Solubility product Questions in English

(B) For precipitation to occur,the ionic product $(Q_{sp})$ must be greater than the solubility product $(K_{sp})$.
When equal volumes are mixed,the concentration of each ion is halved.
$K_{sp} = [Ca^{2+}][F^{-}]^2 = 1.7 \times 10^{-10}$.
For option $A$: $Q_{sp} = (\frac{10^{-4}}{2}) \times (\frac{10^{-4}}{2})^2 = 1.25 \times 10^{-13} < K_{sp}$.
For option $B$: $Q_{sp} = (\frac{10^{-2}}{2}) \times (\frac{10^{-3}}{2})^2 = (0.5 \times 10^{-2}) \times (0.25 \times 10^{-6}) = 1.25 \times 10^{-9} > K_{sp}$.
For option $C$: $Q_{sp} = (\frac{10^{-5}}{2}) \times (\frac{10^{-3}}{2})^2 = 1.25 \times 10^{-12} < K_{sp}$.
For option $D$: $Q_{sp} = (\frac{10^{-3}}{2}) \times (\frac{10^{-5}}{2})^2 = 1.25 \times 10^{-14} < K_{sp}$.
Since $Q_{sp} > K_{sp}$ only in option $B$,a precipitate will form.

(B) For a basic buffer,the Henderson-Hasselbalch equation is: $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]}$.
Since the moles of $NH_4^+$ and $NH_3$ are equal,$\frac{[NH_4^+]}{[NH_3]} = 1$,so $pOH = pK_b = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Therefore,$[OH^-] = K_b = 1.8 \times 10^{-5} \ M$.
The solubility product expression for $Mn(OH)_2$ is $K_{sp} = [Mn^{2+}][OH^-]^2 = S \times [OH^-]^2$.
Substituting the values: $4.5 \times 10^{-14} = S \times (1.8 \times 10^{-5})^2$.
$S = \frac{4.5 \times 10^{-14}}{3.24 \times 10^{-10}} = 1.388 \times 10^{-4} \ M \approx 1.38 \times 10^{-4} \ M$.

(A) The solubility product expression for $AgI$ is $K_{sp} = [Ag^+][I^-] = S^2$.
Given $K_{sp} = 1.0 \times 10^{-16}$,the solubility $S = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \ mol/L$.
The molar mass of $AgI = 108 + 127 = 235 \ g/mol$.
The mass of $AgI$ dissolved in $1.0 \ L$ is $Mass = S \times \text{Molar Mass} = 1.0 \times 10^{-8} \ mol/L \times 235 \ g/mol = 2.35 \times 10^{-6} \ g$.

(A) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$
In a saturated solution,the solubility $s = 1.0 \times 10^{-3} \ M$.
The solubility product $K_{sp}$ is calculated as: $K_{sp} = [Pb^{2+}][Cl^{-}]^2 = (s)(2s)^2 = 4s^3$.
$K_{sp} = 4 \times (1.0 \times 10^{-3})^3 = 4 \times 10^{-9}$.
In a $0.1 \ M \ NaCl$ solution,the concentration of $Cl^-$ ions is dominated by $NaCl$ $([Cl^-] \approx 0.1 \ M)$.
Using the $K_{sp}$ expression: $K_{sp} = [Pb^{2+}][Cl^-]^2$.
$4 \times 10^{-9} = s' \times (0.1)^2$,where $s'$ is the new solubility.
$s' = \frac{4 \times 10^{-9}}{0.01} = 4 \times 10^{-7} \ M$.

(C) The dissociation of $BaF_2$ is given by: $BaF_{2(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2F^{-}_{(aq)}$
Let the solubility of $BaF_2$ be $S \ mol/L$.
Then,$[Ba^{2+}] = S$ and $[F^{-}] = 2S$.
The solubility product expression is: $K_{sp} = [Ba^{2+}][F^{-}]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 1.0 \times 10^{-6}$,we have $4S^3 = 1.0 \times 10^{-6}$.
$S^3 = 0.25 \times 10^{-6} = 250 \times 10^{-9}$.
$S = (250)^{1/3} \times 10^{-3} \approx 6.3 \times 10^{-3} \ mol/L$.
The concentration of $F^{-}$ ions is $[F^{-}] = 2S = 2 \times 6.3 \times 10^{-3} = 1.26 \times 10^{-2} \ M$.

(A) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^{-}_{(aq)}$
Let the solubility be $S \ mol \ L^{-1}$. Then $[Pb^{2+}] = S$ and $[Cl^-] = 2S$.
The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 1.0 \times 10^{-6}$,we have:
$4S^3 = 1.0 \times 10^{-6}$
$S^3 = 0.25 \times 10^{-6} = 250 \times 10^{-9}$
$S = (250)^{1/3} \times 10^{-3} \approx 6.3 \times 10^{-3} \ mol \ L^{-1}$.

(C) The solubility equilibrium for $BaF_2$ is given by:
$BaF_{2(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2F^{-}_{(aq)}$
The solubility product expression is:
$K_{sp} = [Ba^{2+}] [F^{-}]^2$
Given $K_{sp} = 1.0 \times 10^{-6}$ and $[F^{-}] = 0.30 \, M$.
Substituting the values into the expression:
$1.0 \times 10^{-6} = [Ba^{2+}] \times (0.30)^2$
$[Ba^{2+}] = \frac{1.0 \times 10^{-6}}{0.09}$
$[Ba^{2+}] = 1.11 \times 10^{-5} \, M \approx 1.1 \times 10^{-5} \, M$

(B) The dissociation of $AlCl_3$ is given by: $AlCl_3(s) \rightleftharpoons Al^{3+}(aq) + 3Cl^-(aq)$.
Let the solubility of $AlCl_3$ be $S$.
Then,$[Al^{3+}] = S$ and $[Cl^-] = 3S$.
In a solution of $CaCl_2$ with concentration $C$,the dissociation is $CaCl_2 \rightarrow Ca^{2+}(aq) + 2Cl^-(aq)$.
Thus,the concentration of $Cl^-$ ions from $CaCl_2$ is $2C$.
The total concentration of $Cl^-$ ions is $[Cl^-] = (3S + 2C)$.
Since $S$ is very small compared to $C$,we can approximate $[Cl^-] \approx 2C$.
The solubility product expression is $K_{sp} = [Al^{3+}][Cl^-]^3$.
Substituting the values: $K_{sp} = (S)(2C)^3$.
$K_{sp} = S \times 8C^3$.
Therefore,$S = \frac{K_{sp}}{8C^3}$.

(A) $AgCN$ is a sparingly soluble salt. When $KCN$ is added,the $CN^-$ ions react with $AgCN$ to form a soluble complex ion,$[Ag(CN)_2]^-$. The reaction is: $AgCN(s) + CN^-(aq) \rightarrow [Ag(CN)_2]^-(aq)$. This complex formation reduces the concentration of free $Ag^+$ ions,shifting the equilibrium to the right and increasing the solubility of $AgCN$.

(A) The dissociation of $Ag_{2}CO_{3}$ is given by: $Ag_{2}CO_{3(s)} \longleftrightarrow 2Ag^{+}_{(aq)} + C{O_{3}}^{2-}_{(aq)}$
In the presence of $0.1 \ M \ Na_{2}CO_{3}$,the concentration of $CO_{3}^{2-}$ is dominated by the salt,so $[CO_{3}^{2-}] \approx 0.1 \ M$.
Let $s$ be the molar solubility of $Ag_{2}CO_{3}$. Then $[Ag^{+}] = 2s$.
The solubility product expression is $K_{sp} = [Ag^{+}]^{2} [CO_{3}^{2-}]$.
Substituting the values: $4 \times 10^{-13} = (2s)^{2} \times (0.1)$.
$4 \times 10^{-13} = 4s^{2} \times 0.1$.
$10^{-13} = s^{2} \times 10^{-1}$.
$s^{2} = 10^{-12}$.
$s = 10^{-6} \ M$.

(B) The solubility product constant $(K_{sp})$ for $BaSO_4$ is given by: $K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 \times 10^{-10}$.
Let the solubility of $BaSO_4$ in the presence of $Na_2SO_4$ be $S = 1.1 \times 10^{-8} \, M$.
In the solution,$BaSO_4$ dissociates as $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$,so $[Ba^{2+}] = S = 1.1 \times 10^{-8} \, M$.
$Na_2SO_4$ is a strong electrolyte and dissociates completely as $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$. Let the concentration of $Na_2SO_4$ be $C$.
Then,$[SO_4^{2-}] = S + C \approx C$ (since $S$ is very small compared to $C$).
Substituting these into the $K_{sp}$ expression: $1.1 \times 10^{-10} = (1.1 \times 10^{-8}) \times C$.
Solving for $C$: $C = \frac{1.1 \times 10^{-10}}{1.1 \times 10^{-8}} = 10^{-2} = 0.01 \, M$.

(C) The dissociation of $PbCl_2$ is represented as: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.
Let the solubility of $PbCl_2$ be $s \ mol/L$.
Then,$[Pb^{2+}] = s$ and $[Cl^-] = 2s$.
The solubility product constant is given by: $K_{sp} = [Pb^{2+}][Cl^-]^2$.
Substituting the values: $K_{sp} = (s)(2s)^2 = (s)(4s^2) = 4s^3$.
Solving for $s$: $s^3 = K_{sp}/4$,which gives $s = (K_{sp}/4)^{1/3}$.

(A) $1$. Calculate the molar mass of $AgCl$: $M(AgCl) = 108 + 35.5 = 143.5 \, g/mol$.
$2$. Convert solubility from $g/L$ to $mol/L$ (molar solubility,$s$): $s = \frac{1.435 \times 10^{-3} \, g/L}{143.5 \, g/mol} = 1.0 \times 10^{-5} \, mol/L$.
$3$. For $AgCl$,the dissociation is $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
$4$. The solubility product constant $(K_{sp})$ is given by $K_{sp} = [Ag^+][Cl^-] = s \times s = s^2$.
$5$. $K_{sp} = (1.0 \times 10^{-5})^2 = 1.0 \times 10^{-10}$.

(A) The dissociation of calcium fluoride is given by: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^-(aq)$.
Let the solubility of $CaF_2$ be $s \ mol/L$.
Then,$[Ca^{2+}] = s$ and $[F^-] = 2s$.
The solubility product expression is $K_{sp} = [Ca^{2+}][F^-]^2$.
Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$.
So,$4s^3 = 3.2 \times 10^{-11}$.
$s^3 = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
$s = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \ M$.

(A) In an acidic medium,the concentration of $S^{2-}$ ions is very low due to the common ion effect of $H^+$ ions from the strong acid,which suppresses the dissociation of $H_2S$ $(H_2S \rightleftharpoons 2H^+ + S^{2-})$.
The solubility product $(K_{sp})$ of $As_2S_3$ is extremely low compared to that of $ZnS$.
Even with the low concentration of $S^{2-}$ ions,the ionic product of $As^{3+}$ and $S^{2-}$ exceeds the $K_{sp}$ of $As_2S_3$,leading to its precipitation.
However,for $ZnS$,the ionic product remains lower than its $K_{sp}$ due to its relatively higher solubility,so $ZnS$ does not precipitate.

(C) The dissociation of $Ag_2CrO_4$ is given by: $Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$.
Let the solubility of $Ag_2CrO_4$ be $s$. Then $[Ag^{+}] = 2s$ and $[CrO_4^{2-}] = s$.
Given $[Ag^{+}] = 1.5 \times 10^{-4}\,mol\,L^{-1}$.
Since $[Ag^{+}] = 2s$,we have $s = \frac{1.5 \times 10^{-4}}{2} = 0.75 \times 10^{-4}\,mol\,L^{-1}$.
The solubility product $K_{sp}$ is defined as: $K_{sp} = [Ag^{+}]^2 [CrO_4^{2-}] = (2s)^2 (s) = 4s^3$.
Substituting the values: $K_{sp} = (1.5 \times 10^{-4})^2 \times (0.75 \times 10^{-4})$.
$K_{sp} = (2.25 \times 10^{-8}) \times (0.75 \times 10^{-4}) = 1.6875 \times 10^{-12}$.

(A) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.
Given $K_{sp} = 1.8 \times 10^{-10}$.
When equal volumes are mixed,the concentration of each ion is halved: $[Ag^{+}]_{new} = [Ag^{+}]_{initial} / 2$ and $[Cl^{-}]_{new} = [Cl^{-}]_{initial} / 2$.
For option $A$: $[Ag^{+}] = 10^{-4} / 2 = 5 \times 10^{-5} \ M$ and $[Cl^{-}] = 10^{-4} / 2 = 5 \times 10^{-5} \ M$.
$Q_{sp} = [Ag^{+}][Cl^{-}] = (5 \times 10^{-5}) \times (5 \times 10^{-5}) = 25 \times 10^{-10} = 2.5 \times 10^{-9}$.
Since $2.5 \times 10^{-9} > 1.8 \times 10^{-10}$,precipitation will occur.

(A) The precipitation of $AgCl$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.
For the reaction $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$,the solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.
Given $K_{sp} = 1.8 \times 10^{-10}$ and $[Ag^+] = 4 \times 10^{-3} \ M$.
To find the minimum concentration of $Cl^-$ required for precipitation,we set the ionic product equal to $K_{sp}$:
$[Cl^-] = \frac{K_{sp}}{[Ag^+]} = \frac{1.8 \times 10^{-10}}{4 \times 10^{-3}} = 0.45 \times 10^{-7} \ M = 4.5 \times 10^{-8} \ M$.

(D) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
The solubility product constant is $K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10}$.
In a $0.01 \ M \ HCl$ solution,$HCl$ dissociates completely as $HCl \rightarrow H^+ + Cl^-$,so $[Cl^-] = 0.01 \ M = 10^{-2} \ M$.
Let the solubility of $AgCl$ be $s$. Then $[Ag^+] = s$ and $[Cl^-] = (s + 0.01) \approx 0.01 \ M$ (since $s$ is very small).
Substituting these values into the $K_{sp}$ expression: $1.8 \times 10^{-10} = s \times (0.01)$.
$s = \frac{1.8 \times 10^{-10}}{10^{-2}} = 1.8 \times 10^{-8} \ M$.

(B) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
Let the solubility of $AgCl$ in $0.2 \ M \ NaCl$ be $s$.
In the presence of $0.2 \ M \ NaCl$,the concentration of $Cl^-$ ions is $[Cl^-] = (s + 0.2) \ M \approx 0.2 \ M$ (since $s$ is very small).
The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.
Substituting the values: $1.8 \times 10^{-10} = (s)(0.2)$.
Solving for $s$: $s = \frac{1.8 \times 10^{-10}}{0.2} = 9 \times 10^{-10} \ M$.

(D) $1$. $ZnS$ and $MnS$ are metal sulfides that react with dilute $HCl$ to release $H_2S$ gas,making them soluble.
$2$. $BaCO_3$ is a carbonate that reacts with dilute $HCl$ to form $BaCl_2$,$CO_2$,and $H_2O$,making it soluble.
$3$. $BaSO_4$ (Barium sulfate) is a salt of a strong acid $(H_2SO_4)$ and a strong base $(Ba(OH)_2)$. It is highly insoluble in water and dilute acids like $HCl$ due to its very low solubility product $(K_{sp})$.
$4$. Therefore,$BaSO_4$ is the correct answer.

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

(B) The precipitation of a salt occurs when the ionic product exceeds its solubility product $(K_{sp})$.
For salts with the same stoichiometry (like $BaSO_4$ and $CaSO_4$),the salt with the lowest $K_{sp}$ precipitates first.
Comparing the given values:
$K_{sp}(BaSO_4) = 10^{-11}$
$K_{sp}(CaSO_4) = 10^{-6}$
$K_{sp}(Ag_2SO_4) = 10^{-5}$
Since $BaSO_4$ has the minimum $K_{sp}$ value,it will precipitate first.

(A) The solubility of $PbF_2$ in water is $S = 10^{-3}\, M$.
The solubility product constant $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$.
$K_{sp} = 4 \times (10^{-3})^3 = 4 \times 10^{-9}$.
In $0.05\, M\, NaF$ solution,$NaF$ dissociates completely to provide $[F^-] = 0.05\, M$.
Let the solubility of $PbF_2$ in this solution be $S'\, M$. Then $[Pb^{2+}] = S'$ and $[F^-] = (2S' + 0.05)\, M$.
$K_{sp} = S'(2S' + 0.05)^2 = 4 \times 10^{-9}$.
Since $S'$ is very small,we assume $2S' \ll 0.05$,so $(2S' + 0.05) \approx 0.05$.
$S' \times (0.05)^2 = 4 \times 10^{-9}$.
$S' \times 2.5 \times 10^{-3} = 4 \times 10^{-9}$.
$S' = \frac{4 \times 10^{-9}}{2.5 \times 10^{-3}} = 1.6 \times 10^{-6}\, M$.

(C) The dissociation of the salt is given by: $AB \rightleftharpoons A^{+} + B^{-}$
The solubility product expression is: $K_{sp} = [A^{+}][B^{-}]$
Precipitation occurs when the ionic product exceeds the solubility product: $[A^{+}][B^{-}] > K_{sp}$
Given $[A^{+}] = 10^{-3} \ M$ and $K_{sp} = 1 \times 10^{-8}$,we substitute these values:
$(10^{-3})[B^{-}] > 1 \times 10^{-8}$
$[B^{-}] > \frac{1 \times 10^{-8}}{10^{-3}}$
$[B^{-}] > 1 \times 10^{-5} \ M$

(A) The precipitation of metal sulfides depends on the solubility product $(K_{sp})$ and the concentration of sulfide ions $(S^{2-})$.
In an acidic medium,the dissociation of $H_2S$ is suppressed due to the common ion effect of $H^+$ ions,resulting in a very low concentration of $S^{2-}$ ions.
Since the $K_{sp}$ of $As_2S_3$ is extremely low,even this low concentration of $S^{2-}$ is sufficient to exceed the ionic product of $As_2S_3$,causing it to precipitate.
Conversely,the $K_{sp}$ of $ZnS$ is relatively higher,so the low concentration of $S^{2-}$ in the acidic medium is insufficient to exceed the $K_{sp}$ of $ZnS$,preventing its precipitation.

(C) For a salt of the type $M(OH)_2$,the solubility product is given by $K_{sp} = [M^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$,we have $4S^3 = 3.2 \times 10^{-11}$,which implies $S^3 = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
Thus,$S = 2 \times 10^{-4} \ M$.
The concentration of hydroxide ions is $[OH^-] = 2S = 2 \times (2 \times 10^{-4}) = 4 \times 10^{-4} \ M$.
$pOH = -\log[OH^-] = -\log(4 \times 10^{-4}) = 4 - \log 4 = 4 - 0.60 = 3.40$.
Since $pH + pOH = 14$,we have $pH = 14 - 3.40 = 10.60$.

(D) To determine the most and least soluble compounds,we calculate the molar solubility $(S)$ for each:
$1$. For $AgCl$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{1.1 \times 10^{-10}} \approx 1.05 \times 10^{-5} \, M$
$2$. For $AgI$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \, M$
$3$. For $PbCrO_4$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{4.0 \times 10^{-14}} = 2.0 \times 10^{-7} \, M$
$4$. For $Ag_2CO_3$ ($2:1$ type): $K_{sp} = 4S^3$ $\Rightarrow S = \sqrt[3]{K_{sp}/4} = \sqrt[3]{8.0 \times 10^{-12} / 4} = \sqrt[3]{2.0 \times 10^{-12}} \approx 1.26 \times 10^{-4} \, M$
Comparing the solubilities: $1.26 \times 10^{-4} > 1.05 \times 10^{-5} > 2.0 \times 10^{-7} > 1.0 \times 10^{-8}$.
Thus,$Ag_2CO_3$ is the most soluble and $AgI$ is the least soluble.

(B) The dissociation of $Mg(OH)_2$ is given by: $Mg(OH)_2(s) \leftrightarrow Mg^{2+}(aq) + 2OH^{-}(aq)$.
The solubility product expression is $K_{sp} = [Mg^{2+}][OH^{-}]^2$.
Given $K_{sp} = 1.0 \times 10^{-11}$ and $[Mg^{2+}] = 0.001\, M = 10^{-3}\, M$.
Substituting the values: $1.0 \times 10^{-11} = (10^{-3})[OH^{-}]^2$.
$[OH^{-}]^2 = \frac{1.0 \times 10^{-11}}{10^{-3}} = 10^{-8}$.
$[OH^{-}] = \sqrt{10^{-8}} = 10^{-4}\, M$.
Now,$pOH = -\log[OH^{-}] = -\log(10^{-4}) = 4$.
Since $pH + pOH = 14$ at $25\,^{\circ}C$,we have $pH = 14 - 4 = 10$.

(C) The solubility product constant $(K_{sp})$ values at $25^{\circ}C$ are approximately $K_{sp}(AgCl) \approx 1.8 \times 10^{-10}$ and $K_{sp}(AgBr) \approx 5.0 \times 10^{-13}$.
Since $K_{sp}(AgBr) < K_{sp}(AgCl)$,$AgBr$ requires a lower concentration of $Ag^+$ ions to exceed its solubility product compared to $AgCl$.
Therefore,$AgBr$ will precipitate first.
The Assertion is correct,but the Reason is incorrect because it states $K_{sp}(AgCl) < K_{sp}(AgBr)$,which is false.

(A) Barium carbonate $(BaCO_3)$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(Ba(OH)_2)$.
In water,it has very low solubility.
When $HNO_3$ is added,the carbonate ion $(CO_3^{2-})$ acts as a weak base and reacts with $H^+$ ions from the strong acid to form $H_2CO_3$,which further decomposes into $CO_2$ and $H_2O$.
This reaction removes $CO_3^{2-}$ ions from the equilibrium,shifting the dissolution equilibrium of $BaCO_3$ to the right according to Le Chatelier's principle,thereby increasing its solubility.
The reaction is: $BaCO_3(s) + 2H^+(aq) \to Ba^{2+}(aq) + CO_2(g) + H_2O(l)$.

(A) The dissociation of $Ca(OH)_{2}$ is given by: $Ca(OH)_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2OH^{-}_{(aq)}$
Given $pH = 9$,we know $pOH = 14 - pH = 14 - 9 = 5$.
Therefore,$[OH^{-}] = 10^{-pOH} = 10^{-5} \ M$.
From the stoichiometry,$[OH^{-}] = 2S$,where $S$ is the solubility of $Ca(OH)_{2}$.
So,$2S = 10^{-5} \implies S = 0.5 \times 10^{-5} \ M$.
The solubility product expression is $K_{sp} = [Ca^{2+}][OH^{-}]^{2} = (S)(2S)^{2} = 4S^{3}$.
Substituting the value of $S$: $K_{sp} = 4 \times (0.5 \times 10^{-5})^{3} = 4 \times 0.125 \times 10^{-15} = 0.5 \times 10^{-15}$.

(C) $CaF_2$ dissociates as: $CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^{-}_{(aq)}$
$NaF$ is a strong electrolyte: $NaF_{(aq)} \rightarrow Na^{+}_{(aq)} + F^{-}_{(aq)}$
Given $[NaF] = 0.1 \ M$,so $[F^-] = 0.1 \ M$ from $NaF$.
Let $s$ be the molar solubility of $CaF_2$. Then $[Ca^{2+}] = s$ and total $[F^-] = (2s + 0.1) \approx 0.1 \ M$ (since $s$ is very small).
$K_{sp} = [Ca^{2+}][F^-]^2$
$5.3 \times 10^{-11} = s \times (0.1)^2$
$s = \frac{5.3 \times 10^{-11}}{0.01} = 5.3 \times 10^{-9} \ mol \ L^{-1}$

(A) The solubility $(s)$ in $mol \ L^{-1}$ is calculated by dividing the solubility in $g \ L^{-1}$ by the molar mass of $BaSO_{4}$.
$s = \frac{2.42 \times 10^{-3} \ g \ L^{-1}}{233 \ g \ mol^{-1}} \approx 1.0386 \times 10^{-5} \ mol \ L^{-1}$.
For a salt like $BaSO_{4}$,the dissociation is $BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO_{4}^{2-}(aq)$.
The solubility product $K_{sp} = [Ba^{2+}][SO_{4}^{2-}] = s \times s = s^{2}$.
$K_{sp} = (1.0386 \times 10^{-5})^{2} \approx 1.078 \times 10^{-10} \ mol^{2} \ L^{-2}$.
Rounding to two decimal places,we get $1.08 \times 10^{-10} \ mol^{2} \ L^{-2}$.

(C) From the graph,at $[X] = 1 \times 10^{-3} \ M$,the corresponding $[Y] = 2 \times 10^{-3} \ M$.
Since the concentration of $Y$ is twice that of $X$,the stoichiometry of the salt is $XY_{2}$.
The dissociation equilibrium is: $XY_{2(s)} \rightleftharpoons X_{(aq)}^{2+} + 2Y_{(aq)}^{-}$.
The solubility product constant $K_{sp}$ is given by: $K_{sp} = [X^{2+}][Y^{-}]^{2}$.
Substituting the values: $K_{sp} = (1 \times 10^{-3}) \times (2 \times 10^{-3})^{2}$.
$K_{sp} = (10^{-3}) \times (4 \times 10^{-6}) = 4 \times 10^{-9} \ M^{3}$.

(A) The dissociation of $Cr(OH)_3$ is given by: $Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$.
Let the solubility of $Cr(OH)_3$ be $s \ mol/L$.
Then,$[Cr^{3+}] = s$ and $[OH^-] = 3s$.
The solubility product expression is $K_{sp} = [Cr^{3+}][OH^-]^3$.
Substituting the values: $K_{sp} = (s)(3s)^3 = 27s^4$.
Given $K_{sp} = 6.0 \times 10^{-31}$,we have $27s^4 = 6.0 \times 10^{-31}$.
Thus,$s^4 = \frac{6.0 \times 10^{-31}}{27} = \frac{60 \times 10^{-32}}{27} \approx 2.22 \times 10^{-32}$.
The concentration of hydroxide ions is $[OH^-] = 3s$.
Since $s = (2.22 \times 10^{-32})^{1/4}$,then $[OH^-] = 3 \times (2.22 \times 10^{-32})^{1/4} = (3^4 \times 2.22 \times 10^{-32})^{1/4} = (81 \times 2.22 \times 10^{-32})^{1/4} = (179.82 \times 10^{-32})^{1/4} = (17.982 \times 10^{-31})^{1/4} \approx (18 \times 10^{-31})^{1/4}$.

(B) First,calculate the final concentrations of the ions in the mixture (total volume = $300 \ mL + 100 \ mL = 400 \ mL$):
$[Pb^{2+}] = \frac{300 \ mL \times 0.134 \ M}{400 \ mL} = 0.1005 \ M$
$[Cl^-] = \frac{100 \ mL \times 0.4 \ M}{400 \ mL} = 0.1 \ M$
Next,calculate the reaction quotient $Q$ for the dissociation $PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$:
$Q = [Pb^{2+}][Cl^-]^2$
$Q = (0.1005) \times (0.1)^2$
$Q = 0.1005 \times 0.01 = 1.005 \times 10^{-3}$
Comparing $Q$ with $K_{sp}$ $(1.6 \times 10^{-5})$:
$1.005 \times 10^{-3} > 1.6 \times 10^{-5}$
Therefore,$Q > K_{sp}$.

(A) The dissociation of $A_{2}X_{3}$ is given by: $A_{2}X_{3} \rightarrow 2A^{3+} + 3X^{2-}$
The solubility product expression is: $K_{sp} = [A^{3+}]^{2} [X^{2-}]^{3} = 1.1 \times 10^{-23}$
Let $S$ be the solubility of $A_{2}X_{3}$. Then,$[A^{3+}] = 2S$ and $[X^{2-}] = 3S$.
Substituting these into the $K_{sp}$ expression: $K_{sp} = (2S)^{2} (3S)^{3} = 4S^{2} \times 27S^{3} = 108S^{5}$
Equating to the given value: $108S^{5} = 1.1 \times 10^{-23}$
$S^{5} = \frac{1.1 \times 10^{-23}}{108} \approx 1.0185 \times 10^{-25} \approx 1.0 \times 10^{-25}$
Taking the fifth root: $S = (1.0 \times 10^{-25})^{1/5} = 1.0 \times 10^{-5} \ mol/L$.

(N/A) For $AgCN$: $AgCN \rightleftharpoons Ag^{+} + CN^{-}$. Let solubility be $S_1$. $K_{sp} = S_1^2 = 6 \times 10^{-17}$. Thus,$S_1 = \sqrt{6 \times 10^{-17}} \approx 7.75 \times 10^{-9} \ M$.
For $Ni(OH)_2$: $Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^{-}$. Let solubility be $S_2$. $K_{sp} = (S_2)(2S_2)^2 = 4S_2^3 = 2.0 \times 10^{-15}$.
$S_2^3 = 0.5 \times 10^{-15} = 5 \times 10^{-16}$.
$S_2 = \sqrt[3]{5 \times 10^{-16}} \approx 7.94 \times 10^{-6} \ M$.
Comparing $S_1$ and $S_2$,$S_2 > S_1$,therefore $Ni(OH)_2$ is more soluble.

(A) Let the molar solubility of $Ni(OH)_{2}$ be $S \, mol/L$.
The dissolution reaction is: $Ni(OH)_{2}(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq)$.
From the reaction,$Ni^{2+}$ concentration is $S$ and $OH^{-}$ concentration from $Ni(OH)_{2}$ is $2S$.
Since $NaOH$ is a strong electrolyte,it provides $0.10 \, M$ $OH^{-}$ ions.
Total $[OH^{-}] = (0.10 + 2S) \, M \approx 0.10 \, M$ (since $S$ is very small).
The solubility product expression is $K_{sp} = [Ni^{2+}][OH^{-}]^{2}$.
Substituting the values: $2.0 \times 10^{-15} = (S)(0.10)^{2}$.
$S = \frac{2.0 \times 10^{-15}}{0.01} = 2.0 \times 10^{-13} \, M$.

(A) The solubility product of $CuS$ is given as $K_{sp} = 6 \times 10^{-16}$.
Let $s$ be the solubility of $CuS$ in $mol \ L^{-1}$.
The dissociation equilibrium is: $CuS(s) \leftrightarrow Cu^{2+}(aq) + S^{2-}(aq)$.
The solubility product expression is $K_{sp} = [Cu^{2+}][S^{2-}] = s \times s = s^2$.
Equating the values: $s^2 = 6 \times 10^{-16}$.
Taking the square root: $s = \sqrt{6 \times 10^{-16}} = 2.45 \times 10^{-8} \ mol \ L^{-1}$.
Thus,the maximum molarity of $CuS$ in an aqueous solution is $2.45 \times 10^{-8} \ M$.

$(1)$ Silver chromate $(Ag_{2}CrO_{4})$:
$Ag_{2}CrO_{4} \rightleftharpoons 2Ag^{+} + CrO_{4}^{2-}$
$K_{sp} = [Ag^{+}]^{2} [CrO_{4}^{2-}] = (2s)^{2}(s) = 4s^{3}$
$1.1 \times 10^{-12} = 4s^{3} \Rightarrow s = 0.65 \times 10^{-4} \, M$
$[Ag^{+}] = 1.30 \times 10^{-4} \, M, [CrO_{4}^{2-}] = 0.65 \times 10^{-4} \, M$
$(2)$ Barium chromate $(BaCrO_{4})$:
$BaCrO_{4} \rightleftharpoons Ba^{2+} + CrO_{4}^{2-}$
$K_{sp} = [Ba^{2+}][CrO_{4}^{2-}] = s^{2}$
$1.2 \times 10^{-10} = s^{2} \Rightarrow s = 1.09 \times 10^{-5} \, M$
$[Ba^{2+}] = 1.09 \times 10^{-5} \, M, [CrO_{4}^{2-}] = 1.09 \times 10^{-5} \, M$
$(3)$ Ferric hydroxide $(Fe(OH)_{3})$:
$Fe(OH)_{3} \rightleftharpoons Fe^{3+} + 3OH^{-}$
$K_{sp} = [Fe^{3+}][OH^{-}]^{3} = (s)(3s)^{3} = 27s^{4}$
$1.0 \times 10^{-38} = 27s^{4} \Rightarrow s = 1.39 \times 10^{-10} \, M$
$[Fe^{3+}] = 1.39 \times 10^{-10} \, M, [OH^{-}] = 4.17 \times 10^{-10} \, M$
$(4)$ Lead chloride $(PbCl_{2})$:
$PbCl_{2} \rightleftharpoons Pb^{2+} + 2Cl^{-}$
$K_{sp} = [Pb^{2+}][Cl^{-}]^{2} = (s)(2s)^{2} = 4s^{3}$
$1.6 \times 10^{-5} = 4s^{3} \Rightarrow s = 1.58 \times 10^{-2} \, M$
$[Pb^{2+}] = 1.58 \times 10^{-2} \, M, [Cl^{-}] = 3.16 \times 10^{-2} \, M$
$(5)$ Mercurous iodide $(Hg_{2}I_{2})$:
$Hg_{2}I_{2} \rightleftharpoons Hg_{2}^{2+} + 2I^{-}$
$K_{sp} = [Hg_{2}^{2+}][I^{-}]^{2} = (s)(2s)^{2} = 4s^{3}$
$4.5 \times 10^{-29} = 4s^{3} \Rightarrow s = 2.24 \times 10^{-10} \, M$
$[Hg_{2}^{2+}] = 2.24 \times 10^{-10} \, M, [I^{-}] = 4.48 \times 10^{-10} \, M$

For $Ag_{2}CrO_{4}$:
$Ag_{2}CrO_{4} \longleftrightarrow 2Ag^{+} + CrO_{4}^{2-}$
$K_{sp} = (2s)^{2} \cdot s = 4s^{3} = 1.1 \times 10^{-12}$
$s^{3} = 0.275 \times 10^{-12} = 275 \times 10^{-15}$
$s = (275)^{1/3} \times 10^{-5} \approx 6.5 \times 10^{-5} \, M$
For $AgBr$:
$AgBr \longleftrightarrow Ag^{+} + Br^{-}$
$K_{sp} = (s^{\prime})^{2} = 5.0 \times 10^{-13} = 50 \times 10^{-14}$
$s^{\prime} = \sqrt{50} \times 10^{-7} \approx 7.07 \times 10^{-7} \, M$
Ratio of molarities:
$\frac{s}{s^{\prime}} = \frac{6.5 \times 10^{-5}}{7.07 \times 10^{-7}} \approx 91.9$

(B) When equal volumes of sodium iodate and cupric chlorate solutions are mixed,the concentration of each species is halved,becoming $0.001 \, M$.
$NaIO_3 \rightarrow Na^{+} + IO_3^{-}$
$[IO_3^{-}] = 0.001 \, M$
$Cu(ClO_3)_2 \rightarrow Cu^{2+} + 2ClO_3^{-}$
$[Cu^{2+}] = 0.001 \, M$
The ionic product $(Q_{sp})$ for $Cu(IO_3)_2$ is calculated as:
$Q_{sp} = [Cu^{2+}][IO_3^{-}]^2$
$Q_{sp} = (0.001) \times (0.001)^2$
$Q_{sp} = 10^{-3} \times 10^{-6} = 10^{-9}$
Since $Q_{sp} (1 \times 10^{-9}) < K_{sp} (7.4 \times 10^{-8})$,the ionic product is less than the solubility product constant.
Therefore,precipitation will not occur.

(C) Given $pH = 3.19$,then $[H_{3}O^{+}] = 10^{-3.19} \approx 6.46 \times 10^{-4} \, M$.
For the equilibrium: $C_{6}H_{5}COOH + H_{2}O \longleftrightarrow C_{6}H_{5}COO^{-} + H_{3}O^{+}$.
The ratio of concentrations is $\frac{[C_{6}H_{5}COOH]}{[C_{6}H_{5}COO^{-}]} = \frac{[H_{3}O^{+}]}{K_{a}} = \frac{6.46 \times 10^{-4}}{6.46 \times 10^{-5}} = 10$.
Let the solubility of $C_{6}H_{5}COOAg$ in the buffer be $x \, mol/L$.
Then $[Ag^{+}] = x$ and $[C_{6}H_{5}COOH] + [C_{6}H_{5}COO^{-}] = x$.
Substituting the ratio: $10[C_{6}H_{5}COO^{-}] + [C_{6}H_{5}COO^{-}] = x$,so $[C_{6}H_{5}COO^{-}] = \frac{x}{11}$.
Using $K_{sp} = [Ag^{+}][C_{6}H_{5}COO^{-}] = x \times \frac{x}{11} = 2.5 \times 10^{-13}$.
$x^{2} = 27.5 \times 10^{-13} = 2.75 \times 10^{-12}$,so $x \approx 1.66 \times 10^{-6} \, mol/L$.
In pure water,let solubility be $x^{\prime}$. $K_{sp} = (x^{\prime})^{2} = 2.5 \times 10^{-13}$.
$x^{\prime} = \sqrt{2.5 \times 10^{-13}} \approx 5.0 \times 10^{-7} \, mol/L$.
The ratio of solubility is $\frac{x}{x^{\prime}} = \frac{1.66 \times 10^{-6}}{5.0 \times 10^{-7}} = 3.32$.

(A) Let the maximum concentration of each solution be $x \ mol/L$. After mixing equal volumes,the concentration of each ion in the mixture becomes half of the initial concentration,i.e.,$x/2 \ M$.
$\therefore [Fe^{2+}] = \frac{x}{2} \ M$ and $[S^{2-}] = \frac{x}{2} \ M$.
For no precipitation,the ionic product must be less than or equal to the solubility product $(K_{sp})$.
$Q_{sp} = [Fe^{2+}][S^{2-}] \leq K_{sp}$
$(\frac{x}{2})(\frac{x}{2}) \leq 6.3 \times 10^{-18}$
$\frac{x^2}{4} \leq 6.3 \times 10^{-18}$
$x^2 \leq 25.2 \times 10^{-18}$
$x \leq \sqrt{25.2} \times 10^{-9}$
$x \leq 5.02 \times 10^{-9} \ M$.
Thus,the maximum concentration is $5.02 \times 10^{-9} \ M$.

(D) $CaSO_{4(s)} \leftrightarrow Ca^{2+}_{(aq)} + S{O_{4}}^{2-}_{(aq)}$
$K_{sp} = [Ca^{2+}][SO_{4}^{2-}] = s^2$
$s = \sqrt{K_{sp}} = \sqrt{9.1 \times 10^{-6}} = 3.0166 \times 10^{-3} \, mol/L$
Molar mass of $CaSO_{4} = 40 + 32 + (4 \times 16) = 136 \, g/mol$
Solubility in $g/L = s \times \text{Molar mass} = 3.0166 \times 10^{-3} \times 136 \approx 0.4103 \, g/L$
Volume required for $1 \, g = \frac{1 \, g}{0.4103 \, g/L} \approx 2.44 \, L$.