Solubility product Questions in English

(A) $1$. Convert solubility from $g \ dm^{-3}$ to $mol \ dm^{-3}$ (molarity,$S$):
$S = \frac{1.12 \times 10^{-4} \ g \ dm^{-3}}{112 \ g \ mol^{-1}} = 1.0 \times 10^{-6} \ mol \ dm^{-3}$.
$2$. For a binary salt $(AB)$,the dissociation is $AB \rightleftharpoons A^+ + B^-$.
$3$. The solubility product expression is $K_{sp} = [A^+][B^-] = S \times S = S^2$.
$4$. Substitute the value of $S$: $K_{sp} = (1.0 \times 10^{-6})^2 = 1.0 \times 10^{-12} \ mol^2 \ dm^{-6}$.

(A) The solubility product expression for $PbS$ is given by: $K_{sp} = [Pb^{2+}][S^{2-}]$.
Given $K_{sp} = 8.0 \times 10^{-28}$ and $[S^{2-}] = 1 \times 10^{-11} \ mol \ dm^{-3}$.
Substituting the values into the equation: $8.0 \times 10^{-28} = [Pb^{2+}] \times (1 \times 10^{-11})$.
Therefore,$[Pb^{2+}] = \frac{8.0 \times 10^{-28}}{1 \times 10^{-11}} = 8.0 \times 10^{-17} \ mol \ dm^{-3}$.

(C) The dissociation of the salt $AB_2$ is given by: $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq)$.
Let the solubility of $AB_2$ be $s \ mol \ dm^{-3}$.
Then,$[A^{2+}] = s$ and $[B^-] = 2s$.
The solubility product expression is $K_{sp} = [A^{2+}][B^-]^2$.
Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.
Given $K_{sp} = 2.56 \times 10^{-10}$.
So,$4s^3 = 2.56 \times 10^{-10}$.
$s^3 = 0.64 \times 10^{-10} = 64 \times 10^{-12}$.
$s = \sqrt[3]{64 \times 10^{-12}} = 4 \times 10^{-4} \ mol \ dm^{-3}$.

(C) For a salt of the type $AB$,the solubility product $K_{sp}$ is related to solubility $S$ by the equation: $K_{sp} = S^2$.
Given $K_{sp} = 4.9 \times 10^{-13}$.
Therefore,$S = \sqrt{K_{sp}} = \sqrt{4.9 \times 10^{-13}}$.
$S = \sqrt{49 \times 10^{-14}} = 7 \times 10^{-7} \ mol \ dm^{-3}$.
Thus,the correct option is $C$.

(C) The dissociation of salt $B_2A$ is given by: $B_2A(s) \rightleftharpoons 2B^+(aq) + A^{2-}(aq)$.
Let the solubility of $B_2A$ be $s \ mol \ dm^{-3}$.
Then,$[B^+] = 2s$ and $[A^{2-}] = s$.
The solubility product expression is: $K_{sp} = [B^+]^2 [A^{2-}] = (2s)^2 (s) = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$.
So,$4s^3 = 3.2 \times 10^{-11}$.
$s^3 = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
$s = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \ mol \ dm^{-3}$.

(A) For a sparingly soluble salt of the type $BA$,the dissociation equilibrium is given by: $BA(s) \rightleftharpoons B^+(aq) + A^-(aq)$.
Let the solubility of the salt be $s \ mol \ dm^{-3}$.
Then,$[B^+] = s$ and $[A^-] = s$.
The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [B^+][A^-] = s \times s = s^2$.
Given $K_{sp} = 2.7 \times 10^{-10}$.
Therefore,$s^2 = 2.7 \times 10^{-10}$.
$s = \sqrt{2.7 \times 10^{-10}} = \sqrt{27 \times 10^{-11}} = \sqrt{2.7} \times 10^{-5} \approx 1.643 \times 10^{-5} \ mol \ dm^{-3}$.
Thus,the ionic concentration of each ion is $1.643 \times 10^{-5} \ mol \ dm^{-3}$.

(C) The dissociation of the salt $BA$ is given by: $BA_{(s)} \rightleftharpoons B^{+}_{(aq)} + A^{-}_{(aq)}$
For a salt of type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula: $K_{sp} = S^2$
Given solubility $S = 9.1 \times 10^{-3} \ mol \ dm^{-3}$
$K_{sp} = (9.1 \times 10^{-3})^2$
$K_{sp} = 82.81 \times 10^{-6} = 8.281 \times 10^{-5}$

(A) The salt $BA_3$ ionizes in an aqueous solution as follows:
$BA_3(s) \rightleftharpoons B^{3+}(aq) + 3A^{-}(aq)$
Let the solubility of $BA_3$ be $S \ mol/L$.
Then,$[B^{3+}] = S$ and $[A^{-}] = 3S$.
The solubility product constant $(K_{sp})$ is given by:
$K_{sp} = [B^{3+}][A^{-}]^3$
$K_{sp} = (S)(3S)^3$
$K_{sp} = S \times 27S^3 = 27S^4$
Rearranging for $S$:
$S^4 = K_{sp} / 27$
$S = (K_{sp} / 27)^{1/4}$

(B) The dissociation of the sparingly soluble salt $BA$ is given by:
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
For a salt of the type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula:
$K_{sp} = [B^+][A^-] = S \times S = S^2$
Given,solubility $S = 7.2 \times 10^{-7} \ mol \ dm^{-3}$.
Substituting the value of $S$ in the formula:
$K_{sp} = (7.2 \times 10^{-7})^2$
$K_{sp} = 51.84 \times 10^{-14} = 5.184 \times 10^{-13}$

(C) The dissociation of the salt $BA$ is given by: $BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
Let the solubility of the salt be $S \ mol \ dm^{-3}$.
Then,$[B^{+}] = S$ and $[A^{-}] = S$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$.
Given $K_{sp} = 4.9 \times 10^{-9}$.
Therefore,$S^2 = 4.9 \times 10^{-9} = 49 \times 10^{-10}$.
Taking the square root on both sides: $S = \sqrt{49 \times 10^{-10}} = 7.00 \times 10^{-5} \ mol \ dm^{-3}$.

(C) For a sparingly soluble salt $AB$,the dissociation equilibrium is:
$AB_{(s)} \rightleftharpoons A_{(aq)}^{+} + B_{(aq)}^{-}$
Let the solubility be $S \ mol \ dm^{-3}$.
Then,$[A^{+}] = S$ and $[B^{-}] = S$.
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [A^{+}][B^{-}] = S \times S = S^2$
Given $K_{sp} = 1.6 \times 10^{-5}$.
Therefore,$S^2 = 1.6 \times 10^{-5} = 16 \times 10^{-6}$.
Taking the square root on both sides:
$S = \sqrt{16 \times 10^{-6}} = 4.0 \times 10^{-3} \ mol \ dm^{-3}$.

(A) The dissociation of the salt $B_3A_2$ is given by:
$B_3A_{2(s)} \rightleftharpoons 3B^{2+} + 2A^{3-}$
Let the solubility be $S \ mol \ L^{-1}$.
Then,$[B^{2+}] = 3S$ and $[A^{3-}] = 2S$.
The solubility product expression is:
$K_{sp} = [B^{2+}]^3 [A^{3-}]^2$
Substituting the values:
$K_{sp} = (3S)^3 (2S)^2$
$K_{sp} = (27S^3) \times (4S^2)$
$K_{sp} = 108S^5$
Solving for $S$:
$S^5 = \frac{K_{sp}}{108}$
$S = \left( \frac{K_{sp}}{108} \right)^{\frac{1}{5}}$

(A) For a sparingly soluble salt $BA$,the dissociation equilibrium is represented as:
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
Let the solubility of $BA$ be $S \ mol \ dm^{-3}$.
Then,$[B^{+}] = S$ and $[A^{-}] = S$.
The solubility product constant $K_{sp}$ is given by:
$K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$
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:
$S = \sqrt{49 \times 10^{-14}} = 7.0 \times 10^{-7} \ mol \ dm^{-3}$.

(A) The dissociation of the sparingly soluble salt $BA$ is given by:
$BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$
For a salt of the type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the expression:
$K_{sp} = [B^{+}][A^{-}] = S \times S = S^2$
Given the solubility $S = 1.8 \times 10^{-5} \ mol \ dm^{-3}$,we calculate $K_{sp}$ as:
$K_{sp} = (1.8 \times 10^{-5})^2 = 3.24 \times 10^{-10} \ mol^2 \ dm^{-6}$

(A) The dissociation of the salt is given by: $A_2B_{3(s)} \rightleftharpoons 2A^{3+}_{(aq)} + 3B^{2-}_{(aq)}$
Here,the stoichiometric coefficients are $x=2$ and $y=3$.
The solubility product expression is $K_{sp} = [A^{3+}]^2 [B^{2-}]^3 = (2S)^2 (3S)^3$.
$K_{sp} = 4S^2 \times 27S^3 = 108S^5$.
Given solubility $S = 1 \times 10^{-3} \ mol \ dm^{-3}$.
$K_{sp} = 108 \times (1 \times 10^{-3})^5 = 108 \times 10^{-15} = 1.08 \times 10^{-13}$.

(A) For $PbI_2$,the dissociation equilibrium is:
$PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^{-}_{(aq)}$
Let the solubility be $S$. Then $[Pb^{2+}] = S$ and $[I^-] = 2S$.
The solubility product expression is:
$K_{sp} = [Pb^{2+}][I^-]^2 = (S)(2S)^2 = 4S^3$
Given $K_{sp} = 7.0 \times 10^{-9}$.
$4S^3 = 7.0 \times 10^{-9}$
$S^3 = \frac{7.0 \times 10^{-9}}{4} = 1.75 \times 10^{-9}$
$S = \sqrt[3]{1.75 \times 10^{-9}} = 1.205 \times 10^{-3} \ mol \ L^{-1} \approx 1.21 \times 10^{-3} \ mol \ L^{-1}$.

(B) The dissociation of silver bromide is given by: $AgBr_{(s)} \rightleftharpoons Ag_{(aq)}^{+} + Br_{(aq)}^{-}$
For a salt of the type $AB$,the solubility product is $K_{sp} = S^2$,where $S$ is the solubility.
Given $K_{sp} = 6.4 \times 10^{-13}$.
Therefore,$S = \sqrt{K_{sp}} = \sqrt{6.4 \times 10^{-13}}$.
$S = \sqrt{64 \times 10^{-14}} = 8 \times 10^{-7} \ mol \ L^{-1}$.

(C) The dissociation of silver chromate $(Ag_2CrO_4)$ in water is represented as:
$Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+{(aq)} + CrO_4^{2-}{(aq)}$
If the solubility is $S \ mol \ L^{-1}$,the concentration of ions at equilibrium will be:
$[Ag^+] = 2S \ mol \ L^{-1}$
$[CrO_4^{2-}] = S \ mol \ L^{-1}$
The solubility product $(K_{sp})$ is defined as:
$K_{sp} = [Ag^+]^2 [CrO_4^{2-}]$
Substituting the values:
$K_{sp} = (2S)^2 \times (S)$
$K_{sp} = 4S^2 \times S = 4S^3$

(A) The dissociation of $AgCl$ is given by: $AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$
Let the solubility be $S \ M$.
Then,$[Ag^{+}] = S$ and $[Cl^{-}] = S$.
The solubility product constant is $K_{sp} = [Ag^{+}][Cl^{-}] = S \times S = S^2$.
Given $K_{sp} = 1.6 \times 10^{-10}$.
Therefore,$S^2 = 1.6 \times 10^{-10}$.
$S = \sqrt{1.6 \times 10^{-10}} = 1.26 \times 10^{-5} \ M$.

(C) For $Al(OH)_3$,the dissociation equation is: $Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^{-}$.
If the solubility is $s$,then $[Al^{3+}] = s$ and $[OH^{-}] = 3s$.
The solubility product $K_{sp} = [Al^{3+}][OH^{-}]^3 = (s)(3s)^3 = 27s^4$.
Solving for $s$: $s^4 = \frac{K_{sp}}{27} \implies s = \sqrt[4]{\frac{K_{sp}}{27}}$.

(A) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2OH^{-}_{(aq)}$
Let the solubility be $S \ mol \ dm^{-3}$.
Then,$[Mg^{2+}] = S$ and $[OH^-] = 2S$.
The solubility product expression is: $K_{sp} = [Mg^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 1.8 \times 10^{-11}$.
$4S^3 = 1.8 \times 10^{-11}$
$S^3 = \frac{1.8 \times 10^{-11}}{4} = 0.45 \times 10^{-11} = 4.5 \times 10^{-12}$.
$S = \sqrt[3]{4.5 \times 10^{-12}} = 1.650 \times 10^{-4} \ mol \ dm^{-3}$.

(B) The solubility equilibrium for $PbCl_2$ is given by:
$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$
Let the solubility be $S \ mol \ dm^{-3}$.
Then,$[Pb^{2+}] = S$ and $[Cl^-] = 2S$.
The solubility product expression is:
$K_{sp} = [Pb^{2+}][Cl^-]^2 = (S)(2S)^2 = 4S^3$
Given $K_{sp} = 3.2 \times 10^{-5}$.
$4S^3 = 3.2 \times 10^{-5}$
$S^3 = \frac{3.2 \times 10^{-5}}{4} = 0.8 \times 10^{-5} = 8 \times 10^{-6}$
$S = \sqrt[3]{8 \times 10^{-6}} = 2 \times 10^{-2} \ mol \ dm^{-3}$

(D) The solubility product expression is $K_{sp} = [Mg^{2+}][OH^-]^2$.
Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.1 \ M$.
Substituting these values: $1.0 \times 10^{-11} = (0.1)[OH^-]^2$.
$[OH^-]^2 = \frac{1.0 \times 10^{-11}}{0.1} = 10^{-10}$.
$[OH^-] = \sqrt{10^{-10}} = 10^{-5} \ M$.
Now,calculate $pOH$: $pOH = -\log[OH^-] = -\log(10^{-5}) = 5$.
Finally,calculate $pH$: $pH = 14 - pOH = 14 - 5 = 9$.

(D) The dissociation of $AgCl$ is given by: $AgCl_{(s)} \rightleftharpoons Ag^+_{(aq)} + Cl^-_{(aq)}$.
For a $1:1$ electrolyte,the solubility product $K_{sp} = s^2$,where $s$ is the solubility.
Given $s = 7.2 \times 10^{-7} \ mol \ dm^{-3}$.
$K_{sp} = (7.2 \times 10^{-7})^2$.
$K_{sp} = 51.84 \times 10^{-14}$.
$K_{sp} = 5.184 \times 10^{-13} \approx 5.18 \times 10^{-13}$.

(A) The dissociation of $AgCl$ is given by: $AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$
For a $1:1$ electrolyte,the solubility product $K_{sp}$ is related to solubility $S$ by the equation: $K_{sp} = S^2$
Given $S = 1.25 \times 10^{-5} \ mol \ dm^{-3}$
$K_{sp} = (1.25 \times 10^{-5})^2 = 1.5625 \times 10^{-10} \approx 1.56 \times 10^{-10}$

(C) The dissociation of the salt is given by: $AX_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2X^{-}_{(aq)}$
Let $S$ be the solubility of the salt.
At equilibrium,$[A^{2+}] = S$ and $[X^{-}] = 2S$.
The solubility product expression is: $K_{sp} = [A^{2+}][X^{-}]^2$
$K_{sp} = (S)(2S)^2 = 4S^3$
Given $K_{sp} = 3.2 \times 10^{-8}$.
$4S^3 = 3.2 \times 10^{-8}$
$S^3 = \frac{3.2}{4} \times 10^{-8} = 0.8 \times 10^{-8} = 8 \times 10^{-9}$
$S = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3} \ mol \ dm^{-3}$

(D) The dissociation of $AgBr$ is given by: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$.
Let the solubility be $S \ mol \ dm^{-3}$.
Then,$[Ag^+] = S$ and $[Br^-] = S$.
The solubility product expression is $K_{sp} = [Ag^+][Br^-] = S \times S = S^2$.
Given $K_{sp} = 4.9 \times 10^{-13}$.
So,$S^2 = 4.9 \times 10^{-13}$.
$S = \sqrt{4.9 \times 10^{-13}} = \sqrt{49 \times 10^{-14}} = 7.0 \times 10^{-7} \ mol \ dm^{-3}$.

(B) The dissociation of silver oxalate is given by: $Ag_2C_2O_{4(s)} \rightleftharpoons 2Ag^+{(aq)} + C_2O_4^{2-}{(aq)}$
Let the solubility be $S = 2 \times 10^{-4} \ mol \ L^{-1}$.
Then,$[Ag^+] = 2S$ and $[C_2O_4^{2-}] = S$.
The solubility product expression is: $K_{sp} = [Ag^+]^2 [C_2O_4^{2-}]$
Substituting the values: $K_{sp} = (2S)^2 \times (S) = 4S^3$
Calculating the value: $K_{sp} = 4 \times (2 \times 10^{-4})^3 = 4 \times (8 \times 10^{-12}) = 3.2 \times 10^{-11}$

(C) The dissociation of the salt is given by: $AB_{2(s)} \rightleftharpoons A^{2+}_{(aq)} + 2B^-_{(aq)}$
Let the solubility be $S = 1.0 \times 10^{-4} \ mol \ dm^{-3}$.
At equilibrium,the concentration of $[A^{2+}] = S$ and $[B^-] = 2S$.
The solubility product expression is: $K_{sp} = [A^{2+}][B^-]^2$.
Substituting the values: $K_{sp} = (S) \cdot (2S)^2 = 4S^3$.
$K_{sp} = 4 \times (1.0 \times 10^{-4})^3 = 4 \times 10^{-12}$.

(A) The dissociation of calcium phosphate in water is given by the equation: $Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}{(aq)} + 2PO_4^{3-}{(aq)}$
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,$K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$.

(A) The dissociation of $Ag_{2}CrO_{4}$ is given by: $Ag_{2}CrO_{4} \rightleftharpoons 2 Ag^{+} + CrO_{4}^{2-}$.
Let the solubility be $s$. Then $[Ag^{+}] = 2s$ and $[CrO_{4}^{2-}] = s$.
The solubility product expression is $K_{sp} = [Ag^{+}]^{2} [CrO_{4}^{2-}]$.
Substituting the values: $K_{sp} = (2s)^{2} (s) = 4s^{3}$.
Given $K_{sp} = 32 \times 10^{-12}$,we have $4s^{3} = 32 \times 10^{-12}$.
$s^{3} = 8 \times 10^{-12}$.
$s = (8 \times 10^{-12})^{1/3} = 2 \times 10^{-4} \ M$.
Since $[CrO_{4}^{2-}] = s$,the concentration is $2 \times 10^{-4} \ M$.

(A) The dissociation of $Ca(OH)_{2}$ in water is represented as:
$Ca(OH)_{2}(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^{-}(aq)$
Let the solubility be $s \ mol \ L^{-1}$.
At equilibrium,$[Ca^{2+}] = s$ and $[OH^{-}] = 2s$.
The solubility product expression is:
$K_{sp} = [Ca^{2+}][OH^{-}]^{2}$
Substituting the values:
$K_{sp} = (s)(2s)^{2} = s \times 4s^{2} = 4s^{3}$

(C) The solubility product constant $(K_{sp})$ of $AgCl$ is calculated as:
$K_{sp} = (\text{solubility})^2 = (1 \times 10^{-5})^2 = 1 \times 10^{-10}$.
Let the solubility of $AgCl$ in $0.1 \ M \ NaCl$ be $x \ mol/L$.
In the presence of $0.1 \ M \ NaCl$,the concentration of chloride ions is $[Cl^-] = (x + 0.1) \ M$.
Since $x$ is very small compared to $0.1$,we can approximate $[Cl^-] \approx 0.1 \ M$.
The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.
Substituting the values: $1 \times 10^{-10} = x \times 0.1$.
Solving for $x$: $x = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \ mol/L$.

(D) The dissociation of $CaCO_3$ is given by: $CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2-}_{3(aq)}$.
Let the solubility be $S = 7 \times 10^{-5} \ mol \ dm^{-3}$.
The solubility product expression is $K_{sp} = [Ca^{2+}][CO^{2-}_{3}] = S \times S = S^2$.
Substituting the value of $S$:
$K_{sp} = (7 \times 10^{-5})^2 = 49 \times 10^{-10} = 4.9 \times 10^{-9}$.

(C) The dissolution of $CuS$ in water is represented as: $CuS(s) \rightleftharpoons Cu^{2+}(aq) + S^{2-}(aq)$.
Let the solubility of $CuS$ be $s \ mol/L$.
Then,$[Cu^{2+}] = s$ and $[S^{2-}] = s$.
The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [Cu^{2+}][S^{2-}] = s \times s = s^2$.
Given $K_{sp} = 9 \times 10^{-16}$.
Therefore,$s^2 = 9 \times 10^{-16}$.
Taking the square root on both sides,$s = \sqrt{9 \times 10^{-16}} = 3 \times 10^{-8} \ M$.
Thus,the maximum molarity of $CuS$ in an aqueous solution is $3 \times 10^{-8} \ M$.

(C) For the equilibrium: $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$
The solubility product expression is: $K_{sp} = [Fe^{3+}][OH^{-}]^3$
Let the initial concentrations be $[Fe^{3+}]_1$ and $[OH^{-}]_1$.
$K_{sp} = [Fe^{3+}]_1 [OH^{-}]_1^3$ ... $(i)$
If the new concentration $[OH^{-}]_2 = \frac{1}{4} [OH^{-}]_1$, then:
$K_{sp} = [Fe^{3+}]_2 [OH^{-}]_2^3 = [Fe^{3+}]_2 (\frac{1}{4} [OH^{-}]_1)^3$
$K_{sp} = [Fe^{3+}]_2 \times \frac{1}{64} [OH^{-}]_1^3$ ... $(ii)$
From $(i)$ and $(ii)$:
$[Fe^{3+}]_1 [OH^{-}]_1^3 = [Fe^{3+}]_2 \times \frac{1}{64} [OH^{-}]_1^3$
$[Fe^{3+}]_2 = 64 \times [Fe^{3+}]_1$
Therefore, the concentration of $Fe^{3+}$ increases by $64$ times.

(A) Given,$K_{sp} = 4 \times 10^{-9} \ (mol \ L^{-1})^2$.
For the dissociation of $CaC_2O_4$:
$CaC_2O_4(s) \rightleftharpoons Ca^{2+}(aq) + C_2O_4^{2-}(aq)$.
Let the solubility be $S \ mol \ L^{-1}$.
Then,$[Ca^{2+}] = S$ and $[C_2O_4^{2-}] = S$.
$K_{sp} = [Ca^{2+}][C_2O_4^{2-}] = S \times S = S^2$.
$S^2 = 4 \times 10^{-9}$.
$S = \sqrt{4 \times 10^{-9}} = \sqrt{40 \times 10^{-10}} = 6.32 \times 10^{-5} \ mol \ L^{-1}$.
Thus,the solubility of $CaC_2O_4$ is $6.3 \times 10^{-5} \ mol \ L^{-1}$.

(B) The ionic product $(IP)$ for $ZnS$ is $[Zn^{2+}][S^{2-}] = 0.1 \times 8.1 \times 10^{-19} = 8.1 \times 10^{-20}$. Since $8.1 \times 10^{-20} > 3 \times 10^{-22}$ ($K_{sp}$ of $ZnS$),$ZnS$ precipitates.
The ionic product $(IP)$ for $CuS$ is $[Cu^{2+}][S^{2-}] = 0.01 \times 8.1 \times 10^{-19} = 8.1 \times 10^{-21}$. Since $8.1 \times 10^{-21} > 8 \times 10^{-36}$ ($K_{sp}$ of $CuS$),$CuS$ precipitates.
Therefore,both $CuS$ and $ZnS$ precipitate.

(C) For precipitation to occur,the ionic product must exceed the solubility product $(K_{sp})$.
For $Ag_{2}CrO_{4}$:
Ionic Product $(IP) = [Ag^{+}]^{2}[CrO_{4}^{2-}] = (10^{-4})^{2}(10^{-5}) = 10^{-13}$.
Given $K_{sp}(Ag_{2}CrO_{4}) = 4 \times 10^{-12}$.
Since $IP < K_{sp}$,$Ag_{2}CrO_{4}$ will not precipitate.
For $AgCl$:
Ionic Product $(IP) = [Ag^{+}][Cl^{-}] = (10^{-4})(10^{-5}) = 10^{-9}$.
Given $K_{sp}(AgCl) = 1 \times 10^{-10}$.
Since $IP > K_{sp}$,$AgCl$ will precipitate.
Therefore,silver chloride gets precipitated first.

(C) precipitate of $AgCl$ forms when the ionic product $[Ag^{+}][Cl^{-}]$ exceeds the solubility product constant $K_{sp} = 1 \times 10^{-10} \ M^2$.
When mixing $1 \ L$ of each solution,the final volume becomes $2 \ L$,so the concentrations are halved.
For $(iv)$ and $(v)$:
$[Ag^{+}] = \frac{10^{-3} \ M}{2} = 0.5 \times 10^{-3} \ M$
$[Cl^{-}] = \frac{10^{-5} \ M}{2} = 0.5 \times 10^{-5} \ M$
Ionic product = $(0.5 \times 10^{-3}) \times (0.5 \times 10^{-5}) = 0.25 \times 10^{-8} = 2.5 \times 10^{-9}$.
Since $2.5 \times 10^{-9} > 1 \times 10^{-10}$,a precipitate will form.

(C) The precipitation of a salt occurs when the ionic product exceeds its solubility product $(K_{sp})$.
For a salt $AgX$,the precipitation starts when $[Ag^{+}] = \frac{K_{sp}(AgX)}{[X^{-}]}$.
Given $[Cl^{-}] = [Br^{-}] = [I^{-}] = 1 \times 10^{-8} \ M$.
For $AgCl$: $[Ag^{+}] = \frac{1.8 \times 10^{-10}}{10^{-8}} = 1.8 \times 10^{-2} \ M$.
For $AgBr$: $[Ag^{+}] = \frac{5 \times 10^{-13}}{10^{-8}} = 5 \times 10^{-5} \ M$.
For $AgI$: $[Ag^{+}] = \frac{8.3 \times 10^{-17}}{10^{-8}} = 8.3 \times 10^{-9} \ M$.
Since the concentration of $Ag^{+}$ required for precipitation is lowest for $AgI$,it will precipitate first,followed by $AgBr$,and finally $AgCl$.
The order of precipitation is $AgI > AgBr > AgCl$.

(C) The dissociation of $AgBr$ is given by: $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$.
Let the solubility of $AgBr$ in $0.1 \ M$ $KBr$ be $s$.
In the presence of $0.1 \ M$ $KBr$,the concentration of $Br^-$ ions is $[Br^-] = (s + 0.1) \approx 0.1 \ M$ (since $s$ is very small).
The concentration of $Ag^+$ ions is $[Ag^+] = s$.
The solubility product expression is $K_{sp} = [Ag^+][Br^-]$.
Substituting the values: $4 \times 10^{-13} = s \times 0.1$.
Solving for $s$: $s = \frac{4 \times 10^{-13}}{0.1} = 4 \times 10^{-12} \ M$.

(C) The dissociation of $PbI_2$ is given by: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$.
Let the molar solubility of $PbI_2$ be $S \ mol/L$.
The concentration of $Pb^{2+}$ ions from $Pb(NO_3)_2$ is $0.2 \ M$.
Total concentration of $[Pb^{2+}] = (0.2 + S) \approx 0.2 \ M$ (since $S$ is very small).
Total concentration of $[I^-] = 2S$.
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 $S$: $S^2 = \frac{K_{sp}}{0.8}$,therefore $S = (\frac{K_{sp}}{0.8})^{1/2}$.

(A) For salt $MX_2$: $MX_2 \rightleftharpoons M^{2+} + 2X^-$. Let solubility be $S_1$. Then $K_{sp} = (S_1)(2S_1)^2 = 4S_1^3$. Given $4S_1^3 = 5 \times 10^{-13}$,so $S_1^3 = 1.25 \times 10^{-13} = 125 \times 10^{-15}$. Thus,$S_1 = 5 \times 10^{-5} \ M$.
For salt $MX$: $MX \rightleftharpoons M^+ + X^-$. Let solubility be $S_2$. Then $K_{sp} = S_2^2$. Given $S_2^2 = 1.6 \times 10^{-11} = 16 \times 10^{-12}$. Thus,$S_2 = 4 \times 10^{-6} \ M$.
The ratio of molar solubility is $\frac{S_1}{S_2} = \frac{5 \times 10^{-5}}{4 \times 10^{-6}} = \frac{50}{4} = 12.5$.

(C) The dissolution of $AX$ is given by: $AX(s) \rightleftharpoons A^+(aq) + X^-(aq)$.
$K_{sp} = [A^+][X^-] = 10^{-10}$.
In $0.1$ $M$ $HX$ solution,$HX$ is a strong acid (assuming complete dissociation),so $[H^+] = 0.1$ $M$ and $[X^-] = 0.1$ $M$.
Let $S$ be the molar solubility of $AX$ in the presence of $0.1$ $M$ $HX$.
Then $[A^+] = S$ and $[X^-] = 0.1 + S \approx 0.1$ (since $S$ is very small compared to $0.1$).
Substituting these into the $K_{sp}$ expression:
$10^{-10} = S \times 0.1$.
$S = \frac{10^{-10}}{10^{-1}} = 10^{-9}$ $M$.

(B) The dissociation of barium phosphate is: $Ba_3(PO_4)_2 \rightleftharpoons 3Ba^{2+} + 2PO_4^{3-}$.
Solubility in $g/100 \ mL$ is $x$,so in $g/L$ it is $10x$.
Solubility in $mol/L$ $(S)$ = $\frac{10x}{M} = 10 \times (x/M)$.
$K_{sp} = [Ba^{2+}]^3 [PO_4^{3-}]^2 = (3S)^3 (2S)^2 = 27S^3 \times 4S^2 = 108S^5$.
Substituting $S = 10(x/M)$:
$K_{sp} = 108 \times (10(x/M))^5 = 108 \times 10^5 \times (x/M)^5$.
$K_{sp} = 1.08 \times 10^2 \times 10^5 \times (x/M)^5 = 1.08 \times (x/M)^5 \times 10^7$.
Comparing with $1.08 \times (x/M)^a \times 10^b$,we get $a = 5$ and $b = 7$.

(D) The precipitation of a metal sulphide $MS$ occurs when the ionic product $[M^{2+}][S^{2-}]$ exceeds the solubility product $K_{sp}$.
Given that the concentrations of all metal ions are equal,the ion with the lowest $K_{sp}$ will require the lowest concentration of $S^{2-}$ to exceed its $K_{sp}$ and precipitate first.
Comparing the $K_{sp}$ values:
$K_{sp}(HgS) = 4 \times 10^{-53}$
$K_{sp}(CdS) = 8 \times 10^{-27}$
$K_{sp}(ZnS) = 1.6 \times 10^{-24}$
$K_{sp}(NiS) = 4.7 \times 10^{-5}$
Since $HgS$ has the lowest $K_{sp}$,$Hg^{2+}$ will precipitate first.
Since $NiS$ has the highest $K_{sp}$,$Ni^{2+}$ will precipitate last.
Therefore,the first and last ions to precipitate are $Hg^{2+}$ and $Ni^{2+}$ respectively.

(C) $K_{sp}$ of $MCl = 1 \times 10^{-10}$.
Let the molar solubility of $MCl$ in $0.1 \ M \ NaCl$ be $S \ mol \ L^{-1}$.
$MCl \rightleftharpoons M^{+} + Cl^{-}$
The concentration of $Cl^{-}$ will be $(S + 0.1) \ mol \ L^{-1}$,as $0.1 \ mol \ L^{-1}$ is provided by $0.1 \ M \ NaCl$.
$K_{sp} = [M^{+}][Cl^{-}]$
$K_{sp} = S \times (S + 0.1) = 1 \times 10^{-10}$
Since $K_{sp}$ is very small,$S \ll 0.1$,so $S$ can be neglected in the term $(S + 0.1)$.
$S \times 0.1 = 1 \times 10^{-10} \Rightarrow S = 10^{-9} \ mol \ L^{-1}$.

(B) The solubility product $(K_{sp})$ of $AgBr$ is $5 \times 10^{-10}$.
For the dissociation reaction: $AgBr_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Br^{-}_{(aq)}$.
The expression for $K_{sp}$ is $K_{sp} = [Ag^{+}][Br^{-}]$.
In $0.2 \ M$ $NaBr$ solution,the concentration of $Br^{-}$ ions is $0.2 \ M$ due to the common ion effect.
Let the solubility of $AgBr$ be $s$.
Then $[Ag^{+}] = s$ and $[Br^{-}] = 0.2 + s \approx 0.2$ (since $s$ is very small).
$5 \times 10^{-10} = s \times 0.2$.
$s = \frac{5 \times 10^{-10}}{0.2} = 25 \times 10^{-10} \ M$.

(D) The dissociation of $Ag_2CO_3$ is given by: $Ag_2CO_3(s) \rightleftharpoons 2Ag^{+}(aq) + CO_3^{2-}(aq)$.
Given that $[Ag^{+}] = 1.20 \times 10^{-4} \ mol \ L^{-1}$.
From the stoichiometry of the reaction,$[CO_3^{2-}] = \frac{1}{2} [Ag^{+}] = \frac{1}{2} \times 1.20 \times 10^{-4} = 0.60 \times 10^{-4} \ mol \ L^{-1}$.
The solubility product expression is $K_{sp} = [Ag^{+}]^2 [CO_3^{2-}]$.
Substituting the values: $K_{sp} = (1.20 \times 10^{-4})^2 \times (0.60 \times 10^{-4})$.
$K_{sp} = (1.44 \times 10^{-8}) \times (0.60 \times 10^{-4}) = 0.864 \times 10^{-12} = 8.64 \times 10^{-13}$.
Wait,re-evaluating based on standard solubility $S$: If $[Ag^{+}] = 2S = 1.20 \times 10^{-4}$,then $S = 0.60 \times 10^{-4}$.
$K_{sp} = 4S^3 = 4 \times (0.60 \times 10^{-4})^3 = 4 \times 0.216 \times 10^{-12} = 0.864 \times 10^{-12} = 8.64 \times 10^{-13}$.
Given the options provided,there is a discrepancy. If we assume the question implies $[Ag^{+}] = 2S$ and the result matches option $D$ $(6.90 \times 10^{-12})$,it suggests $S = 1.20 \times 10^{-4}$ was intended as the solubility $S$ itself,not the concentration of $Ag^{+}$.
Assuming $S = 1.20 \times 10^{-4} \ mol \ L^{-1}$,then $K_{sp} = 4S^3 = 4 \times (1.20 \times 10^{-4})^3 = 4 \times 1.728 \times 10^{-12} = 6.912 \times 10^{-12} \approx 6.90 \times 10^{-12}$.