Solubility product Questions in English

(D) $AgCl_{(s)} \leftrightarrow Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$
Initially: $[Cl^{-}] = 4 \times 10^{-5} \ M$
Final: $[Cl^{-}] = 10^{-5} \ M$
Decrease in $[Cl^{-}] = 3 \times 10^{-5} \ M$
Moles of $Cl^{-}$ precipitated $= 3 \times 10^{-5} \ M \times 0.1 \ L = 3 \times 10^{-6} \ mol$
Since $AgCl$ precipitates in $1:1$ ratio,moles of $Ag^{+}$ used for precipitation $= 3 \times 10^{-6} \ mol$
Remaining $Ag^{+}$ in solution to maintain equilibrium $= K_{sp} / [Cl^{-}]_{final} = 10^{-10} / 10^{-5} = 10^{-5} \ M$
Moles of $Ag^{+}$ remaining in solution $= 10^{-5} \ M \times 0.1 \ L = 10^{-6} \ mol$
Total moles of $Ag^{+}$ to be added $= 3 \times 10^{-6} + 10^{-6} = 4 \times 10^{-6} \ mol$

(A) The concentration of $Mg^{2+}$ ions is $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.
The solubility product expression for $Mg(OH)_2$ is $K_{sp} = [Mg^{2+}][OH^-]^2$.
Substituting the given values: $1 \times 10^{-12} = (10^{-2}) \times [OH^-]^2$.
Solving for $[OH^-]$: $[OH^-]^2 = 10^{-10}$,so $[OH^-] = 10^{-5} \ M$.
Calculating $pOH$: $pOH = -\log[OH^-] = -\log(10^{-5}) = 5$.
Finally,calculating $pH$: $pH = 14 - pOH = 14 - 5 = 9$.

(C) $AgCl$ is a sparingly soluble salt in water. However,it dissolves in aqueous ammonia $(NH_3)$ due to the formation of a soluble complex compound,diamminesilver$(I)$ chloride. The reaction is: $AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]Cl(aq)$.

(B) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-] = S^2$.
Given $K_{sp} = 4.9 \times 10^{-13} = 49 \times 10^{-14}$.
Thus,solubility $S = \sqrt{49 \times 10^{-14}} = 7 \times 10^{-7} \, mol/L$.
To find the mass dissolved in $20 \, L$,we use the formula: $\text{Mass} = S \times \text{Volume} \times \text{Molar Mass}$.
$\text{Mass} = (7 \times 10^{-7} \, mol/L) \times (20 \, L) \times (188 \, g/mol)$.
$\text{Mass} = 140 \times 10^{-7} \times 188 \, g = 14 \times 10^{-6} \times 188 \, g$.

(D) For $MX$: $K_{sp} = S^2 = 4 \times 10^{-8} \implies S = 2 \times 10^{-4} \, M$.
For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14} \implies S^3 = 0.8 \times 10^{-14} = 8 \times 10^{-15} \implies S = 2 \times 10^{-5} \, M$.
For $M_3X$: $K_{sp} = 27S^4 = 2.7 \times 10^{-15} \implies S^4 = 10^{-16} \implies S = 10^{-4} \, M$.
Comparing the solubilities: $2 \times 10^{-4} > 10^{-4} > 2 \times 10^{-5}$.
Thus,the order is $MX > M_3X > MX_2$.

(C) The dissociation reaction for $Hg_2I_2$ is given by:
$Hg_2I_2(s) \rightleftharpoons Hg_2^{2+}(aq) + 2I^{-}(aq)$
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 in the balanced chemical equation.
Therefore,$K_{sp} = [Hg_2^{2+}][I^{-}]^2$.

(A) For the dissolution of the salts in equilibrium with a common $[Ag^{+}]$ ion concentration:
$K_{sp}(AgCl) = [Ag^{+}][Cl^{-}] = 10^{-10}$
$K_{sp}(AgBr) = [Ag^{+}][Br^{-}] = 10^{-13}$
$K_{sp}(AgI) = [Ag^{+}][I^{-}] = 10^{-16}$
Dividing the expressions,we get:
$\frac{[Cl^{-}]}{[Br^{-}]} = \frac{K_{sp}(AgCl)}{K_{sp}(AgBr)} = \frac{10^{-10}}{10^{-13}} = 10^3$
$\frac{[Br^{-}]}{[I^{-}]} = \frac{K_{sp}(AgBr)}{K_{sp}(AgI)} = \frac{10^{-13}}{10^{-16}} = 10^3$
Thus,$[Cl^{-}] = 10^3 [Br^{-}]$ and $[I^{-}] = 10^{-3} [Br^{-}]$.
Substituting these into the ratio $[Cl^{-}] : [Br^{-}] : [I^{-}]$:
$10^3 [Br^{-}] : [Br^{-}] : 10^{-3} [Br^{-}] = 10^3 : 1 : 10^{-3} = 10^6 : 10^3 : 1$.

(A) $BiOCl$ is insoluble in water but dissolves in concentrated $HCl$ to form soluble $BiCl_3$ complex: $BiOCl + 2HCl \rightleftharpoons BiCl_3 + H_2O$.
$Hg_2Cl_2$ (calomel) is insoluble in water and does not dissolve in concentrated $HCl$ (it reacts with $NH_3$ to form black $Hg + Hg(NH_2)Cl$).
$AgCl$ is insoluble in water but dissolves in excess $Cl^-$ ions (from $HCl$) to form the soluble complex $[AgCl_2]^-$.
Comparing the options,the pair $BiOCl$ and $Hg_2Cl_2$ fits the criteria where $BiOCl$ dissolves in concentrated $HCl$ and $Hg_2Cl_2$ does not.

(B) The precipitation order depends on the $K_{sp}$ values. Since $[Cl^-] = [Br^-] = [I^-] = 0.1 \ M$,the ion requiring the lowest $[Ag^+]$ will precipitate first.
$1$. For $AgI$: $[Ag^+] = \frac{K_{sp}(AgI)}{[I^-]} = \frac{9 \times 10^{-17}}{0.1} = 9 \times 10^{-16} \ M$.
$2$. For $AgBr$: $[Ag^+] = \frac{K_{sp}(AgBr)}{[Br^-]} = \frac{5 \times 10^{-13}}{0.1} = 5 \times 10^{-12} \ M$.
$3$. For $AgCl$: $[Ag^+] = \frac{K_{sp}(AgCl)}{[Cl^-]} = \frac{2 \times 10^{-10}}{0.1} = 2 \times 10^{-9} \ M$.
Thus,the order of precipitation is $I^- > Br^- > Cl^-$. The third ion to precipitate is $Cl^-$. When $Cl^-$ starts precipitating,$[Ag^+] = 2 \times 10^{-9} \ M$.
At this point,the concentration of the first ion $(I^-)$ is:
$[I^-] = \frac{K_{sp}(AgI)}{[Ag^+]} = \frac{9 \times 10^{-17}}{2 \times 10^{-9}} = 4.5 \times 10^{-8} \ M$.

(C) The dissolution of $MX$ is represented as: $MX(s) \rightleftharpoons M^{+}(aq) + X^{-}(aq)$,with $K_{sp} = [M^{+}][X^{-}]$.
In an acidic solution,$X^{-}$ reacts with $H^{+}$ to form the weak acid $HX$: $X^{-}(aq) + H^{+}(aq) \rightleftharpoons HX(aq)$.
The equilibrium constant for this reaction is $K = \frac{1}{K_a}$.
The total concentration of the anion $X$ in the solution is $[X]_{total} = [X^{-}] + [HX]$.
From the acid dissociation equilibrium,$[HX] = \frac{[H^{+}][X^{-}]}{K_a}$.
Thus,$[X]_{total} = [X^{-}] \left( 1 + \frac{[H^{+}]}{K_a} \right)$.
Since $[M^{+}] = [X]_{total} = S$ (solubility),we have $[X^{-}] = \frac{S}{1 + \frac{[H^{+}]}{K_a}}$.
Substituting into the $K_{sp}$ expression: $K_{sp} = S \cdot \frac{S}{1 + \frac{[H^{+}]}{K_a}} = \frac{S^2}{1 + \frac{[H^{+}]}{K_a}}$.
Solving for $S$,we get $S = \sqrt{K_{sp} \cdot \left( 1 + \frac{[H^{+}]}{K_a} \right)}$.

(A) $Al(OH)_3$,$Cr(OH)_3$,and $Zn(OH)_2$ are amphoteric hydroxides. They react with strong bases like $NaOH$ to form soluble complex salts (e.g.,$[Al(OH)_4]^-$,$[Cr(OH)_4]^-$,and $[Zn(OH)_4]^{2-}$).
$Fe(OH)_3$ is a basic hydroxide and does not exhibit amphoteric character. Therefore,it does not dissolve in $NaOH$ solution.
The reaction for $Fe(OH)_3$ solubility is governed by its $K_{sp}$ value $(6.3 \times 10^{-38})$,which is extremely low,confirming it is insoluble in basic media.

(B) The solubility equilibrium for $Al(OH)_3$ is given by: $Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + 3OH^-(aq)$.
Given $pH = 8.0$,we know $pOH = 14 - pH = 14 - 8 = 6.0$.
Therefore,$[OH^-] = 10^{-pOH} = 10^{-6} \ M$.
The solubility product expression is $K_{sp} = [Al^{3+}][OH^-]^3$.
Let the solubility of $Al(OH)_3$ be $S$. Then $[Al^{3+}] = S$.
Substituting the values: $2.7 \times 10^{-27} = S \times (10^{-6})^3$.
$2.7 \times 10^{-27} = S \times 10^{-18}$.
$S = \frac{2.7 \times 10^{-27}}{10^{-18}} = 2.7 \times 10^{-9} \ M$.

(A) The solubility product $K_{sp}$ for a salt $A_x B_y$ is given by the formula: $K_{sp} = x^x \cdot y^y \cdot (S)^{x+y}$,where $S$ is the solubility.
Given $S = 1.4 \times 10^{-4} \ M$ and $K_{sp} = 1.1 \times 10^{-11}$.
Substituting the values: $1.1 \times 10^{-11} = x^x \cdot y^y \cdot (1.4 \times 10^{-4})^{x+y}$.
If we assume $x+y = 3$,then $(1.4 \times 10^{-4})^3 = 2.744 \times 10^{-12}$.
Then $x^x \cdot y^y = \frac{1.1 \times 10^{-11}}{2.744 \times 10^{-12}} \approx 4$.
For $x=1, y=2$,$x^x \cdot y^y = 1^1 \cdot 2^2 = 4$.
Thus,the values are $x=1, y=2$.

(D) For salt $MX$,the solubility product expression is $K_{sp} = s^2$,where $s$ is the molar solubility.
Given $K_{sp} = 4 \times 10^{-10}$,so $s = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5} \ mol \ L^{-1}$.
For salt $MX_3$,the dissociation is $MX_3 \rightleftharpoons M^{3+} + 3X^-$. The solubility product expression is $K_{sp} = (s)(3s)^3 = 27s^4$.
Substituting the value of $s = 2 \times 10^{-5} \ mol \ L^{-1}$ into the expression for $MX_3$:
$K_{sp} = 27 \times (2 \times 10^{-5})^4 = 27 \times 16 \times 10^{-20} = 432 \times 10^{-20} = 4.32 \times 10^{-18}$.

(B) The solubility $s$ in $g/L$ is calculated as: $s = (6.9 \times 10^{-2} \ g / 100 \ mL) \times (1000 \ mL / 1 \ L) = 0.69 \ g/L$.
Convert solubility to $mol/L$: $s = 0.69 \ g/L / 690 \ g/mol = 10^{-3} \ mol/L$.
The dissociation of $Ba_3(AsO_4)_2$ is: $Ba_3(AsO_4)_2(s) \rightleftharpoons 3Ba^{2+}(aq) + 2AsO_4^{3-}(aq)$.
$K_{sp} = [Ba^{2+}]^3 [AsO_4^{3-}]^2 = (3s)^3 (2s)^2 = 27s^3 \times 4s^2 = 108s^5$.
Substituting $s = 10^{-3}$: $K_{sp} = 108 \times (10^{-3})^5 = 108 \times 10^{-15} = 1.08 \times 10^{-13}$.

(B) $1$. Calculate the molar solubility $(S)$ of $Ag_2CrO_4$ in $mol/L$:
$S = \frac{\text{solubility in } g/L}{\text{molar mass in } g/mol} = \frac{1.992 \times 10^{-2}}{332} = 6 \times 10^{-5} \ mol/L$.
$2$. The dissociation of silver chromate is: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$.
$3$. From the stoichiometry,the concentration of $Ag^{+}$ is $2 \times S$.
$4$. $[Ag^{+}] = 2 \times (6 \times 10^{-5} \ M) = 1.2 \times 10^{-4} \ M$.

(B) First,calculate the molar solubility $(S)$ of $PbF_2$ in $mol/L$.
$S = \frac{\text{solubility in } g/L}{\text{molar mass}} = \frac{0.46 \ g/L}{245 \ g/mol} \approx 1.877 \times 10^{-3} \ mol/L$.
For the dissociation $PbF_2(s) \rightleftharpoons Pb^{2+}(aq) + 2F^-(aq)$,the solubility product expression is $K_{sp} = [Pb^{2+}][F^-]^2$.
Given the stoichiometry,$[Pb^{2+}] = S$ and $[F^-] = 2S$.
Therefore,$K_{sp} = (S)(2S)^2 = 4S^3$.
Substituting $S = 1.877 \times 10^{-3}$:
$K_{sp} = 4 \times (1.877 \times 10^{-3})^3 \approx 4 \times 6.61 \times 10^{-9} \approx 2.64 \times 10^{-8}$.

(B) The solubility product expression is $K_{sp} = [Mg^{2+}][OH^-]^2$.
Given $K_{sp} = 1 \times 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M$.
Substituting the values: $1 \times 10^{-12} = 0.01 \times [OH^-]^2$.
$[OH^-]^2 = \frac{1 \times 10^{-12}}{10^{-2}} = 1 \times 10^{-10}$.
$[OH^-] = 1 \times 10^{-5} \ M$.
Calculating $pOH$: $pOH = -\log[OH^-] = -\log(1 \times 10^{-5}) = 5$.
Calculating $pH$: $pH = 14 - pOH = 14 - 5 = 9$.
Therefore,precipitation will occur when $pH > 9$.

(A) The dissociation of the salt $MX_4$ is represented as:
$MX_{4(s)} \rightleftharpoons M_{(aq)}^{4+} + 4X_{(aq)}^{-}$
If the solubility of the salt is $S$,then the concentration of $M^{4+}$ is $S$ and the concentration of $X^{-}$ is $4S$.
The solubility product constant $K_{sp}$ is given by the product of the concentrations of the ions raised to their stoichiometric coefficients:
$K_{sp} = [M^{4+}][X^{-}]^4$
Substituting the values:
$K_{sp} = (S)(4S)^4$
$K_{sp} = S \times 256S^4$
$K_{sp} = 256S^5$

(B) For $Ag_2CO_3$: The dissociation is $Ag_2CO_3(s) \rightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$.
Let solubility be $S_1$. Then $K_{sp1} = (2S_1)^2(S_1) = 4S_1^3$.
Given $K_{sp1} = 4 \times 10^{-12}$,so $4S_1^3 = 4 \times 10^{-12} \implies S_1^3 = 10^{-12} \implies S_1 = 10^{-4} \, M$.
For $FeCO_3$: The dissociation is $FeCO_3(s) \rightleftharpoons Fe^{2+}(aq) + CO_3^{2-}(aq)$.
Let solubility be $S_2$. Then $K_{sp2} = (S_2)(S_2) = S_2^2$.
Given $K_{sp2} = 2.5 \times 10^{-11} = 25 \times 10^{-12}$,so $S_2^2 = 25 \times 10^{-12} \implies S_2 = 5 \times 10^{-6} \, M$.
Now,$S_1/S_2 = (10^{-4}) / (5 \times 10^{-6}) = 100 / 5 = 20$.
Thus,$S_1/S_2 = 20 : 1$.

(B) Precipitation occurs when the ionic product $(IP)$ exceeds the solubility product $(K_{sp})$.
When equal volumes are mixed,the concentration of each ion is halved.
For option $B$,the new concentrations are $[Ag^{+}] = \frac{10^{-3}}{2} \ M$ and $[Cl^{-}] = \frac{10^{-3}}{2} \ M$.
$IP = [Ag^{+}][Cl^{-}] = (0.5 \times 10^{-3}) \times (0.5 \times 10^{-3}) = 0.25 \times 10^{-6} = 2.5 \times 10^{-7}$.
Since $2.5 \times 10^{-7} > 1.8 \times 10^{-10}$,the condition $IP > K_{sp}$ is satisfied,and precipitation will occur.

(A) The complex $[Cu(CN)_4]^{3-}$ is highly stable,so it does not provide enough $Cu^{2+}$ ions to exceed the solubility product of $CuS$ when $H_2S$ is passed.
Conversely,$[Cd(CN)_4]^{2-}$ is less stable and dissociates to provide sufficient $Cd^{2+}$ ions.
These $Cd^{2+}$ ions react with $H_2S$ to form a yellow precipitate of $CdS$ $(Cd^{2+} + H_2S \rightarrow CdS \downarrow + 2H^+)$.

(D) The solubility of ionic compounds in water depends on the balance between lattice energy and hydration energy.
$LiCl$ and $NaCl$ are alkali metal halides and are highly ionic.
$AgCl$ and $PbCl_2$ are sparingly soluble due to high lattice energy and covalent character.
Among $LiCl$ and $NaCl$,$LiCl$ has a high degree of covalent character due to Fajan's rule,whereas $NaCl$ is highly ionic and has a very high hydration energy,making it highly soluble in water.

(D) The solubility $S$ in $mol/L$ is given by $S = \sqrt{K_{sp}}$.
Given $K_{sp} = 1.5625 \times 10^{-10}$.
$S = \sqrt{1.5625 \times 10^{-10}} = 1.25 \times 10^{-5} \ mol/L$.
The molar mass of $AgCl$ is $107.87 + 35.45 = 143.32 \ g/mol$ (approximately $143.5 \ g/mol$).
Solubility in $g/L = S \times \text{Molar Mass} = 1.25 \times 10^{-5} \ mol/L \times 143.5 \ g/mol = 1.79375 \times 10^{-3} \ g/L$.
Thus,the solubility is approximately $1.79 \times 10^{-3} \ g/L$.

(B) The precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product constant $(K_{sp})$.
$K_{sp} = [Ba^{2+}][CO_3^{2-}]$
Given $K_{sp} = 5.1 \times 10^{-9}$ and $[CO_3^{2-}] = 1.0 \times 10^{-4} \, M$.
Substituting the values:
$5.1 \times 10^{-9} = [Ba^{2+}] \times 1.0 \times 10^{-4}$
$[Ba^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}$
$[Ba^{2+}] = 5.1 \times 10^{-5} \, M$

(D) The dissociation of mercurous iodide $(Hg_2I_2)$ in water is represented by the following equilibrium equation:
$Hg_2I_2(s) \rightleftharpoons Hg_2^{2+}(aq) + 2I^{-}(aq)$
Note that the mercurous ion exists as a dimer,$Hg_2^{2+}$.
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 in the balanced chemical equation.
Therefore,the expression for $K_{sp}$ is:
$K_{sp} = [Hg_2^{2+}] [I^{-}]^2$
Comparing this with the given options,the correct expression is represented by option $D$.

(A) The dissolution equilibrium of $Pb(OH)_2$ is given by: $Pb(OH)_2(s) \rightleftharpoons Pb^{2+}(aq) + 2OH^-(aq)$.
Adding $HCl$ introduces $H^+$ ions,which react with $OH^-$ ions to form water: $H^+(aq) + OH^-(aq) \longrightarrow H_2O(l)$.
This removal of $OH^-$ ions from the solution shifts the equilibrium to the right according to Le Chatelier's Principle to replace the consumed $OH^-$ ions.
Consequently,more $Pb(OH)_2$ dissolves,increasing its solubility.
Adding $Pb(NO_3)_2$ or $NaOH$ would introduce common ions ($Pb^{2+}$ or $OH^-$ respectively),which would decrease the solubility due to the common ion effect.

(C) For a salt of type $M_2X$,the solubility product is $K_{sp} = (2s)^2(s) = 4s^3$,so $s = (K_{sp}/4)^{1/3}$.
For a salt of type $MX$,the solubility product is $K_{sp} = (s)(s) = s^2$,so $s = (K_{sp})^{1/2}$.
For a salt of type $MX_3$,the solubility product is $K_{sp} = (s)(3s)^3 = 27s^4$,so $s = (K_{sp}/27)^{1/4}$.
Assuming $K_{sp} = 1$ for comparison:
For $M_2X$,$s = (1/4)^{1/3} \approx 0.63$.
For $MX$,$s = (1)^{1/2} = 1.0$.
For $MX_3$,$s = (1/27)^{1/4} \approx 0.44$.
Comparing the values,the order of solubility is $MX > M_2X > MX_3$.

(A) The precipitation of $BaSO_4$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.
For precipitation to start,the ionic product must be equal to $K_{sp}$.
$K_{sp} = [Ba^{2+}][SO_4^{2-}]$
Given: $K_{sp} = 4 \times 10^{-10}$ and $[Ba^{2+}] = 10^{-4} \ mol/L$.
Substituting the values: $4 \times 10^{-10} = (10^{-4}) \times [SO_4^{2-}]$
$[SO_4^{2-}] = \frac{4 \times 10^{-10}}{10^{-4}} = 4 \times 10^{-6} \ mol/L$.
Therefore,the minimum concentration of sulphate ion required is $4 \times 10^{-6} \ mol/L$.

(B) The molar solubility $S$ is given by $S = \frac{x}{M} \ mol/L$.
The dissociation of the salt is $A_3B_2 \rightleftharpoons 3A^{+2} + 2B^{-3}$.
The concentration of $B^{-3}$ ions is $[B^{-3}] = 2S = 2(\frac{x}{M})$.
The solubility product $K_{sp}$ is given by $K_{sp} = [A^{+2}]^3 [B^{-3}]^2 = (3S)^3 (2S)^2 = 27S^3 \times 4S^2 = 108S^5$.
Substituting $S = \frac{x}{M}$,we get $K_{sp} = 108(\frac{x}{M})^5$.
The ratio $\frac{K_{sp}}{[B^{-3}]} = \frac{108S^5}{2S} = 54S^4$.
Substituting $S = \frac{x}{M}$,the ratio is $54(\frac{x}{M})^4$.

(D) For a sparingly soluble salt $A_x B_y$,the solubility product constant is given by $K_{sp} = (xS)^x (yS)^y = x^x y^y S^{(x+y)}$,where $S$ is the molar solubility.
From this relation,$S = [K_{sp} / (x^x y^y)]^{1/(x+y)}$.
As the total number of ions produced $(x+y)$ increases,the denominator $(x^x y^y)$ increases significantly,which leads to a decrease in the value of $S$ for a given $K_{sp}$.
Therefore,the salt that produces the maximum number of ions will have the lowest solubility.

(D) The dissolution of $CaF_2$ is given by: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)$.
Let $s$ be the solubility of $CaF_2$ in the presence of $Ca(NO_3)_2$.
From the stoichiometry of the reaction,for every $1 \ mol$ of $CaF_2$ that dissolves,$2 \ mol$ of $F^{-}$ ions are produced.
Therefore,the concentration of fluoride ions in the solution is $[F^{-}] = 2s$.
Rearranging this for solubility $s$,we get $s = \frac{1}{2}[F^{-}]$.

(C) To find the solubility $(S)$ of each salt,we use the $K_{sp}$ expressions:
$1$. For $AgCl$ ($1:1$ type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{2 \times 10^{-10}} \approx 1.41 \times 10^{-5} \ M$.
$2$. For $AgBr$ ($1:1$ type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{5 \times 10^{-13}} \approx 7.07 \times 10^{-7} \ M$.
$3$. For $Ag_2CO_3$ ($2:1$ type): $K_{sp} = 4S^3 \implies S = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{8 \times 10^{-12}}{4}} = \sqrt[3]{2 \times 10^{-12}} \approx 1.26 \times 10^{-4} \ M$.
$4$. For $AgI$ ($1:1$ type): $K_{sp} = S^2 \implies S = \sqrt{K_{sp}} = \sqrt{8 \times 10^{-17}} \approx 8.94 \times 10^{-9} \ M$.
Comparing the values,$1.26 \times 10^{-4} > 1.41 \times 10^{-5} > 7.07 \times 10^{-7} > 8.94 \times 10^{-9}$. Thus,$Ag_2CO_3$ has the highest solubility.

(D) The solubility equilibrium for $Fe(OH)_2$ is: $Fe(OH)_2(s) \rightleftharpoons Fe^{2+}(aq) + 2OH^-(aq)$.
Given $pH = 13.0$,we calculate $pOH = 14.0 - 13.0 = 1.0$.
Thus,$[OH^-] = 10^{-pOH} = 10^{-1.0} = 0.1 \ M$.
The solubility product expression is $K_{sp} = [Fe^{2+}][OH^-]^2$.
Substituting the values: $8.0 \times 10^{-16} = [Fe^{2+}](0.1)^2$.
$[Fe^{2+}] = \frac{8.0 \times 10^{-16}}{0.01} = 8.0 \times 10^{-14} \ M$.
Since the molar solubility $S = [Fe^{2+}]$,the solubility is $8.0 \times 10^{-14} \ M$.

(D) The dissociation of $InF_3$ is given by: $InF_3(s) \rightleftharpoons In^{3+}(aq) + 3F^{-}(aq)$
Let the solubility of $InF_3$ be $S \ M$.
Then,$[In^{3+}] = S$ and $[F^{-}] = 3S$.
$K_{sp} = [In^{3+}][F^{-}]^3$
$7.9 \times 10^{-10} = (S)(3S)^3$
$7.9 \times 10^{-10} = 27S^4$
$S^4 = \frac{7.9 \times 10^{-10}}{27} \approx 2.926 \times 10^{-11}$
$S = (2.926 \times 10^{-11})^{1/4} \approx 2.32 \times 10^{-3} \ M$
The concentration of $F^{-}$ is $3S = 3 \times 2.32 \times 10^{-3} \ M = 6.96 \times 10^{-3} \ M \approx 7.0 \times 10^{-3} \ M$.

(D) The molar mass of $PbCl_2 = 207 + 2 \times 35.5 = 278 \ g/mol$.
The solubility equilibrium is $PbCl_2(s) \leftrightarrow Pb^{2+}(aq) + 2Cl^-(aq)$.
The solubility product expression is $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-8}$,we have $4s^3 = 3.2 \times 10^{-8}$,so $s^3 = 0.8 \times 10^{-8} = 8 \times 10^{-9}$.
Thus,solubility $s = 2 \times 10^{-3} \ mol/L$.
The number of moles of $PbCl_2$ is $n = \frac{0.1 \ g}{278 \ g/mol} \approx 3.597 \times 10^{-4} \ mol$.
Since $s = \frac{n}{V}$,the volume $V = \frac{n}{s} = \frac{3.597 \times 10^{-4}}{2 \times 10^{-3}} \approx 0.1798 \ L \approx 0.18 \ L$.

(C) The dissociation reaction is: $[Zr_3(PO_4)_4] \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$
If $S$ is the molar solubility,the concentration of $Zr^{4+}$ is $3S$ and $PO_4^{3-}$ is $4S$.
The solubility product expression is: $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$
Substituting the values: $K_{sp} = (3S)^3 (4S)^4$
$K_{sp} = (27S^3) \times (256S^4)$
$K_{sp} = 6912S^7$
Therefore,$S = (K_{sp} / 6912)^{1/7}$

(A) Given concentration of $Na_2CO_3 = 1.0 \times 10^{-4} \ M$.
Since $Na_2CO_3$ is a strong electrolyte,$[CO_3^{2-}] = 1.0 \times 10^{-4} \ M$.
Precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product $(K_{sp})$.
At the point of precipitation,$[Ba^{2+}][CO_3^{2-}] = K_{sp} = 5.1 \times 10^{-9}$.
Therefore,$[Ba^{2+}] = \frac{K_{sp}}{[CO_3^{2-}]} = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} = 5.1 \times 10^{-5} \ M$.

(B) The solubility $(s)$ of a sparingly soluble salt depends on its $K_{sp}$ and its stoichiometric dissociation.
For $BaSO_4$ ($1:1$ type): $K_{sp} = s^2 \implies s = (K_{sp})^{1/2}$.
For $Hg_2Cl_2$ ($1:2$ type): $K_{sp} = (s)(2s)^2 = 4s^3 \implies s = (K_{sp}/4)^{1/3}$.
For $CrCl_3$ ($1:3$ type): $K_{sp} = (s)(3s)^3 = 27s^4 \implies s = (K_{sp}/27)^{1/4}$.
For $Cr_2(SO_4)_3$ ($2:3$ type): $K_{sp} = (2s)^2(3s)^3 = 108s^5 \implies s = (K_{sp}/108)^{1/5}$.
Comparing the powers of solubility,the order of solubility for these salts is $BaSO_4 > Hg_2Cl_2 > CrCl_3 > Cr_2(SO_4)_3$.

(A) precipitate forms when the ionic product of the ions in the solution exceeds the solubility product constant $(K_{sp})$.
For the dissociation of $CaF_2$: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^{-}$.
The ionic product is given by: $Q_{sp} = [Ca^{2+}][F^{-}]^2$.
For option $(A)$: $[Ca^{2+}] = 1 \times 10^{-2}\,M$ and $[F^{-}] = 1 \times 10^{-3}\,M$.
$Q_{sp} = (1 \times 10^{-2}) \times (1 \times 10^{-3})^2 = (1 \times 10^{-2}) \times (1 \times 10^{-6}) = 1 \times 10^{-8}$.
Since $1 \times 10^{-8} > 1.7 \times 10^{-10}$,the ionic product exceeds $K_{sp}$,leading to the formation of a precipitate.
Therefore,option $(A)$ is correct.

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)$.
First,calculate the molar solubility $(s)$ in $mol\, L^{-1}$:
$s = \frac{\text{solubility in } g\, L^{-1}}{\text{molar mass}} = \frac{0.7}{461.2} \approx 1.5178 \times 10^{-3} \, mol\, L^{-1}$.
The solubility product expression is $K_{sp} = [Pb^{2+}][I^{-}]^2 = (s)(2s)^2 = 4s^3$.
Substituting the value of $s$:
$K_{sp} = 4 \times (1.5178 \times 10^{-3})^3 \approx 4 \times 3.5 \times 10^{-9} = 14.0 \times 10^{-9}$.

(C) The dissociation of $Ag_2CO_3$ is given by: $Ag_2CO_3(s) \rightleftharpoons 2Ag^+(aq) + CO_3^{2-}(aq)$.
Let the solubility of $Ag_2CO_3$ in $0.1 \ M \ AgNO_3$ be $S' \ M$.
The concentration of $Ag^+$ ions from $AgNO_3$ is $0.1 \ M$,and from $Ag_2CO_3$ is $2S'$.
Total $[Ag^+] = (0.1 + 2S') \approx 0.1 \ M$ (since $S'$ is very small).
$[CO_3^{2-}] = S'$.
The solubility product expression is $K_{sp} = [Ag^+]^2 [CO_3^{2-}]$.
Substituting the values: $8 \times 10^{-12} = (0.1)^2 \times S'$.
$S' = \frac{8 \times 10^{-12}}{0.01} = 8 \times 10^{-10} \ M$.

(A) The dissociation of $Zr_3(PO_4)_4$ is given by: $Zr_3(PO_4)_{4(s)} \rightleftharpoons 3Zr^{4+}{(aq)} + 4PO_4^{3-}{(aq)}$.
Let the molar solubility be $S$. Then the concentrations of the ions are $[Zr^{4+}] = 3S$ and $[PO_4^{3-}] = 4S$.
The solubility product expression is $K_{SP} = [Zr^{4+}]^3 [PO_4^{3-}]^4$.
Substituting the values: $K_{SP} = (3S)^3 \cdot (4S)^4 = (27S^3) \cdot (256S^4) = 6912S^7$.
Solving for $S$: $S^7 = K_{SP} / 6912$,which gives $S = (K_{SP} / 6912)^{1/7}$.

(D) The dissociation of $Al(OH)_3$ is given by: $Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-$
$K_{sp} = [Al^{3+}][OH^-]^3$
In $0.2 \ M \ NaOH$,the concentration of $OH^-$ is $0.2 \ M$ (assuming complete dissociation of $NaOH$ and neglecting the contribution from $Al(OH)_3$ due to the common ion effect).
Let $s$ be the molar solubility of $Al(OH)_3$.
Then $[Al^{3+}] = s$ and $[OH^-] = 0.2 + 3s \approx 0.2$.
Substituting these values into the $K_{sp}$ expression:
$2.4 \times 10^{-24} = s(0.2)^3$
$2.4 \times 10^{-24} = s(0.008)$
$s = \frac{2.4 \times 10^{-24}}{0.008} = 3 \times 10^{-22} \ M$.

(D) The solubility product constant $K_{sp}$ for $Cd(OH)_2$ is given by $K_{sp} = 4s^3$,where $s$ is the molar solubility in water.
$K_{sp} = 4 \times (1.84 \times 10^{-5})^3$.
Given $pH = 12$,we have $pOH = 14 - 12 = 2$.
Therefore,$[OH^-] = 10^{-2} \ M$.
Using the expression $K_{sp} = [Cd^{2+}][OH^-]^2$,we substitute the values:
$4 \times (1.84 \times 10^{-5})^3 = [Cd^{2+}] \times (10^{-2})^2$.
$[Cd^{2+}] = \frac{4 \times (1.84)^3 \times 10^{-15}}{10^{-4}}$.
$[Cd^{2+}] = 4 \times 6.2295 \times 10^{-11} \approx 2.49 \times 10^{-10} \ M$.

(C) The precipitation reaction of $Ag^{\oplus}$ ions with $Cl^-$ ions to form $AgCl(s)$ is a spontaneous process.
For any spontaneous process occurring at constant temperature and pressure,the change in Gibbs free energy,$\Delta G,$ must be negative $(\Delta G < 0)$.

(C) The dissolution of $CaF_2$ is represented by the equilibrium: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{\Theta}(aq)$.
Let the solubility of $CaF_2$ be $s \ mol/L$.
In the presence of $0.1 \ M \ Ca(NO_3)_2$,the concentration of $Ca^{2+}$ ions is $[Ca^{2+}] = 0.1 + s$.
The concentration of $F^{\Theta}$ ions is $[F^{\Theta}] = 2s$.
Since $s$ is very small compared to $0.1$,we can approximate $[Ca^{2+}] \approx 0.1 \ M$.
From the stoichiometry of the reaction,for every $1 \ mol$ of $CaF_2$ that dissolves,$2 \ mol$ of $F^{\Theta}$ ions are produced.
Therefore,the solubility $s = \frac{[F^{\Theta}]}{2}$.

(D) The solubility $S$ of $AgCl$ is given by $K_{sp} = S^2$.
$S = \sqrt{10^{-10}} = 10^{-5} \, mol/L$.
Molar mass of $AgCl = 143.5 \, g/mol$.
Mass of $AgCl = 14.35 \, mg = 14.35 \times 10^{-3} \, g$.
Number of moles of $AgCl = \frac{14.35 \times 10^{-3} \, g}{143.5 \, g/mol} = 10^{-4} \, mol$.
Since $S = \frac{n}{V}$,we have $V = \frac{n}{S} = \frac{10^{-4} \, mol}{10^{-5} \, mol/L} = 10 \, L$.
Converting $10 \, L$ to $m^3$: $10 \, L = 10 \times 10^{-3} \, m^3 = 0.01 \, m^3$.

(A) For a salt of the type $A_2B_3$,the dissociation is given by: $A_2B_3(s) \rightleftharpoons 2A^{3+}(aq) + 3B^{2-}(aq)$.
If $S$ is the solubility,then $[A^{3+}] = 2S$ and $[B^{2-}] = 3S$.
The solubility product $K_{sp}$ is given by: $K_{sp} = [A^{3+}]^2 [B^{2-}]^3$.
$K_{sp} = (2S)^2 \times (3S)^3 = 4S^2 \times 27S^3 = 108S^5$.
Given $S = 10^{-3} \ M$,substituting the value:
$K_{sp} = 108 \times (10^{-3})^5 = 108 \times 10^{-15} = 1.08 \times 10^{-13}$.

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