Solubility product Questions in English

(D) The solubility of $MnS$ in dilute $HCl$ occurs because the sulphide ions $(S^{2-})$ react with the $H^+$ ions from $HCl$ to form the weak acid $H_2S$.
$MnS(s) + 2H^+(aq) \rightleftharpoons Mn^{2+}(aq) + H_2S(aq)$.
This reaction removes $S^{2-}$ ions from the solution,shifting the equilibrium to the right according to Le Chatelier's principle,thereby dissolving the $MnS$.

(C) The precipitation of a metal sulfide occurs when the ionic product of the metal ions and sulfide ions exceeds the solubility product $(K_{sp})$ of the respective metal sulfide.
Since the $K_{sp}$ of $CuS$ is significantly lower than the $K_{sp}$ of $ZnS$,the ionic product of $Cu^{2+}$ and $S^{2-}$ reaches the $K_{sp}$ value of $CuS$ much earlier than that of $ZnS$.
Therefore,$CuS$ precipitates first.

(A) $BaSO_4$ has a very high lattice energy due to the strong electrostatic forces between the $Ba^{2+}$ and $SO_4^{2-}$ ions.
Since the hydration energy released upon dissolution is not sufficient to overcome this high lattice energy,$BaSO_4$ exhibits low solubility in water.

(D) The solubility rules state that most salts of alkali metals (like $Na^{+}$) and ammonium $(NH_4^{+})$ are soluble in water.
$Fe^{3+}$ reacts with $PO_4^{3-}$ to form $FePO_4$,which is an insoluble salt and forms a precipitate.

(D) $AgCl$ is a white precipitate that is insoluble in water due to its low solubility product constant $(K_{sp})$.
It is soluble in $NH_4OH$ because it forms a soluble complex,diamminesilver$(I)$ chloride.
The reaction is: $AgCl(s) + 2NH_4OH(aq) \to [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$.

(D) $AgI$ does not dissolve in ammonium hydroxide solution because of its high covalent character and very low solubility product $(K_{sp})$,which prevents the formation of the soluble complex $[Ag(NH_3)_2]^+$. The strong lattice energy of $AgI$ overcomes the stabilization energy provided by the formation of the ammine complex.

(C) The compound $Hg_2Cl_2$ (Mercurous chloride) is insoluble in water.
In contrast,$Hg_2(NO_3)_2$ (Mercurous nitrate),$Hg(NO_3)_2$ (Mercuric nitrate),and $Hg_2(ClO_4)_2$ (Mercurous perchlorate) are soluble in water.

(A) The solubility of silver halides in ammonia depends on the stability of the complex formed,$[Ag(NH_3)_2]^+$.
$AgCl$ has the highest solubility product $(K_{sp})$ among the given silver halides,making it the most soluble in aqueous ammonia.
The order of solubility in ammonia is $AgCl > AgBr > AgI$.
Therefore,$AgCl$ is the most soluble.

(A) $HgS$ is insoluble in hot dilute $HNO_3$ because the solubility product of $HgS$ is extremely low. However,it is soluble in aqua regia ($HCl + HNO_3$ in $3:1$ ratio) due to the formation of a stable complex $[HgCl_4]^{2-}$,which shifts the equilibrium forward.

(D) The solubility of a salt is determined by its solubility product constant $(K_{sp})$. $A$ lower $K_{sp}$ value indicates lower solubility.
Comparing the solubility products:
$K_{sp}(AgCl) \approx 1.8 \times 10^{-10}$
$K_{sp}(AgBr) \approx 5.0 \times 10^{-13}$
$K_{sp}(AgI) \approx 8.3 \times 10^{-17}$
$K_{sp}(Ag_2S) \approx 6.0 \times 10^{-51}$
Since $Ag_2S$ has the lowest $K_{sp}$ value among the given options,it is the least soluble in water.

(D) The solubility of silver salts depends on the lattice energy and hydration energy.
$Ag_2S$ has an extremely low solubility product constant $(K_{sp})$ compared to the silver halides $(AgCl, AgBr, AgI)$.
Therefore,the order of solubility is $AgCl > AgBr > AgI > Ag_2S$.
Thus,$Ag_2S$ is the least soluble in water.

(C) $CaC_2O_4$ (calcium oxalate) is the salt of oxalic acid.
Since oxalic acid is a stronger acid than acetic acid,$CaC_2O_4$ does not dissolve in acetic acid.
In contrast,$CaCO_3$,$CaO$,and $Ca(OH)_2$ react with acetic acid to form soluble calcium acetate.

(C) The precipitation of $AgCl$ from $Ag^{+}$ and $Cl^{-}$ ions is a spontaneous process.
For any spontaneous process,the change in Gibbs free energy,$\Delta G$,must be negative $(\Delta G < 0)$.

(A) The dissociation reaction is: $Mg(OH)_{2(s)} \to Mg^{2+}_{(aq)} + 2OH^-_{(aq)}$
Calculate the standard Gibbs free energy change $(\Delta G^o)$:
$\Delta G^o = [\Delta G_f^o(Mg^{2+}) + 2 \times \Delta G_f^o(OH^-)] - [\Delta G_f^o(Mg(OH)_2)]$
$\Delta G^o = [-456.0 + 2(-157.3)] - (-833.9)$
$\Delta G^o = [-456.0 - 314.6] + 833.9 = 63.3 \ kJ \ mol^{-1} = 63300 \ J \ mol^{-1}$
Using the relation $\Delta G^o = -RT \ln K_{sp}$ or $\Delta G^o = -2.303 RT \log K_{sp}$:
$63300 = -2.303 \times 8.314 \times 298 \times \log K_{sp}$
$63300 = -5705.85 \times \log K_{sp}$
$\log K_{sp} = -11.0938$
$K_{sp} = 10^{-11.0938} \approx 8.06 \times 10^{-12} \ M^3$
Rounding to the nearest provided option,the correct answer is $8.4 \times 10^{-12} \ M^3$.

(C) Calcium oxalate $(CaC_2O_4)$ is a salt of a weak acid (oxalic acid) and is insoluble in weak acids like acetic acid $(CH_3COOH)$.
It only dissolves in strong mineral acids like hydrochloric acid $(HCl)$.

(C) $AgCl$ is soluble in $NH_4OH$ due to the formation of a soluble complex compound,diamminesilver$(I)$ chloride.
The reaction is: $AgCl(s) + 2NH_4OH(aq) \rightarrow [Ag(NH_3)_2]Cl(aq) + 2H_2O(l)$.
Other compounds like $PbCl_2$,$PbSO_4$,and $CaCO_3$ do not form such soluble complexes with $NH_4OH$.

(B) When $SnS$ is dissolved in yellow ammonium sulfide $(NH_4)_2S_x$,it forms a soluble complex,ammonium thiostannate,$(NH_4)_2SnS_3$. The reaction is: $SnS + (NH_4)_2S_2 \rightarrow (NH_4)_2SnS_3$.
When $HCl$ is added to this solution,the complex decomposes to precipitate $SnS_2$ (tin$(IV)$ sulfide) as a yellow precipitate. The reaction is: $(NH_4)_2SnS_3 + 2HCl \rightarrow SnS_2 + 2NH_4Cl + H_2S$.

(C) Acetic acid $(CH_3COOH)$ is a weak acid.
$CaO$,$CaCO_3$,and $Ca(OH)_2$ are basic or amphoteric in nature and react with acetic acid to form soluble calcium acetate.
$CaO + 2CH_3COOH \rightarrow (CH_3COO)_2Ca + H_2O$
$CaCO_3 + 2CH_3COOH \rightarrow (CH_3COO)_2Ca + H_2O + CO_2$
$Ca(OH)_2 + 2CH_3COOH \rightarrow (CH_3COO)_2Ca + 2H_2O$
Calcium oxalate $(CaC_2O_4)$ is a salt of a strong acid (oxalic acid) and is insoluble in weak acids like acetic acid.

(D) Precipitation occurs when the combination of ions forms an insoluble salt.
$1$. $K^{+}$ and $SO_4^{2-}$ form $K_2SO_4$,which is soluble in water.
$2$. $Na^{+}$ and $S^{2-}$ form $Na_2S$,which is soluble in water.
$3$. $Ag^{+}$ and $NO_3^{-}$ form $AgNO_3$,which is soluble in water.
$4$. $Al^{3+}$ and $OH^{-}$ react to form $Al(OH)_3$,which is an insoluble white precipitate.
Therefore,the correct pair is $Al^{3+}$ and $OH^{-}$.

(D) The solubility of $MnS$ in dilute $HCl$ is governed by the following equilibrium reaction:
$MnS(s) + 2H^+(aq) \rightleftharpoons Mn^{2+}(aq) + H_2S(aq)$
In this reaction,$H^+$ ions from $HCl$ react with $S^{2-}$ ions from $MnS$ to form the weak acid $H_2S$.
According to Le Chatelier's principle,the removal of $S^{2-}$ ions from the solution to form $H_2S$ shifts the equilibrium to the right,thereby increasing the solubility of $MnS$ in the acidic medium.

(A) For a salt of the type $AB$,the solubility $S$ is given by $S = \sqrt{K_{sp}}$.
For $AgCl$: $S_{AgCl} = \sqrt{1.2 \times 10^{-10}} \approx 1.095 \times 10^{-5} \ M$.
For $AgBr$: $S_{AgBr} = \sqrt{3.5 \times 10^{-13}} = \sqrt{35 \times 10^{-14}} \approx 5.916 \times 10^{-7} \ M$.
Comparing the two values,$S_{AgCl} > S_{AgBr}$,which means $S$ of $AgBr < S$ of $AgCl$.

(A) For a salt of type $AB$ like $CaSO_4$,the solubility product $K_{sp}$ is given by $K_{sp} = s^2$,where $s$ is the solubility.
Given $K_{sp} = 2.5 \times 10^{-5} = 25 \times 10^{-6}$.
Therefore,$s = \sqrt{K_{sp}} = \sqrt{25 \times 10^{-6}} = 5 \times 10^{-3} \ mol/L$.

(B) When equal volumes are mixed,the concentration of each ion is halved.
For $CaF_2$,the ionic product $Q_{sp} = [Ca^{2+}][F^-]^2$.
Precipitation occurs if $Q_{sp} > K_{sp}$.
For option $B$: $[Ca^{2+}] = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, M$ and $[F^-] = \frac{10^{-3}}{2} = 5 \times 10^{-4} \, M$.
$Q_{sp} = (5 \times 10^{-3}) \times (5 \times 10^{-4})^2 = 5 \times 10^{-3} \times 25 \times 10^{-8} = 1.25 \times 10^{-9}$.
Since $1.25 \times 10^{-9} > 1.7 \times 10^{-10}$,precipitation occurs.

(B) The dissociation of calcium phosphate is given by the equation:
$Ca_3(PO_4)_2(s) ⇌ 3Ca^{+2}(aq) + 2PO_4^{-3}(aq)$
The solubility product constant $(K_{sp})$ is defined as the product of the concentrations of the ions raised to the power of their stoichiometric coefficients:
$K_{sp} = [Ca^{+2}]^3 [PO_4^{-3}]^2$

(B) The solubility product constant $(K_{sp})$ is a constant value for a specific salt at a given temperature.
It does not change with the addition of common ions or changes in concentration of the constituent ions.
Therefore,adding $10^{-5} \ mol$ of $Ag^{+}$ ions will not change the $K_{sp}$ value of $AgCl$.
The value remains $1.8 \times 10^{-10}$.

(A) The molar mass of $CaCO_3$ is $40 + 12 + (3 \times 16) = 100 \ g/mol$.
Given mass of residue = $7.0 \ g$ in $1 \ L$ of solution.
Solubility $(S)$ in $mol/L = \frac{\text{mass}}{\text{molar mass}} = \frac{7.0 \ g}{100 \ g/mol} = 0.07 \ mol/L$.
For $CaCO_3 \rightleftharpoons Ca^{2+} + CO_3^{2-}$,the solubility product $K_{sp} = [Ca^{2+}][CO_3^{2-}] = S \times S = S^2$.
$K_{sp} = (0.07)^2 = 0.0049 = 4.9 \times 10^{-3}$.

(A) Let the metal hydroxide be $M(OH)_3$.
The dissociation reaction is: $M(OH)_3(s) \rightleftharpoons M^{3+}(aq) + 3OH^-(aq)$.
If solubility is $S$,then $[M^{3+}] = S$ and $[OH^-] = 3S$.
The solubility product expression is $K_{sp} = [M^{3+}][OH^-]^3 = (S)(3S)^3 = 27S^4$.
Given $S = 1 \times 10^{-5} \ mol/L$.
Therefore,$K_{sp} = 27 \times (1 \times 10^{-5})^4 = 27 \times 10^{-20}$.

(D) The dissociation of the salt $MX_4$ is given by:
$MX_4(s) \rightleftharpoons M^{4+}(aq) + 4X^{-}(aq)$
Let the solubility be $s \ mol \ L^{-1}$.
Then,$[M^{4+}] = s$ and $[X^{-}] = 4s$.
The solubility product expression is:
$K_{sp} = [M^{4+}][X^{-}]^4$
$K_{sp} = (s)(4s)^4$
$K_{sp} = s \times 256s^4$
$K_{sp} = 256s^5$
Rearranging for $s$:
$s^5 = \frac{K_{sp}}{256}$
$s = (\frac{K_{sp}}{256})^{1/5}$

(C) The dissociation reaction of $SnS_2$ is:
$SnS_2(s) ⇌ Sn^{4+}(aq) + 2S^{2-}(aq)$
The solubility product constant $(K_{sp})$ is defined as the product of the concentrations of the ions raised to the power of their stoichiometric coefficients.
Therefore,$K_{sp} = [Sn^{4+}] [S^{2-}]^2$.

(D) For $PQ$ (type $AB$): $K_{sp} = s^2$. So,$s = \sqrt{K_{sp}} = \sqrt{4.0 \times 10^{-20}} = 2.0 \times 10^{-10} \ M$.
For $PQ_2$ (type $AB_2$): $K_{sp} = 4s^3$. So,$s = (K_{sp}/4)^{1/3} = (3.2 \times 10^{-14} / 4)^{1/3} = (8.0 \times 10^{-15})^{1/3} = 2.0 \times 10^{-5} \ M$.
For $PQ_3$ (type $AB_3$): $K_{sp} = 27s^4$. So,$s = (K_{sp}/27)^{1/4} = (2.7 \times 10^{-35} / 27)^{1/4} = (1.0 \times 10^{-36})^{1/4} = 1.0 \times 10^{-9} \ M$.
Comparing the molar solubilities: $2.0 \times 10^{-5} (2) > 1.0 \times 10^{-9} (3) > 2.0 \times 10^{-10} (1)$.
Thus,the decreasing order is $2 > 3 > 1$.

(B) For precipitation to occur,the ionic product $(Q_{sp})$ must exceed the solubility product $(K_{sp})$.
For $ZnS$: $Q_{sp} = [Zn^{2+}][S^{2-}] = (0.01)(8.1 \times 10^{-21}) = 8.1 \times 10^{-23}$. Since $8.1 \times 10^{-23} < 3.0 \times 10^{-22}$ $(Q_{sp} < K_{sp})$,$ZnS$ will not precipitate.
For $CuS$: $Q_{sp} = [Cu^{2+}][S^{2-}] = (0.01)(8.1 \times 10^{-21}) = 8.1 \times 10^{-23}$. Since $8.1 \times 10^{-23} > 8.0 \times 10^{-36}$ $(Q_{sp} > K_{sp})$,$CuS$ will precipitate.

(C) For $AgBrO_3$ (type $AB$):
$K_{sp} = S^2$
$S = \sqrt{K_{sp}} = \sqrt{5.5 \times 10^{-5}} \approx 7.41 \times 10^{-3} \ M$.
For $Ag_2SO_4$ (type $A_2B$):
$K_{sp} = 4S^3$
$S = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{2 \times 10^{-5}}{4}} = \sqrt[3]{0.5 \times 10^{-5}} = \sqrt[3]{5 \times 10^{-6}} \approx 1.71 \times 10^{-2} \ M$.
Comparing the two values,$S_{Ag_2SO_4} > S_{AgBrO_3}$ or $S_{AgBrO_3} < S_{Ag_2SO_4}$.

(B) The solubility product expression for $PbS$ is given by: $K_{sp} = [Pb^{2+}][S^{2-}]$.
Given $K_{sp} = 3.4 \times 10^{-28}$ and $[Pb^{2+}] = 1 \times 10^{-2} \ mol/L$.
To precipitate $PbS$,the ionic product must exceed the solubility product $(Q_{sp} > K_{sp})$.
Therefore,$[S^{2-}] > \frac{K_{sp}}{[Pb^{2+}]}$.
$[S^{2-}] > \frac{3.4 \times 10^{-28}}{1 \times 10^{-2}}$.
$[S^{2-}] > 3.4 \times 10^{-26} \ mol/L$.
Thus,the minimum concentration required is $3.4 \times 10^{-26} \ mol/L$.

(B) The solubility $(S)$ of a salt depends on its $K_{sp}$ value and its stoichiometry.
For salts of the same type ($AB$ type),the solubility is directly proportional to the square root of $K_{sp}$ $(S = \sqrt{K_{sp}})$.
Comparing the given values:
$HgS: K_{sp} = 1.6 \times 10^{-54}$
$PbSO_4: K_{sp} = 1.3 \times 10^{-8}$
$ZnS: K_{sp} = 7.0 \times 10^{-26}$
$AgCl: K_{sp} = 1.7 \times 10^{-10}$
Since all these salts are of the $1:1$ type ($AB$ type),the salt with the highest $K_{sp}$ value will have the highest solubility.
Therefore,$PbSO_4$ has the maximum solubility.

(D) The dissociation of the salt $M_2X_3$ is given by:
$M_2X_3 (s) \rightleftharpoons 2M^{3+} (aq) + 3X^{2-} (aq)$
If the solubility is $y \ mol \ m^{-3}$,then the concentration of ions are $[M^{3+}] = 2y$ and $[X^{2-}] = 3y$.
The solubility product $K_{sp}$ is calculated as:
$K_{sp} = [M^{3+}]^2 [X^{2-}]^3$
$K_{sp} = (2y)^2 \times (3y)^3$
$K_{sp} = (4y^2) \times (27y^3)$
$K_{sp} = 108y^5 \ (mol \ m^{-3})^5$

(C) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$
Let the solubility be $S \ mol/L$.
Then,$[Pb^{2+}] = S$ and $[Cl^-] = 2S$.
The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2$
$K_{sp} = (S)(2S)^2 = 4S^3$
Solving for $S$: $S^3 = \frac{K_{sp}}{4}$
Therefore,$S = \sqrt[3]{\frac{K_{sp}}{4}}$.

(D) The dissociation of the salt is given by: $Li_3Na_3(AlF_6)_2 \rightleftharpoons 3Li^+ + 3Na^+ + 2[AlF_6]^{3-}$.
Let the solubility be $S \ mol/L$.
The concentrations of the ions are: $[Li^+] = 3S$,$[Na^+] = 3S$,and $[[AlF_6]^{3-}] = 2S$.
The solubility product expression is: $K_{sp} = [Li^+]^3 \times [Na^+]^3 \times [[AlF_6]^{3-}]^2$.
Substituting the values: $K_{sp} = (3S)^3 \times (3S)^3 \times (2S)^2$.
$K_{sp} = (27S^3) \times (27S^3) \times (4S^2)$.
$K_{sp} = 2916 \ S^8$.

(A) For $1:1$ electrolytes like $AgCl$,$AgBr$,and $AgI$,the solubility $S = \sqrt{K_{sp}}$.
For $AgCl$: $S = \sqrt{2 \times 10^{-10}} \approx 1.41 \times 10^{-5} \ M$.
For $AgBr$: $S = \sqrt{5 \times 10^{-13}} \approx 7.07 \times 10^{-7} \ M$.
For $AgI$: $S = \sqrt{8 \times 10^{-17}} \approx 8.94 \times 10^{-9} \ M$.
For $Ag_2CO_3$ ($1:2$ electrolyte),$K_{sp} = 4S^3$,so $S = (K_{sp}/4)^{1/3} = (8 \times 10^{-12} / 4)^{1/3} = (2 \times 10^{-12})^{1/3} \approx 1.26 \times 10^{-4} \ M$.
Comparing the values,$Ag_2CO_3$ has the highest solubility.

(B) The solubility $s$ is the concentration in $mol/L$.
Given mass of $AgCl = 0.001435 \ g$ in $100 \ mL$.
Mass in $1000 \ mL (1 \ L) = 0.001435 \times 10 = 0.01435 \ g/L$.
Molar mass of $AgCl = 107.87 + 35.45 \approx 143.5 \ g/mol$.
Solubility $s = \frac{0.01435 \ g/L}{143.5 \ g/mol} = 10^{-4} \ mol/L$.
For $AgCl \rightleftharpoons Ag^+ + Cl^-$,$K_{sp} = [Ag^+][Cl^-] = s^2$.
$K_{sp} = (10^{-4})^2 = 10^{-8}$.

(C) The solubility $S$ in $mol/L$ is calculated as follows:
$S = \frac{1.43 \times 10^{-4} \, g}{143 \, g/mol} \times \frac{1000 \, mL}{100 \, mL} = 1 \times 10^{-5} \, mol/L$
For $AgCl$,the dissociation is $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
$K_{sp} = [Ag^+][Cl^-] = S \times S = S^2$
$K_{sp} = (1 \times 10^{-5})^2 = 1 \times 10^{-10}$

(A) The dissociation of $Gd(OH)_3$ is given by: $Gd(OH)_3 \rightleftharpoons Gd^{3+} + 3OH^-$.
The solubility product expression is $K_{sp} = [Gd^{3+}][OH^-]^3$.
Assuming the concentration of $Gd^{3+}$ is $1 \ M$ for the onset of precipitation,we have $2.7 \times 10^{-23} = (1)[OH^-]^3$.
$[OH^-]^3 = 27 \times 10^{-24}$.
Taking the cube root,$[OH^-] = 3 \times 10^{-8} \ M$.
Now,$pOH = -\log[OH^-] = -\log(3 \times 10^{-8}) = 8 - \log 3 = 8 - 0.48 = 7.52$.
Finally,$pH = 14 - pOH = 14 - 7.52 = 6.48$.
Note: Based on standard calculations for $1 \ M$ concentration,the result is $6.48$. Given the options,$6.08$ is the closest provided value.

(C) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-]$.
Given $[Ag^+] = [AgNO_3] = 0.05 \, M$.
Substituting the values into the $K_{sp}$ expression:
$5.0 \times 10^{-13} = (0.05) \times [Br^-]$.
$[Br^-] = \frac{5.0 \times 10^{-13}}{0.05} = 1.0 \times 10^{-11} \, M$.
Since $[Br^-] = [KBr]$,the concentration of $KBr$ required is $1.0 \times 10^{-11} \, mol \, L^{-1}$.
For $1 \, L$ of solution,the number of moles of $KBr$ is $1.0 \times 10^{-11} \, mol$.
Mass of $KBr = \text{moles} \times \text{molar mass} = (1.0 \times 10^{-11} \, mol) \times (120 \, g \, mol^{-1}) = 1.2 \times 10^{-9} \, g$.

(A) For a salt of type $AB_2$,the dissociation is $AB_2 \rightleftharpoons A^{2+} + 2B^-$.
Let the solubility be $s$.
Then,$[A^{2+}] = s$ and $[B^-] = 2s$.
The solubility product constant is $K_{sp} = [A^{2+}][B^-]^2 = (s)(2s)^2 = 4s^3$.
Therefore,$s^3 = \frac{K_{sp}}{4}$,which gives $s = (\frac{K_{sp}}{4})^{1/3}$.

(A) The dissociation of $BaCl_2$ is given by: $BaCl_2(s) \rightleftharpoons Ba^{2+}(aq) + 2Cl^-(aq)$.
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 $K_{sp} = 4 \times 10^{-9}$.
So,$4S^3 = 4 \times 10^{-9}$.
$S^3 = 10^{-9}$.
$S = (10^{-9})^{1/3} = 10^{-3} \ mol/L$.

(D) For an electrolyte of type $MX_2$,the solubility product $K_{sp}$ is given by the expression: $K_{sp} = 4S^3$,where $S$ is the solubility.
Given $S = 0.5 \times 10^{-4} \ mol/L$.
Substituting the value of $S$ into the formula:
$K_{sp} = 4 \times (0.5 \times 10^{-4})^3$
$K_{sp} = 4 \times (0.125 \times 10^{-12})$
$K_{sp} = 0.5 \times 10^{-12} = 5 \times 10^{-13}$.

(C) For $BaCl_2$,the dissociation is: $BaCl_2 \rightleftharpoons Ba^{2+} + 2Cl^-$.
Given solubility $s = 4 \times 10^{-6} \ M$.
$K_{sp} = [Ba^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$.
$K_{sp} = 4 \times (4 \times 10^{-6})^3 = 4 \times 64 \times 10^{-18} = 256 \times 10^{-18}$.
In the presence of $10^{-2} \ M \ Ba(OH)_2$,the common ion is $Ba^{2+}$.
$[Ba^{2+}] = 10^{-2} \ M$ (from $Ba(OH)_2$) and $[Cl^-] = 2s'$.
$K_{sp} = [Ba^{2+}][Cl^-]^2 = (10^{-2})(2s')^2 = 4 \times 10^{-2} \times (s')^2$.
$256 \times 10^{-18} = 4 \times 10^{-2} \times (s')^2$.
$(s')^2 = 64 \times 10^{-16}$.
$s' = 8 \times 10^{-8} \ M$.

(B) For $M(OH)_3$,$K_{sp} = [M^{3+}][OH^-]^3 = 10^{-23}$.
For $M(OH)_2$,$K_{sp} = [M^{2+}][OH^-]^2 = 10^{-14}$.
Precipitation occurs when the ionic product exceeds the solubility product $(K_{sp})$.
For $M^{3+}$,$[OH^-]^3 = 10^{-23} / [M^{3+}]$.
For $M^{2+}$,$[OH^-]^2 = 10^{-14} / [M^{2+}]$.
Since $M(OH)_3$ has a much lower $K_{sp}$ value,it requires a much lower concentration of $OH^-$ ions to exceed its solubility product compared to $M(OH)_2$.
Therefore,$M^{3+}$ will precipitate first.

(B) For $CaCO_3$,the dissociation is $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$.
Let the solubility be $S \ mol/L$.
Then,$K_{sp} = [Ca^{2+}][CO_3^{2-}] = S \times S = S^2$.
Given $K_{sp} = 5 \times 10^{-9}$.
$S = \sqrt{K_{sp}} = \sqrt{5 \times 10^{-9}} = \sqrt{50 \times 10^{-10}} \approx 7.07 \times 10^{-5} \ mol/L$.
Rounding to the nearest provided option,$S = 7 \times 10^{-5} \ mol/L$.

(A) For a sparingly soluble salt $A_pB_q$,the dissociation reaction is:
$A_pB_q(s) \rightleftharpoons pA^{q+}(aq) + qB^{p-}(aq)$
If the solubility of the salt is $S$,then the concentration of $A^{q+}$ is $pS$ and the concentration of $B^{p-}$ is $qS$.
The solubility product expression is:
$K_{sp} = [A^{q+}]^p [B^{p-}]^q$
Substituting the concentrations:
$K_{sp} = (pS)^p (qS)^q$
$K_{sp} = p^p S^p q^q S^q$
$K_{sp} = p^p q^q S^{p+q}$

(B) The dissociation of the base is given by: $M(OH)_3 \rightleftharpoons M^{3+} + 3OH^-$
Let the solubility be $s$. Then $[M^{3+}] = s$ and $[OH^-] = 3s$.
The solubility product expression is: $K_{sp} = [M^{3+}][OH^-]^3$
Substituting the values: $2.7 \times 10^{-11} = (s)(3s)^3$
$2.7 \times 10^{-11} = 27s^4$
$s^4 = \frac{2.7 \times 10^{-11}}{27} = 0.1 \times 10^{-11} = 10^{-12}$
$s = (10^{-12})^{1/4} = 10^{-3} \ M$
The concentration of $OH^-$ is $[OH^-] = 3s = 3 \times 10^{-3} \ M$.