Solubility product Questions in English

(D) Let the solubility of $PbBr_2$ be $S \ M$.
The dissociation reaction is: $PbBr_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Br^-_{(aq)}$.
Given the degree of dissociation $\alpha = 0.8$,the concentrations of ions in the saturated solution are:
$[Pb^{2+}] = \alpha S = 0.8S$
$[Br^-] = 2 \alpha S = 2 \times 0.8S = 1.6S$
The solubility product expression is $K_{sp} = [Pb^{2+}][Br^-]^2$.
Substituting the values: $8 \times 10^{-5} = (0.8S) \times (1.6S)^2$.
$8 \times 10^{-5} = 0.8S \times 2.56S^2 = 2.048S^3$.
$S^3 = \frac{8 \times 10^{-5}}{2.048} \approx 3.906 \times 10^{-5} = 39.06 \times 10^{-6}$.
$S = \sqrt[3]{39.06} \times 10^{-2} \approx 3.39 \times 10^{-2} \ M \approx 3.4 \times 10^{-2} \ M$.

(B) The solubility $(S)$ of $BaSO_4$ is given by $S = \sqrt{K_{sp}} = \sqrt{1.1 \times 10^{-10}} = 1.0488 \times 10^{-5} \ M$.
The molar mass of $BaSO_4$ is $137 + 32 + (4 \times 16) = 233 \ g/mol$.
The solubility in $g/L$ is $S \times \text{Molar Mass} = 1.0488 \times 10^{-5} \times 233 = 2.4437 \times 10^{-3} \ g/L$.
The volume of water required to dissolve $1 \ g$ of $BaSO_4$ is $V = \frac{1 \ g}{2.4437 \times 10^{-3} \ g/L} \approx 409.2 \ L \approx 410 \ L$.

(A) First,calculate the final concentration of ions after mixing:
Total volume = $150 \, mL + 50 \, mL = 200 \, mL$.
For $[Ca^{2+}]$: $M_2 = \frac{M_1 V_1}{V_2} = \frac{0.04 \times 50}{200} = 0.01 \, M$.
For $[SO_4^{2-}]$: $M_2 = \frac{M_1 V_1}{V_2} = \frac{0.0008 \times 150}{200} = 0.0006 \, M$.
Ionic product $(Q) = [Ca^{2+}] [SO_4^{2-}] = (0.01) \times (0.0006) = 6 \times 10^{-6}$.
Comparing with $K_{SP} = 2.4 \times 10^{-5}$,we find $6 \times 10^{-6} < 2.4 \times 10^{-5}$.
Therefore,$Q < K_{SP}$.

(B) The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^{-}]^2$
Rearrange the above expression to solve for $[OH^-]$: $[OH^{-}] = \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}$
Substitute the given values ($K_{sp} = 1 \times 10^{-12}$ and $[Mg^{2+}] = 0.01 \ M$): $[OH^{-}] = \sqrt{\frac{1 \times 10^{-12}}{0.01}} = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \ M$
Calculate $pOH$ from the hydroxide ion concentration: $pOH = -\log [OH^{-}] = -\log (1 \times 10^{-5}) = 5$
Calculate $pH$ from $pOH$: $pH = 14 - pOH = 14 - 5 = 9$
Thus,precipitation will occur at $pH = 9$.

(B) The solubility product expression for $BaSO_4$ is $K_{sp} = [Ba^{2+}][SO_4^{2-}]$.
Given $K_{sp} = 1.1 \times 10^{-10}$.
Let the new solubility be $s = 1.1 \times 10^{-8} \ M$.
In the presence of $Na_2SO_4$,the concentration of $Ba^{2+}$ ions is $s = 1.1 \times 10^{-8} \ M$.
The total concentration of $SO_4^{2-}$ ions is $s + [Na_2SO_4] \approx [Na_2SO_4]$ because $s$ is very small compared to the concentration of $Na_2SO_4$.
Substituting into the $K_{sp}$ expression: $1.1 \times 10^{-10} = (1.1 \times 10^{-8}) \times [Na_2SO_4]$.
$[Na_2SO_4] = \frac{1.1 \times 10^{-10}}{1.1 \times 10^{-8}} = 10^{-2} = 0.01 \ M$.

(D) The solubility of an ionic salt depends on the balance between its lattice enthalpy and hydration enthalpy.
For alkaline earth metal sulphates,the hydration enthalpy decreases as the size of the cation increases from $Mg^{2+}$ to $Ba^{2+}$.
However,the lattice enthalpy remains relatively constant because the sulphate ion is very large.
Since the hydration enthalpy of $BaSO_4$ is significantly lower than its lattice enthalpy,it is the least soluble among the given salts.

(A) When $Na_2CO_3$ (sodium carbonate) is added to a solution of $ZnSO_4$ (zinc sulphate),a double displacement reaction occurs.
The reaction is as follows:
$ZnSO_4(aq) + Na_2CO_3(aq) \rightarrow ZnCO_3(s) + Na_2SO_4(aq)$
Here,$ZnCO_3$ (zinc carbonate) is formed as a white precipitate,while $Na_2SO_4$ remains in the solution.
Therefore,the correct reagent is $Na_2CO_3$.

(B) The molar mass of $AgCl$ is $143 \ g \ mol^{-1}$.
Solubility in $mol \ L^{-1}$ $(S)$ is calculated as: $S = \frac{\text{solubility in } g \ L^{-1}}{\text{molar mass}} = \frac{1.43 \times 10^{-3} \ g \ L^{-1}}{143 \ g \ mol^{-1}} = 10^{-5} \ mol \ L^{-1}$.
For $AgCl$,the dissociation is $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
The solubility product constant is $K_{sp} = [Ag^+][Cl^-] = S \times S = S^2$.
$K_{sp} = (10^{-5})^2 = 10^{-10} \ M^2$.

(C) Let the solubility of $AB_2$ be $S \ M$.
The dissociation reaction is: $AB_2 \rightleftharpoons A^{2+} + 2B^-$.
Given the degree of dissociation $\alpha = 0.80$.
Concentration of $A^{2+} = \alpha S = 0.8S$.
Concentration of $B^- = 2 \alpha S = 2 \times 0.8S = 1.6S$.
$K_{sp} = [A^{2+}][B^-]^2$.
$8 \times 10^{-5} = (0.8S)(1.6S)^2$.
$8 \times 10^{-5} = 0.8S \times 2.56S^2$.
$8 \times 10^{-5} = 2.048S^3$.
$S^3 = \frac{8 \times 10^{-5}}{2.048} \approx 3.906 \times 10^{-5} = 39.06 \times 10^{-6}$.
$S = \sqrt[3]{39.06 \times 10^{-6}} \approx 3.39 \times 10^{-2} \ M \approx 0.034 \ M$.

(A) The dissociation of $Th_3(PO_4)_4$ is: $Th_3(PO_4)_4 \rightleftharpoons 3Th^{4+} + 4PO_4^{3-}$.
Let solubility be $s \ mol/L$.
Given solubility $= w \ g / 100 \ mL = 10w \ g / L = \frac{10w}{m} \ mol/L$.
So,$s = \frac{10w}{m}$.
$K_{sp} = [Th^{4+}]^3 [PO_4^{3-}]^4 = (3s)^3 (4s)^4$.
$K_{sp} = 27s^3 \times 256s^4 = 6912s^7$.
Substituting $s = \frac{10w}{m}$:
$K_{sp} = 6912 \times \left( \frac{10w}{m} \right)^7 = 6912 \times 10^7 \times \left( \frac{w}{m} \right)^7$.
$K_{sp} = 6.912 \times 10^{10} \left( \frac{w}{m} \right)^7$.

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.
In the presence of $0.2 \ M \ Pb(NO_3)_2$,the concentration of $Pb^{2+}$ ions is dominated by the strong electrolyte $Pb(NO_3)_2$.
$[Pb^{2+}] = 0.2 \ M + S \approx 0.2 \ M$ (since $S << 0.2$).
$[I^-] = 2S$.
The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2$.
Substituting the values: $K_{sp} = (0.2)(2S)^2$.
$K_{sp} = 0.2 \times 4S^2 = 0.8S^2$.
Solving for $S$: $S^2 = K_{sp}/0.8$.
Therefore,$S = \sqrt{K_{sp}/0.8}$.

(A) Precipitation of $AgCl$ occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.
When equal volumes are mixed,the concentration of each ion is halved.
$a)$ $Q_{sp} = (10^{-4}/2) \times (10^{-4}/2) = 0.25 \times 10^{-8} = 2.5 \times 10^{-9} \ M^2$. Since $2.5 \times 10^{-9} > 1.8 \times 10^{-10}$,precipitation occurs.
$b)$ $Q_{sp} = (10^{-5}/2) \times (10^{-5}/2) = 2.5 \times 10^{-11} \ M^2$. Since $2.5 \times 10^{-11} < 1.8 \times 10^{-10}$,no precipitation.
$c)$ $Q_{sp} = 2.5 \times 10^{-13} \ M^2 < 1.8 \times 10^{-10}$,no precipitation.
$d)$ $Q_{sp} = 2.5 \times 10^{-21} \ M^2 < 1.8 \times 10^{-10}$,no precipitation.
Therefore,option $(A)$ is correct.

(B) The precipitation of $BaCO_3$ starts when the ionic product exceeds the solubility product constant $(K_{sp})$.
The expression for $K_{sp}$ of $BaCO_3$ is: $K_{sp} = [Ba^{+2}][CO_3^{-2}]$.
Given that $[CO_3^{-2}] = 1 \times 10^{-4} \ M$ and $K_{sp} = 5.1 \times 10^{-9}$.
Substituting the values: $[Ba^{+2}] \times (10^{-4}) = 5.1 \times 10^{-9}$.
Solving for $[Ba^{+2}]$: $[Ba^{+2}] = \frac{5.1 \times 10^{-9}}{10^{-4}} = 5.1 \times 10^{-5} \ M$.

(A) Let the solubility of $PbBr_2$ be $S \, mol/L$.
Given that the salt is $80\%$ dissociated,the concentration of ions in the solution will be:
$[Pb^{2+}] = 0.8 \times S$
$[Br^-] = 2 \times 0.8 \times S = 1.6 \times S$
The solubility product expression is $K_{sp} = [Pb^{2+}][Br^-]^2$.
Substituting the values: $8 \times 10^{-5} = (0.8 \times S)(1.6 \times S)^2$.
$8 \times 10^{-5} = 0.8 \times S \times 2.56 \times S^2$.
$8 \times 10^{-5} = 2.048 \times S^3$.
$S^3 = \frac{8 \times 10^{-5}}{2.048} = \frac{8 \times 10^{-5}}{0.8 \times 1.6 \times 1.6} = \frac{10^{-4}}{1.6 \times 1.6}$.
Therefore,$S = [\frac{10^{-4}}{1.6 \times 1.6}]^{1/3}$.

(D) For $MX$: $K_{sp} = S^2 = 4 \times 10^{-8} \Rightarrow S = 2 \times 10^{-4} \ M$.
For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14}$ $\Rightarrow S^3 = 0.8 \times 10^{-14} = 8 \times 10^{-15}$ $\Rightarrow S = 2 \times 10^{-5} \ M$.
For $M_3X$: $K_{sp} = 27S^4 = 2.7 \times 10^{-15}$ $\Rightarrow S^4 = 0.1 \times 10^{-15} = 1 \times 10^{-16}$ $\Rightarrow S = 10^{-4} \ M$.
Comparing the values: $2 \times 10^{-4} > 1 \times 10^{-4} > 2 \times 10^{-5}$.
Therefore,the order of solubility is $MX > M_3X > MX_2$.

(C) The relationship between equivalent conductance $(\Lambda_{eq})$,conductivity $(\kappa)$,and normality $(N)$ is given by: $\Lambda_{eq} = \frac{\kappa \times 1000}{N}$.
Given $\kappa = 3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and $\Lambda_{eq} = 1.53 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.
Substituting the values: $1.53 = \frac{3.06 \times 10^{-6} \times 1000}{N}$.
$N = \frac{3.06 \times 10^{-3}}{1.53} = 2 \times 10^{-3} \ N$.
For $BaSO_4$,the molarity $(M)$ is equal to the normality $(N)$ because the n-factor is $2$ $(Ba^{2+} + SO_4^{2-})$,but the solubility $(S)$ is the molar concentration of the saturated solution.
Since $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$,the solubility $S = M = 2 \times 10^{-3} \ M$.
The solubility product $K_{sp} = [Ba^{2+}][SO_4^{2-}] = S^2$.
$K_{sp} = (2 \times 10^{-3})^2 = 4 \times 10^{-6} \ M^2$.

(D) The reaction involves calcium oxalate $(CaC_2O_4)$,which is a sparingly soluble salt.
Acetic acid $(CH_3COOH)$ is a weak acid.
For a precipitate to dissolve,the acid must be strong enough to protonate the anion of the salt to form a weak acid or a soluble complex.
Since $CH_3COOH$ is a weak acid,it cannot effectively protonate the oxalate ion $(C_2O_4^{2-})$ to shift the equilibrium of the dissolution of $CaC_2O_4$.
Therefore,no significant reaction occurs,and the precipitate remains undissolved.
Thus,the correct classification is $D$ (For no reaction).

(B) The reaction involves the solid precipitate $BaC_2O_4$ reacting with acetic acid $(AcOH)$ to form soluble barium acetate and oxalic acid.
Since the solid precipitate dissolves into the solution,this is a precipitate dissolution reaction.

(B) The reaction involves the solid precipitate $SrC_2O_4$ reacting with hydrochloric acid $(HCl)$ to form soluble products $SrCl_2$ and $H_2C_2O_4$.
Since the solid precipitate dissolves in the acid,this is a precipitate dissolution reaction.

(B) The reaction involves the solid calcium sulfite $(CaSO_3)$ reacting with sulfur dioxide $(SO_2)$ and water $(H_2O)$ to form soluble calcium bisulfite $(Ca(HSO_3)_2)$.
Since the solid precipitate $(CaSO_3)$ dissolves to form a soluble product,this is a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(B) The reaction involves the solid precipitate $AgCl$ reacting with potassium cyanide $(KCN)$ to form a soluble complex,potassium dicyanoargentate$(I)$ $(K[Ag(CN)_2])$.
Since the solid $AgCl$ dissolves in the solution due to the formation of a stable complex,this is classified as a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(B) The reaction involves the solid mercury$(II)$ sulfide $(HgS)$,which is a precipitate,reacting with sodium sulfide $(Na_2S)$ to form a soluble complex,sodium dithio-mercurate$(II)$ $(Na_2[HgS_2])$.
Since a solid precipitate is dissolving into a soluble complex,this is classified as a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(B) The reaction $FeS \downarrow + 2HCl \longrightarrow FeCl_2 + H_2S \uparrow$ involves the solid iron$(II)$ sulfide $(FeS)$ reacting with hydrochloric acid $(HCl)$ to form soluble iron$(II)$ chloride $(FeCl_2)$ and hydrogen sulfide gas $(H_2S)$.
Since $FeS$ is a solid precipitate that dissolves upon reacting with the acid,this is classified as a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(B) The given reaction involves the solid lead$(II)$ chloride $(PbCl_2)$ dissolving in hot water to form aqueous lead$(II)$ ions $(Pb^{2+})$ and chloride ions $(Cl^-)$.
Since a solid precipitate is being converted into its constituent ions in a solution,this process is classified as a precipitate dissolution reaction.
Therefore,the correct option is $B$.

(A) The reaction between silver nitrate $(AgNO_3)$ and sodium oxalate $(Na_2C_2O_4)$ results in the formation of silver oxalate $(Ag_2C_2O_4)$,which is an insoluble salt that precipitates out of the solution.
The chemical equation is: $2AgNO_3(aq) + Na_2C_2O_4(aq) \longrightarrow Ag_2C_2O_4(s) \downarrow + 2NaNO_3(aq)$.
Since a solid precipitate is formed from the reaction of two aqueous solutions,this is classified as a precipitate formation reaction.

(A) The reaction $CaCl_2 + Na_2C_2O_4 \longrightarrow CaC_2O_4\downarrow + 2NaCl$ involves the mixing of two aqueous solutions,$CaCl_2$ and $Na_2C_2O_4$.
Calcium oxalate $(CaC_2O_4)$ is an insoluble salt that forms a white precipitate.
Since a solid precipitate is formed as a product,this is classified as a precipitate formation reaction.
Therefore,the correct option is $A$.

(A) The reaction between $AlCl_3$ and $NaOH$ results in the formation of aluminum hydroxide,$Al(OH)_3$,which is an insoluble white precipitate.
Since the product $Al(OH)_3$ is formed as a solid precipitate $(downarrow)$,this is classified as a precipitate formation reaction.
Therefore,the correct assignment is $A$.

(B) The reaction involves the solid precipitate $BaCO_3$ reacting with $CO_2$ and $H_2O$ to form a soluble bicarbonate,$Ba(HCO_3)_2$.
Since the solid precipitate is dissolving into the solution,this is a precipitate dissolution reaction.
Therefore,the correct assignment is $B$.

(B) The reaction $ZnS(s) + 2HCl(aq) \longrightarrow ZnCl_2(aq) + H_2S(g)$ involves the dissolution of a solid precipitate $(ZnS)$ by an acid $(HCl)$.
$ZnS$ is an insoluble sulfide that reacts with strong acid to form soluble zinc chloride and hydrogen sulfide gas.
Therefore,this is a precipitate dissolution reaction.

(A) The reaction is $Ba(CH_3COO)_2 + K_2CrO_4 \longrightarrow BaCrO_4 \downarrow + 2CH_3COOK$.
In this reaction,barium acetate reacts with potassium chromate to form barium chromate,which is an insoluble solid (precipitate) and potassium acetate.
Since a solid precipitate $(BaCrO_4)$ is formed as a product,this is a precipitate formation reaction.
Therefore,the correct option is $A$.

(C) The reaction involves the conversion of silver carbonate $(Ag_2CO_3)$,which is a precipitate,into silver chloride $(AgCl)$,which is also a precipitate.
Since the initial precipitate $(Ag_2CO_3)$ is being consumed and a new precipitate $(AgCl)$ is being formed,this is a precipitate exchange reaction.
Therefore,the correct classification is $C$.

(D) The reaction $HgS + HNO_3 \text{ (Conc.)} \longrightarrow \text{No reaction}$ indicates that $HgS$ (mercuric sulfide) is highly insoluble in concentrated nitric acid due to its extremely low solubility product constant $(K_{sp})$.
Since no dissolution occurs,this corresponds to the category of 'No reaction'.

(A) The given reaction is $Ca(OH)_2 + 2HF \longrightarrow CaF_2 \downarrow + 2H_2O$.
In this reaction,calcium hydroxide reacts with hydrofluoric acid to form calcium fluoride,which is an insoluble salt that precipitates out of the solution.
The downward arrow $(\downarrow)$ indicates the formation of a precipitate.
Therefore,this is a precipitate formation reaction.

(A) The given reaction is $Pb(AcO)_2 + H_2S \longrightarrow PbS \downarrow + 2AcOH$.
In this reaction,lead$(II)$ acetate reacts with hydrogen sulfide to produce lead$(II)$ sulfide,which is an insoluble solid that settles as a precipitate.
Since a solid precipitate $(PbS)$ is formed from the reaction of two aqueous solutions,it is classified as a precipitate formation reaction.

(A) The reaction involves the formation of $CdS$ (cadmium sulfide) as a solid precipitate from the complex salt $K_2[Cd(CN)_4]$ upon reaction with $H_2S$.
Since $CdS$ is a highly insoluble salt,it precipitates out of the solution.
Therefore,this is a precipitate formation reaction.

(C) The reaction given is $BaSO_3(s) + SO_2(g) + H_2O(l) \longrightarrow Ba(HSO_3)_2(aq)$.
In this reaction,the solid white precipitate of barium sulfite $(BaSO_3)$ reacts with sulfur dioxide and water to form barium bisulfite $(Ba(HSO_3)_2)$,which is soluble in water.
Since the product $Ba(HSO_3)_2$ is a clear,colourless solution,the correct assignment is $C$.

(D) An ion exchange reaction proceeds in the forward direction if a precipitate is formed,as this removes ions from the solution and drives the equilibrium forward according to Le Chatelier's principle.
$A$. $2AgCl(s) + CaF_2(aq) \rightarrow 2AgF(aq) + CaCl_2(aq)$ (No precipitate formed,reaction does not proceed forward).
$B$. $BaSO_4(s) + 2NaOH(aq) \rightarrow Ba(OH)_2(aq) + Na_2SO_4(aq)$ (No precipitate formed,reaction does not proceed forward).
$C$. $Pb(NO_3)_2(aq) + 2CH_3COONa(aq) \rightarrow Pb(CH_3COO)_2(aq) + 2NaNO_3(aq)$ (All products are soluble,reaction does not proceed forward).
$D$. $Na_2CrO_4(aq) + BaCl_2(aq) \rightarrow BaCrO_4(s) \downarrow + 2NaCl(aq)$ (The formation of the insoluble yellow precipitate $BaCrO_4$ drives the reaction in the forward direction).

(D) $HgS$ is a very insoluble sulfide. It does not dissolve in dilute acids or $NH_3$ solution.
However,it dissolves in a solution of sodium sulfide $(Na_2S)$ due to the formation of a soluble complex:
$HgS + Na_2S(aq.) \rightleftharpoons Na_2[HgS_2] \text{ (soluble)}$

(B) In qualitative analysis,$Cd^{2+}$ belongs to group $II$ and $Ni^{2+}$ belongs to group $IV$.
Group $II$ cations are precipitated as sulfides in the presence of dilute $HCl$ because their solubility product $(K_{sp})$ is very low.
Group $IV$ cations require a higher concentration of sulfide ions $(S^{2-})$ for precipitation,which is achieved in a basic medium.
Since the solution is acidic,the concentration of $S^{2-}$ is very low due to the common ion effect of $H^+$.
Thus,only $CdS$ (which has a much lower $K_{sp}$ than $NiS$) precipitates,while $NiS$ remains in the solution.

(B) The reaction between aqueous copper$(II)$ chloride and ammonium sulfide is a double displacement reaction:
$CuCl_2(aq.) + (NH_4)_2S(aq.) \to CuS(s) \downarrow + 2NH_4Cl(aq.)$
Here,$CuS$ is formed as a black precipitate (solid),while $NH_4Cl$ remains dissolved in the aqueous solution.

(B) $AgCl$ and $AgI$ are both insoluble in dil. $HNO_3$.
Both are soluble in strong complexing agents like $KCN$ and $Na_2S_2O_3$ (Hypo solution).
However,$AgCl$ is soluble in dilute $NH_4OH$ solution due to the formation of the complex $[Ag(NH_3)_2]Cl$,whereas $AgI$ is insoluble even in concentrated $NH_4OH$ solution due to its very low solubility product $(K_{sp})$.
Therefore,dilute $NH_4OH$ can be used to selectively dissolve $AgCl$ and separate it from $AgI$.

(C) The precipitation of $AgBr$ starts when the ionic product exceeds the solubility product $(K_{SP})$.
$AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$
$K_{SP} = [Ag^+][Br^-]$
Given $[Ag^+] = 0.05 \ M = 5 \times 10^{-2} \ M$.
$5 \times 10^{-13} = (5 \times 10^{-2}) \times [Br^-]$
$[Br^-] = \frac{5 \times 10^{-13}}{5 \times 10^{-2}} = 10^{-11} \ M$.
Number of moles of $Br^-$ required $= \text{Molarity} \times \text{Volume in Liters} = 10^{-11} \ mol/L \times 0.5 \ L = 0.5 \times 10^{-11} \ mol$.
Since $KBr$ dissociates as $KBr \rightarrow K^+ + Br^-$,moles of $KBr$ required $= 0.5 \times 10^{-11} \ mol$.
Mass of $KBr = \text{moles} \times \text{molar mass} = 0.5 \times 10^{-11} \ mol \times 119 \ g/mol = 59.5 \times 10^{-11} \ g = 5.95 \times 10^{-10} \ g$.

(A) The precipitation of a salt occurs when the ionic product exceeds its solubility product $(K_{sp})$.
For a given concentration of $Ag^+$ ions,the salt with the lowest $K_{sp}$ value will reach its ionic product limit first and precipitate first.
Comparing the given $K_{sp}$ values:
$K_{sp}(AgI) = 1.7 \times 10^{-16}$
$K_{sp}(AgBr) = 3.5 \times 10^{-13}$
$K_{sp}(AgCl) = 1.2 \times 10^{-10}$
$K_{sp}(AgSCN) = 7.1 \times 10^{-7}$
Since $1.7 \times 10^{-16}$ is the smallest value among the given $K_{sp}$ constants,$AgI$ will precipitate first.
Therefore,$I^-$ will be the first to precipitate.

(A) $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$
The solubility product expression is $K_{sp} = [Ag^{+}][Cl^{-}]$.
Given $K_{sp} = 1.8 \times 10^{-10}$ and $[Ag^{+}] = 4 \times 10^{-3}\ M$.
For precipitation to occur,the ionic product must exceed the solubility product $(Q > K_{sp})$.
$1.8 \times 10^{-10} = (4 \times 10^{-3}) \times [Cl^{-}]$
$[Cl^{-}] = \frac{1.8 \times 10^{-10}}{4 \times 10^{-3}} = 0.45 \times 10^{-7} = 4.5 \times 10^{-8}\ M$.
Therefore,the concentration of $Cl^{-}$ must be greater than $4.5 \times 10^{-8}\ M$ for $AgCl$ to precipitate.

(A) The dissociation equilibrium for silver chromate is:
$Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$
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 in the balanced chemical equation.
Therefore,$K_{sp} = [Ag^{+}]^2 [CrO_4^{2-}]$.

(A) The solubility product $(K_{sp})$ of $Al(OH)_3$ is $8.5 \times 10^{-23}$.
The solubility product $(K_{sp})$ of $Zn(OH)_2$ is $1.8 \times 10^{-14}$.
Precipitation occurs when the ionic product exceeds the solubility product $(K_{sp})$.
Since $K_{sp}(Al(OH)_3) < K_{sp}(Zn(OH)_2)$,$Al(OH)_3$ requires a much lower concentration of $OH^-$ ions to reach its solubility product limit compared to $Zn(OH)_2$.
Therefore,$Al^{3+}$ ions will precipitate first as $Al(OH)_3$ upon the addition of $NH_4OH$.
Thus,option $A$ is correct.

(A) To find which compound precipitates first,we calculate the $[Ag^{+}]$ required for each:
$1. \ AgBr: [Ag^{+}] = \frac{K_{sp}}{[Br^{-}]} = \frac{5 \times 10^{-13}}{0.1} = 5 \times 10^{-12} \ M$
$2. \ AgCl: [Ag^{+}] = \frac{K_{sp}}{[Cl^{-}]} = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \ M$
$3. \ Ag_2CO_3: [Ag^{+}]^2 = \frac{K_{sp}}{[CO_3^{2-}]} = \frac{8.1 \times 10^{-12}}{0.1} = 8.1 \times 10^{-11} \implies [Ag^{+}] = 9 \times 10^{-6} \ M$
$4. \ Ag_3AsO_4: [Ag^{+}]^3 = \frac{K_{sp}}{[AsO_4^{3-}]} = \frac{1 \times 10^{-22}}{0.1} = 10^{-21} \implies [Ag^{+}] = 10^{-7} \ M$
Comparing the values,$5 \times 10^{-12} < 10^{-7} < 9 \times 10^{-6} < 1.8 \times 10^{-9}$.
Thus,$AgBr$ requires the lowest $[Ag^{+}]$ to precipitate.

(C) The solubility product constant of $AgCl$ is given by $(K_{sp})_{AgCl} = [Ag^+][Cl^-]$.
In $0.2 \, M \, NaCl$,the concentration of chloride ions is $[Cl^-] = 0.2 \, M$. Let the solubility be $x$.
Then,$(K_{sp})_{AgCl} = x \times 0.2$,which gives $x = \frac{(K_{sp})_{AgCl}}{0.2} = 5(K_{sp})_{AgCl}$.
In $0.1 \, M \, AgNO_3$,the concentration of silver ions is $[Ag^+] = 0.1 \, M$. Let the solubility be $y$.
Then,$(K_{sp})_{AgCl} = 0.1 \times y$,which gives $y = \frac{(K_{sp})_{AgCl}}{0.1} = 10(K_{sp})_{AgCl}$.
Comparing the two values,$5(K_{sp})_{AgCl} < 10(K_{sp})_{AgCl}$,therefore $x < y$.

(D) An electrolyte $MX_2$ dissociates as follows:
$MX_2 \rightleftharpoons M^{2+} + 2X^{-}$
If $s$ is the solubility of $MX_2$,then at equilibrium:
$[M^{2+}] = s$
$[X^{-}] = 2s$
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [M^{2+}][X^{-}]^2$
$K_{sp} = (s)(2s)^2 = 4s^3$
Given $s = 0.5 \times 10^{-4} \ mol/L$:
$K_{sp} = 4 \times (0.5 \times 10^{-4})^3$
$K_{sp} = 4 \times (0.125 \times 10^{-12})$
$K_{sp} = 0.5 \times 10^{-12} = 5 \times 10^{-13}$

(B) The solubility equilibrium is $Cu(OH)_2(s) \rightleftharpoons Cu^{2+}(aq) + 2OH^-(aq)$.
Let the solubility be $S$. Then $[Cu^{2+}] = S$ and $[OH^-] = 2S$.
The solubility product expression is $K_{sp} = [Cu^{2+}][OH^-]^2 = S(2S)^2 = 4S^3$.
Given $K_{sp} = 4.0 \times 10^{-6}$,we have $4S^3 = 4.0 \times 10^{-6}$,so $S^3 = 10^{-6}$,which gives $S = 10^{-2} \ M$.
Therefore,$[OH^-] = 2S = 2 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(2 \times 10^{-2}) = 2 - \log(2) = 2 - 0.301 = 1.699 \approx 1.7$.
Finally,$pH = 14 - pOH = 14 - 1.7 = 12.3$.