Solubility product Questions in English

(C) $AX_{2}$ is ionized as follows:
$AX_{2} \rightleftharpoons A^{2+} + 2X^{-}$
Solubility product of $AX_{2}$ is given by:
$K_{sp} = [A^{2+}][X^{-}]^{2} = (S) \times (2S)^{2} = 4S^{3}$
Given $K_{sp} = 3.2 \times 10^{-11}$
$4S^{3} = 3.2 \times 10^{-11}$
$S^{3} = 0.8 \times 10^{-11} = 8 \times 10^{-12}$
$S = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \, mol / L$
Thus,the solubility is $2 \times 10^{-4} \, mol / L$.

(D) Solubility of a salt containing a basic anion or hydroxide increases as the $pH$ decreases (i.e.,in acidic medium) because $H^+$ ions react with the basic anions to form weak acids or water.
$Sr(OH)_2$ is a base that dissociates as: $Sr(OH)_2 \rightarrow Sr^{2+} + 2OH^-$.
In an acidic solution,the high concentration of $H^+$ ions reacts with $OH^-$ ions to form water: $H^+ + OH^- \rightarrow H_2O$.
This removal of $OH^-$ ions shifts the equilibrium to the right according to Le Chatelier's principle,thereby increasing the solubility of $Sr(OH)_2$.

(B) For $A_2X$: The dissociation is $A_2X \rightleftharpoons 2A^+ + X^{2-}$.
$K_{sp} = [2S_1]^2 [S_1] = 4S_1^3$.
$4.0 \times 10^{-12} = 4S_1^3$ $\Rightarrow S_1^3 = 10^{-12}$ $\Rightarrow S_1 = 10^{-4} \text{ mol L}^{-1}$.
For $MX$: The dissociation is $MX \rightleftharpoons M^{2+} + X^{2-}$.
$K_{sp} = S_2^2$.
$4.0 \times 10^{-12} = S_2^2 \Rightarrow S_2 = 2.0 \times 10^{-6} \text{ mol L}^{-1}$.
Ratio $\frac{S(A_2X)}{S(MX)} = \frac{10^{-4}}{2.0 \times 10^{-6}} = \frac{100}{2} = 50$.

(A) In pure water,the solubility product constant $K_{sp}$ is given by $K_{sp} = S^{2} = (8.0 \times 10^{-4})^{2} = 64 \times 10^{-8}$.
In $0.01 \ M \ H_{2}SO_{4}$,the dissociation is $H_{2}SO_{4} \rightarrow 2H^{+} + SO_{4}^{2-}$.
The concentration of $SO_{4}^{2-}$ from $H_{2}SO_{4}$ is $0.01 \ M$.
Let the solubility of $CdSO_{4}$ in this solution be $x \ mol \ L^{-1}$.
$CdSO_{4} \rightleftharpoons Cd^{2+} + SO_{4}^{2-}$.
The equilibrium concentrations are $[Cd^{2+}] = x$ and $[SO_{4}^{2-}] = (x + 0.01) \ M$.
Since $x \ll 0.01$,we can approximate $(x + 0.01) \approx 0.01$.
$K_{sp} = [Cd^{2+}][SO_{4}^{2-}] = x(0.01) = 64 \times 10^{-8}$.
$x = \frac{64 \times 10^{-8}}{0.01} = 64 \times 10^{-6} \ mol \ L^{-1}$.
Thus,the value is $64$.

(B) The dissolution reaction is $AgCN_{(s)} + H^{+}_{(aq)} \rightleftharpoons Ag^{+}_{(aq)} + HCN_{(aq)}$.
The equilibrium constant $K_{eq} = \frac{K_{sp}(AgCN)}{K_a(HCN)} = \frac{2.2 \times 10^{-16}}{6.2 \times 10^{-10}} \approx 3.55 \times 10^{-7}$.
Since $s = [Ag^{+}] = [HCN]$ and $[H^{+}] = 10^{-pH} = 10^{-3}$,we have $K_{eq} = \frac{s^2}{[H^{+}]}$.
$s^2 = K_{eq} \times [H^{+}] = 3.55 \times 10^{-7} \times 10^{-3} = 3.55 \times 10^{-10}$.
$s = \sqrt{3.55 \times 10^{-10}} \approx 1.88 \times 10^{-5} \approx 1.9 \times 10^{-5}$.

(D) The solubility product of $PbI_{2}$ is given by $K_{sp} = [Pb^{2+}][I^{-}]^{2} = 8.0 \times 10^{-9}$.
In $0.1 \ M$ $Pb(NO_{3})_{2}$ solution,$Pb(NO_{3})_{2}$ dissociates completely as $Pb(NO_{3})_{2} \rightarrow Pb^{2+} + 2NO_{3}^{-}$.
Thus,the concentration of $Pb^{2+}$ ions from $Pb(NO_{3})_{2}$ is $0.1 \ M$.
Let the solubility of $PbI_{2}$ be $s \ mol/L$. The dissociation of $PbI_{2}$ is $PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)$.
The total concentration of $Pb^{2+}$ is $(0.1 + s) \approx 0.1 \ M$ (since $s$ is very small) and $[I^{-}] = 2s$.
Substituting these into the $K_{sp}$ expression: $8.0 \times 10^{-9} = (0.1)(2s)^{2}$.
$8.0 \times 10^{-9} = 0.1 \times 4s^{2} = 0.4s^{2}$.
$s^{2} = \frac{8.0 \times 10^{-9}}{0.4} = 20 \times 10^{-9} = 2.0 \times 10^{-8}$.
$s = \sqrt{2.0 \times 10^{-8}} = \sqrt{2} \times 10^{-4} = 1.41 \times 10^{-4} = 141 \times 10^{-6} \ mol/L$.
Comparing this with $x \times 10^{-6} \ mol/L$,we get $x = 141$.

(C) The dissociation of $Ca(OH)_2$ is represented as:
$Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$
If $s$ is the solubility,then $[Ca^{2+}] = s$ and $[OH^-] = 2s$.
The solubility product expression is:
$K_{sp} = [Ca^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$
Given $K_{sp} = 5.5 \times 10^{-6}$,we have:
$4s^3 = 5.5 \times 10^{-6}$
$s^3 = \frac{5.5}{4} \times 10^{-6} = 1.375 \times 10^{-6}$
$s = (1.375 \times 10^{-6})^{1/3} \approx 1.11 \times 10^{-2} \text{ M}$

(D) The dissociation of the salt is given by: $A_{3}B_{2(s)} \rightleftharpoons 3A^{2+}_{(aq)} + 2B^{3-}_{(aq)}$
Let the molar solubility be $s \ mol \ L^{-1}$.
At equilibrium,the concentrations are $[A^{2+}] = 3s$ and $[B^{3-}] = 2s$.
The solubility product expression is $K_{sp} = [A^{2+}]^{3} [B^{3-}]^{2}$.
Substituting the values: $K_{sp} = (3s)^{3} (2s)^{2} = (27s^{3}) (4s^{2}) = 108s^{5}$.
Given that molar solubility $s = \frac{x}{M}$,where $x$ is solubility in $g \ L^{-1}$ and $M$ is molar mass in $g \ mol^{-1}$.
Thus,$K_{sp} = 108(\frac{x}{M})^{5}$.
Comparing this with the given expression $K_{sp} = a(\frac{x}{M})^{5}$,we get $a = 108$.

(C) The dissolution equilibrium of $Zn(OH)_{2}$ is given by:
$Zn(OH)_{2(s)} \rightleftharpoons Zn^{2+}_{(aq)} + 2OH^{-}_{(aq)}$
Let the molar solubility be $S$. The concentration of $OH^{-}$ ions from $NaOH$ is $0.1 \, M$. Since $S$ is very small,the total concentration of $OH^{-}$ is approximately $0.1 \, M$.
$K_{sp} = [Zn^{2+}][OH^{-}]^{2}$
$2 \times 10^{-20} = S \times (0.1)^{2}$
$2 \times 10^{-20} = S \times 10^{-2}$
$S = \frac{2 \times 10^{-20}}{10^{-2}} = 2 \times 10^{-18} \, M$
Comparing this with $x \times 10^{-18} \, M$,we get $x = 2$.

(A) $(i)$ $\text{Concentration of } [Ag^{+}] \text{ required to precipitate } AgCl_{(s)}:$
$K_{sp} = [Ag^{+}][Cl^{-}] = 1.7 \times 10^{-10}$
$[Ag^{+}] = \frac{1.7 \times 10^{-10}}{0.1} = 1.7 \times 10^{-9} \ M$
$(ii)$ $\text{Concentration of } [Ag^{+}] \text{ required to precipitate } Ag_{2}CrO_{4(s)}:$
$K_{sp} = [Ag^{+}]^{2}[CrO_{4}^{2-}] = 1.9 \times 10^{-12}$
$[Ag^{+}]^{2} = \frac{1.9 \times 10^{-12}}{0.001} = 1.9 \times 10^{-9}$
$[Ag^{+}] = \sqrt{1.9 \times 10^{-9}} \approx 4.36 \times 10^{-5} \ M$
Since the $[Ag^{+}]$ required to precipitate $AgCl$ $(1.7 \times 10^{-9} \ M)$ is lower than that required for $Ag_{2}CrO_{4}$ $(4.36 \times 10^{-5} \ M)$,$AgCl$ will precipitate first.

(A) The dissociation of bismuth sulphide is given by: $Bi_{2}S_{3}(s) \rightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)$.
Let the solubility be $s \ mol \ L^{-1}$. Then the concentrations are $[Bi^{3+}] = 2s$ and $[S^{2-}] = 3s$.
The solubility product expression is $K_{sp} = [Bi^{3+}]^{2} [S^{2-}]^{3}$.
Substituting the values: $K_{sp} = (2s)^{2} (3s)^{3} = 4s^{2} \times 27s^{3} = 108s^{5}$.
Given $K_{sp} = 1.08 \times 10^{-73}$,we have $108s^{5} = 1.08 \times 10^{-73}$.
$s^{5} = \frac{1.08 \times 10^{-73}}{108} = 0.01 \times 10^{-73} = 10^{-75}$.
Taking the fifth root: $s = (10^{-75})^{1/5} = 10^{-15} \ mol \ L^{-1}$.

(D) The solubility of a sparingly soluble salt like $AgCl$ is governed by the common ion effect.
In the presence of common ions like $Cl^-$ (from $KCl$ or $HCl$) or $Ag^+$ (from $AgNO_3$),the solubility of $AgCl$ decreases due to the common ion effect.
In deionized water,there are no common ions present to suppress the dissociation of $AgCl$.
Therefore,the solubility of $AgCl$ is maximum in deionized water.

(A) First,calculate the solubility $S$ in $mol / L$:
$S = \frac{2.34 \times 10^{-3} \, g}{78 \, g/mol} \times \frac{1}{0.1 \, L} = \frac{0.03 \times 10^{-3}}{0.1} = 3 \times 10^{-4} \, mol / L$.
The dissociation of $CaF_{2}$ is $CaF_{2} \rightleftharpoons Ca^{2+} + 2F^-$.
$K_{sp} = [Ca^{2+}][F^-]^{2} = (S)(2S)^{2} = 4S^{3}$.
$K_{sp} = 4 \times (3 \times 10^{-4})^{3} = 4 \times 27 \times 10^{-12} = 108 \times 10^{-12} = 0.0108 \times 10^{-8} \, (mol / L)^{3}$.
Thus,$x = 0.0108$.

(B) The dissociation of $PbS$ is given by: $PbS(s) \rightleftharpoons Pb^{2+}(aq) + S^{2-}(aq)$.
The solubility product expression is $K_{sp} = [Pb^{2+}][S^{2-}] = S \times S = S^{2}$.
Given $K_{sp} = 8 \times 10^{-28}$.
Therefore,$S = \sqrt{8 \times 10^{-28}} = \sqrt{4 \times 2 \times 10^{-28}} = 2 \sqrt{2} \times 10^{-14} \ mol \ L^{-1}$.
Using the given value $\sqrt{2} = 1.41$,we get $S = 2 \times 1.41 \times 10^{-14} = 2.82 \times 10^{-14} \ mol \ L^{-1}$.
To express this in the form $x \times 10^{-16} \ mol \ L^{-1}$,we write $2.82 \times 10^{-14} = 282 \times 10^{-16}$.
Thus,the value of $x$ is $282$.

(B) For $CuI_{(s)} \rightleftharpoons Cu^{+}_{(aq)} + I^{-}_{(aq)}$:
$K_{sp} = [Cu^{+}][I^{-}] = x^{2} = 4 \times 10^{-12}$
$x = \sqrt{4 \times 10^{-12}} = 2 \times 10^{-6} \ M$
For $Ag_{2}CrO_{4(s)} \rightleftharpoons 2Ag^{+}_{(aq)} + Cr{O_{4}}^{2-}_{(aq)}$:
$K_{sp} = [Ag^{+}]^{2} [CrO_{4}^{2-}] = (2y)^{2}(y) = 4y^{3} = 4 \times 10^{-12}$
$y^{3} = 10^{-12} \implies y = 10^{-4} \ M$
Ratio of solubilities $x/y = (2 \times 10^{-6}) / (10^{-4}) = 2 \times 10^{-2} = 0.02$.

(B) The dissolution of $Ag_{2}CrO_{4}$ is given by the equilibrium: $Ag_{2}CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + Cr{O_{4}}^{2-}_{(aq)}$.
The solubility product expression is $K_{sp} = [Ag^+]^2 [CrO_{4}^{2-}]$.
Given $K_{sp} = 1.1 \times 10^{-12}$ and the common ion concentration $[CrO_{4}^{2-}] = 5 \times 10^{-3} \ M$.
Substituting the values: $1.1 \times 10^{-12} = [Ag^+]^2 (5 \times 10^{-3})$.
$[Ag^+]^2 = \frac{1.1 \times 10^{-12}}{5 \times 10^{-3}} = 0.22 \times 10^{-9} = 2.2 \times 10^{-10}$.
$[Ag^+] = \sqrt{2.2 \times 10^{-10}} \approx 1.48 \times 10^{-5} \ M$.
This is closest to $1.5 \times 10^{-5} \ M$.

(C) The solubility product expression for $Mg(OH)_2$ is given by:
$K_{sp} = [Mg^{2+}][OH^{-}]^2$
Given $K_{sp} = 5.6 \times 10^{-12}$ and $[Mg^{2+}] = 10^{-10} \ M$.
Substituting the values:
$5.6 \times 10^{-12} = (10^{-10}) \times [OH^{-}]^2$
$[OH^{-}]^2 = \frac{5.6 \times 10^{-12}}{10^{-10}} = 5.6 \times 10^{-2}$
$[OH^{-}] = \sqrt{5.6 \times 10^{-2}} \approx 0.2366 \ M \approx 0.24 \ M$
Thus,the concentration of $OH^{-}$ is $0.24 \ M$.

(D) For the precipitation of $Mg(OH)_2$,the ionic product must exceed the solubility product $(K_{sp})$.
The equilibrium is: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$
$K_{sp} = [Mg^{2+}][OH^-]^2 = 1.0 \times 10^{-12}$
Given $[Mg^{2+}] = 0.01 \ M = 10^{-2} \ M$.
Substituting the values: $10^{-2} \times [OH^-]^2 = 10^{-12}$
$[OH^-]^2 = 10^{-10}$
$[OH^-] = 10^{-5} \ M$
$pOH = -\log[OH^-] = -\log(10^{-5}) = 5$
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - 5 = 9$.

(D) The dissociation of $BaSO_4$ is given by: $BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$
Let the solubility be $S \, mol \, L^{-1}$.
Then,$K_{sp} = [Ba^{2+}][SO_4^{2-}] = S \times S = S^2$.
Given $K_{sp} = 1.0 \times 10^{-10}$,we have $S^2 = 1.0 \times 10^{-10}$,so $S = 1.0 \times 10^{-5} \, mol \, L^{-1}$.
To convert solubility from $mol \, L^{-1}$ to $g \, L^{-1}$,multiply by the molar mass of $BaSO_4$ $(233 \, g \, mol^{-1})$:
Solubility in $g \, L^{-1} = (1.0 \times 10^{-5} \, mol \, L^{-1}) \times (233 \, g \, mol^{-1}) = 233 \times 10^{-5} \, g \, L^{-1} = 2.33 \times 10^{-3} \, g \, L^{-1}$.
Thus,the value is closest to $2.3 \times 10^{-3} \, g \, L^{-1}$.

(C) The molar mass of $AgCl = 107.9 + 35.5 = 143.4 \ g \ mol^{-1}$.
Solubility in $mol \ L^{-1}$ $(S)$ = $\frac{1.434 \times 10^{-3} \ g \ L^{-1}}{143.4 \ g \ mol^{-1}} = 10^{-5} \ mol \ L^{-1}$.
For $AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$,the solubility product $K_{sp} = [Ag^{+}][Cl^{-}] = S^2$.
$K_{sp} = (10^{-5})^2 = 10^{-10}$.
Therefore,$-\log K_{sp} = -\log(10^{-10}) = 10$.

(A) $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$
$0.1 \ M \quad \quad \quad 0.2 \ M \quad 0.1 \ M$
For $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$:
$K_{sp} = [Ba^{2+}][SO_4^{2-}]$
$1 \times 10^{-10} = S \times (S + 0.1)$
Since $S$ is very small,$S + 0.1 \approx 0.1$.
$1 \times 10^{-10} = S \times 0.1$
$S = 10^{-9} \ mol \ L^{-1}$
Solubility in $g \ L^{-1} = S \times \text{Molar mass} = 10^{-9} \times 233 \ g \ L^{-1} = 233 \times 10^{-9} \ g \ L^{-1}$.
Thus,the value is $233$.

(C) Step $1$: Calculate the final concentrations of ions after mixing.
Total volume $= 25.0 \ mL + 25.0 \ mL = 50.0 \ mL$.
$[Ba^{2+}] = \frac{25.0 \times 0.050}{50.0} = 0.025 \ M$.
$[F^{-}] = \frac{25.0 \times 0.020}{50.0} = 0.010 \ M$.
Step $2$: Calculate the ionic product $Q = [Ba^{2+}][F^{-}]^2$.
$Q = (0.025) \times (0.010)^2 = 0.025 \times 10^{-4} = 2.5 \times 10^{-6}$.
Step $3$: Calculate the ratio $\frac{Q}{K_{sp}}$.
Given $K_{sp} = 0.5 \times 10^{-6} = 5 \times 10^{-7}$.
Ratio $= \frac{2.5 \times 10^{-6}}{0.5 \times 10^{-6}} = \frac{2.5}{0.5} = 5$.

(B) Precipitation begins when the ionic product $Q_{sp}$ equals the solubility product $K_{sp}$.
For the reaction $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$,the expression is $Q_{sp} = [Mg^{2+}][OH^-]^2$.
Given $[Mg^{2+}] = 0.10 \ M$ and $K_{sp} = 1 \times 10^{-11}$.
Setting $Q_{sp} = K_{sp}$:
$0.10 \times [OH^-]^2 = 1 \times 10^{-11}$
$[OH^-]^2 = 10^{-10}$
$[OH^-] = 10^{-5} \ M$.
Now,calculate $pOH$:
$pOH = -\log[OH^-] = -\log(10^{-5}) = 5$.
Finally,calculate $pH$:
$pH + pOH = 14$
$pH = 14 - 5 = 9$.

(B) The solubility $S$ in $mol \ L^{-1}$ is calculated as: $S = \frac{W \times 1000}{M \times 100} = \frac{10W}{M} \ mol \ L^{-1}$.
The dissociation of calcium phosphate is: $Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)$.
The solubility product expression is: $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 = (3S)^3 (2S)^2$.
$K_{sp} = 27S^3 \times 4S^2 = 108S^5$.
Substituting $S = \frac{10W}{M}$: $K_{sp} = 108 \times \left(\frac{10W}{M}\right)^5 = 108 \times 10^5 \times \left(\frac{W}{M}\right)^5 = 1.08 \times 10^7 \left(\frac{W}{M}\right)^5$.
Rounding to the nearest order of magnitude,the value is approximately $10^7 \left(\frac{W}{M}\right)^5$.

(B) The dissociation of the salt is given by: $AB_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2B^{-}_{(aq)}$
The expression for the solubility product constant $(K_{sp})$ is: $K_{sp} = [A^{2+}] [B^{-}]^2$
Given concentrations: $[A^{2+}] = 1.2 \times 10^{-4} \ M$ and $[B^{-}] = 0.24 \times 10^{-3} \ M = 2.4 \times 10^{-4} \ M$
Substituting the values: $K_{sp} = (1.2 \times 10^{-4}) \times (2.4 \times 10^{-4})^2$
$K_{sp} = (1.2 \times 10^{-4}) \times (5.76 \times 10^{-8})$
$K_{sp} = 6.912 \times 10^{-12} \ M^3$
Thus,the correct option is $B$.

(D) For $MX$: $K_{sp} = s^2 \implies s = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \text{ mol } dm^{-3}$.
For $MX_2$: $K_{sp} = 4s^3 \implies s = \sqrt[3]{(3.2 \times 10^{-14}) / 4} = \sqrt[3]{8.0 \times 10^{-15}} = 2.0 \times 10^{-5} \text{ mol } dm^{-3}$.
For $M_3X$: $K_{sp} = 27s^4 \implies s = \sqrt[4]{(2.7 \times 10^{-15}) / 27} = \sqrt[4]{1.0 \times 10^{-16}} = 1.0 \times 10^{-4} \text{ mol } dm^{-3}$.
Comparing the values: $2.0 \times 10^{-4} > 1.0 \times 10^{-4} > 2.0 \times 10^{-5}$.
Therefore,the order of solubility is $MX > M_3X > MX_2$.

(A) This is a case of simultaneous solubility of salts with a common ion $(Cl^{-})$.
Since $K_{sp}(CuCl) \gg K_{sp}(AgCl)$,the concentration of $Cl^{-}$ ions in the solution is primarily determined by the dissolution of $CuCl$.
For $CuCl \rightleftharpoons Cu^{+} + Cl^{-}$,let $s$ be the solubility.
$K_{sp}(CuCl) = s^2 = 1.0 \times 10^{-6} \Rightarrow s = 10^{-3} \ M$.
Thus,$[Cl^{-}] \approx 10^{-3} \ M$.
Now,for $AgCl \rightleftharpoons Ag^{+} + Cl^{-}$,the solubility product expression is:
$K_{sp}(AgCl) = [Ag^{+}][Cl^{-}] = 1.6 \times 10^{-10}$.
Substituting the value of $[Cl^{-}]$:
$[Ag^{+}] \times 10^{-3} = 1.6 \times 10^{-10}$.
$[Ag^{+}] = 1.6 \times 10^{-7} \ M$.
Comparing this with $1.6 \times 10^{-x}$,we get $x = 7$.

(A) The solubility $(S)$ of a salt of a weak acid $AB$ in an acidic medium is given by the formula: $S = \sqrt{K_{sp} \left(1 + \frac{[H^{+}]}{K_{a}}\right)}$
Given: $K_{sp} = 2 \times 10^{-10}$,$K_{a} = 1 \times 10^{-8}$,and $pH = 3$,so $[H^{+}] = 10^{-3} \ M$.
Substituting the values:
$S = \sqrt{2 \times 10^{-10} \left(1 + \frac{10^{-3}}{10^{-8}}\right)}$
$S = \sqrt{2 \times 10^{-10} \left(1 + 10^{5}\right)}$
Since $10^{5} \gg 1$,we can approximate $1 + 10^{5} \approx 10^{5}$.
$S = \sqrt{2 \times 10^{-10} \times 10^{5}} = \sqrt{2 \times 10^{-5}}$
$S = \sqrt{20 \times 10^{-6}} = \sqrt{20} \times 10^{-3}$
Comparing this with $Y \times 10^{-3}$,we get $Y = \sqrt{20} \approx 4.47$.

(C) To prevent the precipitation of $ZnS$,the ionic product $Q_{sp}$ must be less than or equal to the solubility product $K_{sp}$.
$Q_{sp} = [Zn^{2+}][S^{2-}] \leq K_{sp}(ZnS)$
Given: $[Zn^{2+}] = 0.05 \ M$ and $K_{sp}(ZnS) = 1.25 \times 10^{-22}$
$[S^{2-}] \leq \frac{K_{sp}}{[Zn^{2+}]} = \frac{1.25 \times 10^{-22}}{0.05} = 2.5 \times 10^{-21} \ M$
For the dissociation of $H_2S$: $H_2S \rightleftharpoons 2H^{+} + S^{2-}$
The overall dissociation constant is given by:
$K_{NET} = \frac{[H^{+}]^2 [S^{2-}]}{[H_2S]}$
$1 \times 10^{-21} = \frac{[H^{+}]^2 \times (2.5 \times 10^{-21})}{0.1}$
$[H^{+}]^2 = \frac{1 \times 10^{-21} \times 0.1}{2.5 \times 10^{-21}} = \frac{0.1}{2.5} = 0.04$
$[H^{+}] = \sqrt{0.04} = 0.2 \ M$
Thus,the minimum concentration of $H^{+}$ required is $0.20 \ M$.

(B) The dissociation of $Ag_2CrO_4$ is given by: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$.
In the presence of $0.1 \ M \ AgNO_3$,the concentration of $Ag^{+}$ ions is dominated by the strong electrolyte $AgNO_3$,so $[Ag^{+}] \approx 0.1 \ M$.
Let $s$ be the solubility of $Ag_2CrO_4$ in $mol/L$. Then $[CrO_4^{2-}] = s$.
The solubility product expression is $K_{sp} = [Ag^{+}]^2 [CrO_4^{2-}]$.
Substituting the values: $1.1 \times 10^{-12} = (0.1)^2 \times s$.
$1.1 \times 10^{-12} = 0.01 \times s$.
$s = \frac{1.1 \times 10^{-12}}{0.01} = 1.1 \times 10^{-10} \ mol/L$.

(D) $H_2SO_4 \rightarrow H^{+} + HSO_4^{-}$
$Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$
$HSO_4^{-} \rightleftharpoons H^{+} + SO_4^{2-}; \ K_{a2} = 1.2 \times 10^{-2} \ M$
Let the concentration of $SO_4^{2-}$ be $[SO_4^{2-}] = 1.8 \times 10^{-2} - x$,where $x$ is the amount of $SO_4^{2-}$ consumed by $H^{+}$ to form $HSO_4^{-}$.
$K_{a2} = \frac{[H^{+}][SO_4^{2-}]}{[HSO_4^{-}]} = \frac{(1+x)(1.8 \times 10^{-2} - x)}{(1-x)} = 1.2 \times 10^{-2}$
Assuming $x$ is small,$1+x \approx 1$ and $1-x \approx 1$,so $1.8 \times 10^{-2} - x = 1.2 \times 10^{-2} \Rightarrow x = 0.6 \times 10^{-2} \ M$.
Thus,$[SO_4^{2-}] = 1.8 \times 10^{-2} - 0.6 \times 10^{-2} = 1.2 \times 10^{-2} \ M$.
For $PbSO_4 \rightleftharpoons Pb^{2+} + SO_4^{2-}$,$K_{sp} = [Pb^{2+}][SO_4^{2-}] = s(s + 1.2 \times 10^{-2}) = 1.6 \times 10^{-8}$.
Since $s$ is very small,$s + 1.2 \times 10^{-2} \approx 1.2 \times 10^{-2}$.
$s(1.2 \times 10^{-2}) = 1.6 \times 10^{-8} \Rightarrow s = \frac{1.6}{1.2} \times 10^{-6} = 1.33 \times 10^{-6} \ M$.
Comparing with $X \times 10^{-Y} \ M$,we get $Y = 6$.

(A) The dissociation of zirconium phosphate is represented as: $Zr_3(PO_4)_4(s) \rightleftharpoons 3Zr^{4+}(aq) + 4PO_4^{3-}(aq)$
Let the molar solubility be $s$. Then,the concentration of $Zr^{4+}$ is $3s$ and the concentration of $PO_4^{3-}$ is $4s$.
The solubility product expression is: $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$
Substituting the values: $K_{sp} = (3s)^3 (4s)^4$
$K_{sp} = (27s^3) \times (256s^4) = 6912s^7$
Solving for $s$: $s^7 = \frac{K_{sp}}{6912}$
Therefore,$s = (\frac{K_{sp}}{6912})^{\frac{1}{7}}$

(A) The condition for precipitation is $Q_{ip} > K_{sp}$.
For $A(OH)_{2}$:
$[A^{2+}][OH^{-}]^2 > 9 \times 10^{-10}$
Given $[A^{2+}] = 1 \ M$,we have $[OH^{-}]^2 > 9 \times 10^{-10}$,which implies $[OH^{-}] > 3 \times 10^{-5} \ M$.
For $B(OH)_{3}$:
$[B^{3+}][OH^{-}]^3 > 27 \times 10^{-18}$
Given $[B^{3+}] = 1 \ M$,we have $[OH^{-}]^3 > 27 \times 10^{-18}$,which implies $[OH^{-}] > 3 \times 10^{-6} \ M$.
Since the concentration of $OH^{-}$ required for the precipitation of $B(OH)_{3}$ $(3 \times 10^{-6} \ M)$ is lower than that required for $A(OH)_{2}$ $(3 \times 10^{-5} \ M)$,$B(OH)_{3}$ will precipitate first.

(A) The dissociation of $Cr(OH)_3$ is represented as: $Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$
Let the molar solubility be $s \ mol/L$.
At equilibrium,the concentration of $Cr^{3+}$ is $s$ and $OH^-$ is $3s$.
The solubility product expression is: $K_{sp} = [Cr^{3+}][OH^-]^3$
Substituting the values: $K_{sp} = (s)(3s)^3 = 27s^4$
Given $K_{sp} = 1.6 \times 10^{-30}$,we have: $27s^4 = 1.6 \times 10^{-30}$
Solving for $s$: $s^4 = \frac{1.6 \times 10^{-30}}{27}$
$s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}$

(B) The solubility product constant $(K_{sp})$ represents the extent to which a sparingly soluble salt dissolves in water. The values are as follows:
$HgS \approx 4 \times 10^{-53}$
$PbS \approx 8 \times 10^{-28}$
$AgBr \approx 5 \times 10^{-13}$
$Ca(OH)_2 \approx 5.5 \times 10^{-6}$
Comparing these values,the increasing order of $K_{sp}$ is: $HgS < PbS < AgBr < Ca(OH)_2$.

(C) When equal volumes are mixed,the concentration of each species is halved.
The reaction is $AB_2 + 2XY \rightarrow AY_2 + 2XB$.
For precipitation to occur,the ionic product $Q_{sp}$ must be greater than $K_{sp}$.
$Q_{sp} = [A^{2+}][Y^-]^2$.
For option $C$: $[A^{2+}] = (2.0 \times 10^{-2} / 2) = 1.0 \times 10^{-2} \ M$ and $[Y^-] = (2.0 \times 10^{-2} / 2) = 1.0 \times 10^{-2} \ M$.
$Q_{sp} = (1.0 \times 10^{-2}) \times (1.0 \times 10^{-2})^2 = 1.0 \times 10^{-6}$.
Since $1.0 \times 10^{-6} > 5.2 \times 10^{-7}$,a precipitate will form.

(A) The dissociation of $CaC_2O_4$ is given by: $CaC_2O_{4(s)} \rightleftharpoons Ca^{2 }{(aq)} C_2O_4^{2-}{(aq)}$
Let the solubility be $S \ mol/L$. Then $K_{sp} = [Ca^{2 }][C_2O_4^{2-}] = S \times S = S^2$.
Given $K_{sp} = 2.5 \times 10^{-9}$,so $S^2 = 2.5 \times 10^{-9}$.
$S = \sqrt{2.5 \times 10^{-9}} = \sqrt{25 \times 10^{-10}} = 5 \times 10^{-5} \ mol/L$.
To find the mass in grams per litre,multiply the molar solubility by the molecular mass: $\text{Mass} = S \times \text{Molecular mass} = 5 \times 10^{-5} \ mol/L \times 128 \ g/mol$.
$\text{Mass} = 640 \times 10^{-5} \ g/L = 0.0064 \ g/L$.

(C) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$.
In $0.01 \ M \ NaOH$,the concentration of $OH^-$ ions is dominated by the strong electrolyte $NaOH$,so $[OH^-] \approx 0.01 \ M = 10^{-2} \ M$.
Let the molar solubility of $Ni(OH)_2$ be $S$.
Then,$[Ni^{2+}] = S$ and $[OH^-] = (2S + 0.01) \approx 0.01 \ M$.
The solubility product expression is $K_{sp} = [Ni^{2+}][OH^-]^2$.
Substituting the values: $2 \times 10^{-15} = S \times (10^{-2})^2$.
$2 \times 10^{-15} = S \times 10^{-4}$.
$S = \frac{2 \times 10^{-15}}{10^{-4}} = 2 \times 10^{-11} \ M$.

(A) For a salt of the type $AB$ (where $A$ and $B$ are ions),the solubility $S$ is given by $S = \sqrt{K_{sp}}$.
All the given salts are of the $1:1$ electrolyte type $(AB \leftrightarrows A^{+} + B^{-})$.
Therefore,the salt with the highest $K_{sp}$ value will have the maximum solubility.
Comparing the $K_{sp}$ values:
$AgBrO_3: 2.5 \times 10^{-10}$
$MgSO_4: 2.7 \times 10^{-27}$
$CaCO_3: 2.7 \times 10^{-18}$
$FeC_2O_4: 2.4 \times 10^{-15}$
Since $2.5 \times 10^{-10}$ is the largest value among the given $K_{sp}$ values,$AgBrO_3$ has the maximum solubility.

(A) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$.
Let $s$ be the molar solubility of $Ni(OH)_2$.
In $0.10 \ M \ NaOH$,the concentration of $OH^-$ ions is $0.10 \ M$ from $NaOH$ plus $2s$ from $Ni(OH)_2$. Since $s$ is very small,we can approximate $[OH^-] \approx 0.10 \ M$.
The solubility product expression is $K_{sp} = [Ni^{2+}][OH^-]^2$.
Substituting the values: $2 \times 10^{-15} = (s)(0.10)^2$.
$2 \times 10^{-15} = s \times 0.01$.
$s = \frac{2 \times 10^{-15}}{0.01} = 2 \times 10^{-13} \ M$.

(D) The dissociation of $B_2CO_3$ is given by: $B_2CO_3(s) \rightleftharpoons 2B^+(aq) + CO_3^{2-}(aq)$.
Let the solubility of $B_2CO_3$ be $S \ M$.
Then,$[B^+] = 2S$ and $[CO_3^{2-}] = S$.
The solubility product expression is: $K_{sp} = [B^+]^2 [CO_3^{2-}] = (2S)^2(S) = 4S^3$.
Given $K_{sp} = 3.2 \times 10^{-5}$,we have $4S^3 = 3.2 \times 10^{-5}$.
$S^3 = 0.8 \times 10^{-5} = 8 \times 10^{-6}$.
Taking the cube root,$S = 2 \times 10^{-2} \ M$.
The concentration of $B^+$ is $[B^+] = 2S = 2(2 \times 10^{-2} \ M) = 4 \times 10^{-2} \ M$.

(C) For a sparingly soluble salt of the type $AX$,the dissociation is given by: $AX(s) \rightleftharpoons A^+(aq) + X^-(aq)$.
Let the solubility of the salt be $S \ mol \ dm^{-3}$.
Then,$[A^+] = S$ and $[X^-] = S$.
The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [A^+][X^-] = S \times S = S^2$.
Given $K_{sp} = 4.9 \times 10^{-13}$.
Therefore,$S^2 = 4.9 \times 10^{-13} = 49 \times 10^{-14}$.
Taking the square root on both sides: $S = \sqrt{49 \times 10^{-14}} = 7.0 \times 10^{-7} \ mol \ dm^{-3}$.

(D) For a saturated solution of $Ba(OH)_2$,the dissociation is $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$.
Given $pH = 12$,then $pOH = 14 - 12 = 2$.
Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \ M$.
From the stoichiometry,$[Ba^{2+}] = \frac{1}{2} [OH^-] = \frac{1}{2} \times 10^{-2} = 0.5 \times 10^{-2} \ M$.
The solubility product constant $K_{sp}$ is given by $K_{sp} = [Ba^{2+}][OH^-]^2$.
Substituting the values: $K_{sp} = (0.5 \times 10^{-2}) \times (10^{-2})^2 = 0.5 \times 10^{-2} \times 10^{-4} = 0.5 \times 10^{-6} = 5 \times 10^{-7}$.
Thus,the correct option is $D$.

(A) The dissociation of $AgBr$ is given by: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$.
Let the solubility be $s = 7.1 \times 10^{-7} \ mol \ dm^{-3}$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [Ag^+][Br^-] = s \times s = s^2$.
Substituting the value of $s$: $K_{sp} = (7.1 \times 10^{-7})^2$.
$K_{sp} = 50.41 \times 10^{-14} = 5.041 \times 10^{-13}$.

(B) The dissociation of $Ca_{3}(PO_{4})_{2}$ is represented as:
$Ca_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_{4}^{3-}(aq)$
If the solubility is $S \ mol \ dm^{-3}$,then the concentration of ions at equilibrium is:
$[Ca^{2+}] = 3S$
$[PO_{4}^{3-}] = 2S$
The solubility product expression is:
$K_{sp} = [Ca^{2+}]^{3} [PO_{4}^{3-}]^{2}$
Substituting the values:
$K_{sp} = (3S)^{3} (2S)^{2}$
$K_{sp} = (27S^{3}) (4S^{2})$
$K_{sp} = 108S^{5}$

(B) The dissociation of the salt $BA_2$ is given by: $BA_2 (s) \rightleftharpoons B^{2+} (aq) + 2A^- (aq)$.
Let the solubility of the salt be $s = 4 \times 10^{-4} \ mol \ dm^{-3}$.
Then,$[B^{2+}] = s$ and $[A^-] = 2s$.
The solubility product constant $K_{sp}$ is defined as: $K_{sp} = [B^{2+}][A^-]^2$.
Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.
Now,substitute $s = 4 \times 10^{-4}$:
$K_{sp} = 4 \times (4 \times 10^{-4})^3 = 4 \times (64 \times 10^{-12}) = 256 \times 10^{-12} = 2.56 \times 10^{-10}$.

(D) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.
Let the solubility of $PbI_2$ be $s \ mol \ dm^{-3}$.
Then,$[Pb^{2+}] = s$ and $[I^-] = 2s$.
The solubility product expression is $K_{sp} = [Pb^{2+}][I^-]^2$.
Substituting the values: $1.08 \times 10^{-7} = (s)(2s)^2 = 4s^3$.
Solving for $s$: $s^3 = \frac{1.08 \times 10^{-7}}{4} = 0.27 \times 10^{-7} = 27 \times 10^{-9}$.
Taking the cube root: $s = \sqrt[3]{27 \times 10^{-9}} = 3 \times 10^{-3} \ mol \ dm^{-3}$.

(B) The dissociation of calcium carbonate is given by: $CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq)$.
Let the solubility be $s = 6.4 \times 10^{-5} \ mol \ dm^{-3}$.
The solubility product constant is given by: $K_{sp} = [Ca^{2+}][CO_3^{2-}] = s \times s = s^2$.
Substituting the value of $s$: $K_{sp} = (6.4 \times 10^{-5})^2$.
$K_{sp} = 40.96 \times 10^{-10} = 4.096 \times 10^{-9}$.

(B) For a sparingly soluble salt of the type $AX_2$,the dissociation equilibrium is given by:
$AX_2(s) \rightleftharpoons A^{2+}(aq) + 2X^-(aq)$
If $s$ is the solubility of the salt in $mol \ dm^{-3}$,then the concentration of $A^{2+}$ is $s$ and the concentration of $X^-$ is $2s$.
The solubility product expression is:
$K_{sp} = [A^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$
Given $s = 1 \times 10^{-4} \ mol \ dm^{-3}$.
Substituting the value of $s$:
$K_{sp} = 4 \times (1 \times 10^{-4})^3 = 4 \times 10^{-12}$
Therefore,the correct option is $B$.

(B) For a sparingly soluble salt like $NiS$,the dissociation equilibrium is given by:
$NiS(s) \rightleftharpoons Ni^{2+}(aq) + S^{2-}(aq)$
Let the solubility of $NiS$ be $s \ mol \ dm^{-3}$.
Then,$[Ni^{2+}] = s$ and $[S^{2-}] = s$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [Ni^{2+}][S^{2-}] = s \times s = s^2$
Given $K_{sp} = 4.9 \times 10^{-5}$.
$s^2 = 4.9 \times 10^{-5} = 49 \times 10^{-6}$
$s = \sqrt{49 \times 10^{-6}} = 7.0 \times 10^{-3} \ mol \ dm^{-3}$
Therefore,the correct option is $B$.