} = 1.5 \times 10^{-7} \ M$.
Since $1.5 \times 10^{-7} \ M$ is approximately $10^{-7} \ M$,the correct option is $C$.

(C) The dissociation of silver chromate is given by: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^+(aq) + CrO_4^{2-}(aq)$
The solubility product expression is: $K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$
Given that $[Ag^+] = 1.5 \times 10^{-4} \, M$,the concentration of chromate ions is half the concentration of silver ions due to stoichiometry:
$[CrO_4^{2-}] = \frac{[Ag^+]}{2} = \frac{1.5 \times 10^{-4}}{2} = 0.75 \times 10^{-4} \, M$
Substituting these values into the $K_{sp}$ expression:
$K_{sp} = (1.5 \times 10^{-4})^2 \times (0.75 \times 10^{-4})$
$K_{sp} = (2.25 \times 10^{-8}) \times (0.75 \times 10^{-4})$
$K_{sp} = 1.6875 \times 10^{-12}$

(C) The dissociation of $PbCl_2$ is represented as: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.
Let the solubility of $PbCl_2$ be $S \ mol/L$.
Then,$[Pb^{2+}] = S$ and $[Cl^-] = 2S$.
The solubility product constant expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2$.
Substituting the values: $K_{sp} = (S)(2S)^2$.
$K_{sp} = S \times 4S^2 = 4S^3$.
Therefore,$S^3 = \frac{K_{sp}}{4}$,which gives $S = \sqrt[3]{\frac{K_{sp}}{4}}$.

(B) For the dissolution of $AgCl$ in water,the equilibrium is represented as:
$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$
Let the solubility of $AgCl$ be $S \, mol/L$.
Then,$[Ag^{+}] = S$ and $[Cl^{-}] = S$.
The solubility product expression is given by:
$K_{sp} = [Ag^{+}][Cl^{-}] = S \times S = S^2$
Given $K_{sp} = 1.44 \times 10^{-4}$.
$S^2 = 1.44 \times 10^{-4}$
$S = \sqrt{1.44 \times 10^{-4}} = 1.20 \times 10^{-2} \, M$.
Therefore,the solubility of silver chloride is $1.20 \times 10^{-2} \, M$.

(D) For a sparingly soluble salt of the type $BA_2$,the dissociation reaction is:
$BA_2(s) \rightleftharpoons B^{2+}(aq) + 2A^{-}(aq)$
If the solubility is $x \ mol \ L^{-1}$,then the concentration of $B^{2+}$ is $x \ mol \ L^{-1}$ and the concentration of $A^{-}$ is $2x \ mol \ L^{-1}$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [B^{2+}][A^{-}]^2$
$K_{sp} = (x)(2x)^2$
$K_{sp} = (x)(4x^2)$
$K_{sp} = 4x^3$

(A) The dissociation of $Ag_{2}CrO_{4}$ is given by: $Ag_{2}CrO_{4} \rightleftharpoons 2Ag^{+} + CrO_{4}^{2-}$.
Let the solubility be $S \ M$. Then,$[Ag^{+}] = 2S$ and $[CrO_{4}^{2-}] = S$.
The solubility product expression is $K_{sp} = [Ag^{+}]^{2}[CrO_{4}^{2-}]$.
Substituting the values: $K_{sp} = (2S)^{2}(S) = 4S^{3}$.
Given $K_{sp} = 32 \times 10^{-12}$,we have $4S^{3} = 32 \times 10^{-12}$.
$S^{3} = 8 \times 10^{-12}$.
$S = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \ M$.
Since $[CrO_{4}^{2-}] = S$,the concentration of $CrO_{4}^{2-}$ ions is $2 \times 10^{-4} \ M$.

(C) The solubility product $(K_{sp})$ values for the chromates of alkaline earth metals are in the order: $K_{sp}(CaCrO_4) > K_{sp}(SrCrO_4) > K_{sp}(BaCrO_4)$.
Since $BaCrO_4$ has the lowest $K_{sp}$ value,it requires the lowest concentration of $CrO_4^{2-}$ ions to exceed its solubility product.
Therefore,$BaCrO_4$ will precipitate first upon the addition of $CrO_4^{2-}$ ions to the solution containing these cations.
The reaction is: $Ba^{2+}(aq) + CrO_4^{2-}(aq) \to BaCrO_4(s) \downarrow$ (Yellow precipitate).

(D) For a sparingly soluble salt of the type $AB$,the dissociation is given by: $AB(s) \rightleftharpoons A^+(aq) + B^-(aq)$.
Let the molar solubility be $S \ mol/L$.
Then,$K_{sp} = [A^+][B^-] = S \times S = S^2$.
Given $K_{sp} = 1.21 \times 10^{-6}$.
Therefore,$S = \sqrt{K_{sp}} = \sqrt{1.21 \times 10^{-6}} = 1.1 \times 10^{-3} \ M$.
Thus,the correct option is $D$.

(B) Precipitation occurs when the ionic product is greater than the solubility product $(K_{sp})$.
If the ionic product is less than $K_{sp}$,the solution is unsaturated.
If the ionic product is equal to $K_{sp}$,the solution is saturated.

(C) For a binary electrolyte $AB$,the dissociation is represented as $AB(s) \rightleftharpoons A^+(aq) + B^-(aq)$.
Let the solubility be $S \ mol/L$.
Then,$[A^+] = S$ and $[B^-] = S$.
The solubility product is given by $K_{sp} = [A^+][B^-] = S \times S = S^2$.
Therefore,$S = \sqrt{K_{sp}}$.

(D) The solubility product $(K_{sp})$ represents the maximum concentration of ions that can exist in a saturated solution at a given temperature.
When the ionic product $(Q_{ip})$ of a solution is less than $K_{sp}$,the solution is unsaturated.
When $Q_{ip} = K_{sp}$,the solution is saturated.
When $Q_{ip} > K_{sp}$,the solution becomes supersaturated,and the excess solute precipitates out of the solution.

(C) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
$CaCl_2$ is a strong electrolyte and dissociates completely as: $CaCl_2(aq) \rightarrow Ca^{2+}(aq) + 2Cl^-(aq)$.
Since the concentration of $CaCl_2$ is $0.04 \ m$,the concentration of $Cl^-$ ions from $CaCl_2$ is $2 \times 0.04 \ m = 0.08 \ m$.
Let the solubility of $AgCl$ be $s$. Then $[Ag^+] = s$ and $[Cl^-] = (0.08 + s) \approx 0.08 \ m$ (since $s$ is very small).
The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.
$4.0 \times 10^{-10} = s \times 0.08$.
$s = \frac{4.0 \times 10^{-10}}{0.08} = 5.0 \times 10^{-9} \ m$.

(B) The dissociation of $BaSO_4$ is represented as: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$.
Let the solubility be $S \ mol/L$.
Then,$[Ba^{2+}] = S$ and $[SO_4^{2-}] = S$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S \times S = S^2$.
Given $K_{sp} = 1.5 \times 10^{-9}$.
Therefore,$S^2 = 1.5 \times 10^{-9}$.
$S = \sqrt{1.5 \times 10^{-9}} = \sqrt{15 \times 10^{-10}} \approx 3.87 \times 10^{-5} \ mol/L$.
Rounding to two significant figures,$S \approx 3.9 \times 10^{-5} \ mol/L$.

(D) The dissociation of calcium hydroxide is given by: $Ca(OH)_2 \rightleftharpoons Ca^{2+} + 2OH^-$
Let the solubility be $S$. Then $[Ca^{2+}] = S$ and $[OH^-] = 2S$.
The solubility product expression is: $K_{sp} = [Ca^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.
Given that $S = \sqrt{3}$,we substitute this value into the expression:
$K_{sp} = 4 \times (\sqrt{3})^3 = 4 \times 3\sqrt{3} = 12\sqrt{3}$.

(D) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$.
If $S$ is the solubility of $PbCl_2$,then $[Pb^{2+}] = S$ and $[Cl^{-}] = 2S$.
The solubility product constant is $K_{sp} = [Pb^{2+}][Cl^{-}]^2$.
Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.
Given $S = 2 \times 10^{-2} \, mol/L$,we have:
$K_{sp} = 4 \times (2 \times 10^{-2})^3 = 4 \times (8 \times 10^{-6}) = 3.2 \times 10^{-5}$.

(A) The dissociation of silver sulphide is given by: $Ag_2S(s) \rightleftharpoons 2Ag^+(aq) + S^{2-}(aq)$.
Let the solubility be $S \ mol/L$.
Then,$[Ag^+] = 2S$ and $[S^{2-}] = S$.
The solubility product expression is $K_{sp} = [Ag^+]^2 [S^{2-}] = (2S)^2(S) = 4S^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$.
$4S^3 = 3.2 \times 10^{-11}$.
$S^3 = \frac{3.2 \times 10^{-11}}{4} = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
$S = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \ mol/L$.

(D) The dissociation of $CaCO_3$ is given by:
$CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$
For a binary electrolyte of the type $AB$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula:
$K_{sp} = [Ca^{2+}][CO_3^{2-}] = S \times S = S^2$
Given solubility $S = 3.05 \times 10^{-4} \, mol/L$.
$K_{sp} = (3.05 \times 10^{-4})^2$
$K_{sp} = 9.3025 \times 10^{-8} \approx 9.3 \times 10^{-8}$

(C) The dissolution of $BaF_2$ is represented by the equilibrium: $BaF_2(s) \rightleftharpoons Ba^{2+}(aq) + 2F^{-}(aq)$.
Let $s$ be the solubility of $BaF_2$ in the presence of $Ba(NO_3)_2$.
From the stoichiometry of the reaction,the concentration of fluoride ions produced is $[F^{-}] = 2s$.
Rearranging this expression to solve for solubility $s$,we get $s = \frac{1}{2}[F^{-}]$.

(C) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.
Let the solubility of $PbCl_2$ be $S = 6.3 \times 10^{-3} \ mol/L$.
At equilibrium,the concentration of $Pb^{2+}$ is $S$ and the concentration of $Cl^-$ is $2S$.
Therefore,$[Pb^{2+}] = 6.3 \times 10^{-3} \ mol/L$ and $[Cl^-] = 2 \times (6.3 \times 10^{-3}) = 12.6 \times 10^{-3} \ mol/L$.
The solubility product $K_{sp}$ is defined as: $K_{sp} = [Pb^{2+}][Cl^-]^2$.
Substituting the values: $K_{sp} = (6.3 \times 10^{-3}) \times (12.6 \times 10^{-3})^2$.

(A) The dissociation of $Ag_2SO_4$ is given by: $Ag_2SO_4(s) \rightleftharpoons 2Ag^+(aq) + SO_4^{2-}(aq)$.
Let the solubility be $s = 2.5 \times 10^{-2} \ M$.
The solubility product expression is $K_{sp} = [Ag^+]^2 [SO_4^{2-}] = (2s)^2 (s) = 4s^3$.
Substituting the value of $s$: $K_{sp} = 4 \times (2.5 \times 10^{-2})^3$.
$K_{sp} = 4 \times (15.625 \times 10^{-6}) = 62.5 \times 10^{-6}$.

(C) For sodium chloride $(NaCl)$,the dissociation in water is:
$NaCl_{(s)} \rightleftharpoons Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
Let the solubility be $s$. Then,$[Na^{+}] = s$ and $[Cl^{-}] = s$.
The solubility product is:
$K_{sp} = [Na^{+}][Cl^{-}] = s \times s = s^2$
Given $K_{sp} = 36 \text{ mol}^2/\text{L}^2$:
$s^2 = 36$
$s = \sqrt{36} = 6 \text{ mol/L}$.

(D) The dissociation reaction for lead iodide is: $PbI_2(s) ⇌ Pb^{2+}(aq) + 2I^{-}(aq)$.
For a salt of type $AB_2$,the solubility product $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][I^{-}]^2$.
If $S$ is the solubility,then $[Pb^{2+}] = S$ and $[I^{-}] = 2S$.
Therefore,$K_{sp} = (S)(2S)^2 = 4S^3$.
Given $S = 2 \times 10^{-3} \ mol \ L^{-1}$.
Substituting the value of $S$ into the expression: $K_{sp} = 4 \times (2 \times 10^{-3})^3 = 4 \times (8 \times 10^{-9}) = 32 \times 10^{-9}$.

(B) When equal volumes are mixed,the concentration of each ion is halved.
For option $(A)$: $[Ca^{2+}] = 0.5 \times 10^{-4} \ M$,$[F^-] = 0.5 \times 10^{-4} \ M$.
Ionic Product $(IP)$ = $[Ca^{2+}][F^-]^2 = (0.5 \times 10^{-4}) \times (0.5 \times 10^{-4})^2 = 1.25 \times 10^{-13}$.
Since $IP < K_{sp}$,no precipitate is formed.
For option $(B)$: $[Ca^{2+}] = 0.5 \times 10^{-2} \ M$,$[F^-] = 0.5 \times 10^{-3} \ M$.
Ionic Product $(IP)$ = $[Ca^{2+}][F^-]^2 = (0.5 \times 10^{-2}) \times (0.5 \times 10^{-3})^2 = 0.5 \times 10^{-2} \times 0.25 \times 10^{-6} = 1.25 \times 10^{-9}$.
Since $IP > K_{sp}$ $(1.25 \times 10^{-9} > 1.7 \times 10^{-10})$,precipitation occurs.

(C) The precipitation of $BaSO_4$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.
The equilibrium expression is: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$.
The solubility product expression is: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.
Given that $[Ba^{2+}] = 1.0 \times 10^{-4} \ M$ and $K_{sp} = 4 \times 10^{-10}$.
Substituting the values: $4 \times 10^{-10} = (1.0 \times 10^{-4}) \times [SO_4^{2-}]$.
Solving for $[SO_4^{2-}]$: $[SO_4^{2-}] = \frac{4 \times 10^{-10}}{1.0 \times 10^{-4}} = 4 \times 10^{-6} \ M$.

(C) The dissociation of the salt $AB_2$ is given by: $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^{-}(aq)$.
Let the solubility be $S \ mol \ L^{-1}$. Then $[A^{2+}] = S$ and $[B^{-}] = 2S$.
The solubility product expression is $K_{sp} = [A^{2+}][B^{-}]^2$.
Substituting the values: $K_{sp} = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 4 \times 10^{-12}$,we have $4S^3 = 4 \times 10^{-12}$.
$S^3 = 10^{-12}$,which gives $S = (10^{-12})^{1/3} = 10^{-4} \ mol \ L^{-1}$.

(D) For a salt $AB$,the solubility product expression is $K_{sp} = [A][B]$.
Given $K_{sp} = 1 \times 10^{-8}$ and $[A] = 10^{-3} \ M$.
Precipitation occurs when the ionic product exceeds the solubility product $([A][B] > K_{sp})$.
$[B] > \frac{K_{sp}}{[A]} = \frac{1 \times 10^{-8}}{10^{-3}} = 1 \times 10^{-5} \ M$.
Therefore,the salt will precipitate when the concentration of $B$ becomes more than $10^{-5} \ M$.