Common ion effect Questions in English

(C) The reaction between $NaCl$ and $AgNO_3$ is: $NaCl(aq) + AgNO_3(aq) \rightarrow AgCl(s) + NaNO_3(aq)$.
This is a precipitation reaction where $AgCl$ is formed as a white precipitate.
According to Le Chatelier's principle,the reaction can be suppressed by adding a common ion or by changing the solvent properties.
Adding $AgCl$ (common ion effect) shifts the equilibrium backward.
Adding Acetone (a solvent with lower dielectric constant than water) decreases the solubility of the ionic salts,thereby suppressing the reaction.
Therefore,both $(A)$ and $(B)$ are correct.

(D) The solubility of a sparingly soluble salt like $AgCl$ decreases in the presence of a common ion due to the common ion effect,which shifts the equilibrium $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ to the left.
Calculate the concentration of the common ion ($Ag^+$ or $Cl^-$) in each solution:
$1$. $AgNO_3 \ (0.1 \ M) \to [Ag^+] = 0.1 \ M$
$2$. $H_2O \ (l) \to [Ag^+] = 0, [Cl^-] = 0$
$3$. $NaCl \ (0.4 \ M) \to [Cl^-] = 0.4 \ M$
$4$. $BaCl_2 \ (0.3 \ M) \to [Cl^-] = 2 \times 0.3 = 0.6 \ M$
Since the solubility is inversely proportional to the concentration of the common ion,the solubility is minimum where the common ion concentration is maximum.
Comparing the concentrations: $0.6 \ M > 0.4 \ M > 0.1 \ M > 0$.
Therefore,the solubility of $AgCl$ is minimum in $0.3 \ M \ BaCl_2$.

(C) The dissociation of a weak acid like $CH_3COOH$ is suppressed by the common ion effect.
$CH_3COOH \rightleftharpoons CH_3COO^- + H^+$.
$HCl$ provides $H^+$ ions,which suppresses the dissociation of $CH_3COOH$ due to the common ion effect.
$CH_3COOK$ provides $CH_3COO^-$ ions,which also suppresses the dissociation of $CH_3COOH$ due to the common ion effect.
$KCl$ is a neutral salt and does not provide any common ions ($H^+$ or $CH_3COO^-$).
Therefore,the dissociation of $CH_3COOH$ is least suppressed (maximum) in the presence of $0.001 \, M \, KCl$ compared to the other options which contain common ions.

(C) The given equilibrium is $AgCl_{(s)} + 2NH_{3(aq)} \rightleftharpoons [Ag(NH_3)_2]^+_{(aq)} + Cl^-_{(aq)}$.
When $HNO_3$ is added,it reacts with $NH_3$ to form $NH_4^+$ ions: $NH_{3(aq)} + H^+_{(aq)} \to NH_{4(aq)}^+$.
This decreases the concentration of $NH_3$,which shifts the equilibrium to the left according to Le Chatelier's principle.
Consequently,the complex ion $[Ag(NH_3)_2]^+$ dissociates,releasing $Ag^+$ ions,which then react with $Cl^-$ ions to precipitate $AgCl_{(s)}$.
The overall reaction is: $[Ag(NH_3)_2]^+_{(aq)} + Cl^-_{(aq)} + 2H^+_{(aq)} \to AgCl_{(s)} + 2NH_{4(aq)}^+$.
Thus,adding $HNO_3$ causes the white precipitate of $AgCl$ to reappear.

(A) Let the solubility of $AgCl$ be $x \ mol \ L^{-1}$.
The dissociation equilibrium is: $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$.
The solubility product expression is: $K_{sp} = [Ag^{+}][Cl^{-}]$.
In a $0.1 \ M \ KCl$ solution,$KCl$ dissociates completely: $KCl(aq) \rightarrow K^{+}(aq) + Cl^{-}(aq)$.
Thus,the concentration of $Cl^{-}$ from $KCl$ is $0.1 \ M$.
The total concentration of $Cl^{-}$ is $[Cl^{-}] = (x + 0.1) \ M \approx 0.1 \ M$ (since $x$ is very small).
Substituting the values into the $K_{sp}$ expression: $1.0 \times 10^{-10} = (x)(0.1)$.
Solving for $x$: $x = \frac{1.0 \times 10^{-10}}{0.1} = 1.0 \times 10^{-9} \ mol \ L^{-1}$.

(C) The dissolution of $BaF_2$ is given by: $BaF_2(s) \rightleftharpoons Ba^{2+}(aq) + 2F^{-}(aq)$.
In the presence of $Ba(NO_3)_2$ (a strong electrolyte),the concentration of $Ba^{2+}$ ions is increased due to the common ion effect.
Let $s$ be the solubility of $BaF_2$ and $x$ be the concentration of $Ba(NO_3)_2$.
Total $[Ba^{2+}] = s + x$ and $[F^{-}] = 2s$.
From the solubility product expression: $K_{sp} = [Ba^{2+}][F^{-}]^2$.
Substituting the values: $K_{sp} = (s + x)(2s)^2$.
Since $2s = [F^{-}]$,we have $s = \frac{1}{2}[F^{-}]$.
Thus,the solubility of $BaF_2$ is represented by $\frac{1}{2}[F^{-}]$.

(B) The solubility of $AgCl$ in water is governed by its solubility product constant $(K_{sp})$,where $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.
In the presence of concentrated $HCl$,there is a high concentration of $Cl^-$ ions due to the common ion effect.
According to Le Chatelier's principle,the increase in $[Cl^-]$ shifts the equilibrium to the left,decreasing the solubility of $AgCl$.
Therefore,the solubility of $AgCl$ in concentrated $HCl$ is less than in pure water.

(B) When hydrogen chloride gas $(HCl)$ is passed through a saturated solution of $NaCl$,it dissociates into its ions as:
$NaCl \rightleftharpoons Na^{+} + Cl^{-}$
$HCl \rightleftharpoons H^{+} + Cl^{-}$
Due to the ionization of $HCl$,the concentration of $Cl^{-}$ ions increases significantly in the solution.
According to Le Chatelier's principle,the increase in the concentration of the common ion $(Cl^{-})$ shifts the equilibrium of $NaCl$ dissociation to the left.
This phenomenon is known as the common ion effect,which leads to a decrease in the solubility of $NaCl$,causing it to precipitate out.

(D) The solubility of a salt is governed by its solubility product constant $(K_{sp})$.
When $HCl$ is added to a saturated solution of $NaCl$,the concentration of $Cl^-$ ions increases significantly due to the common ion effect.
According to the solubility equilibrium $NaCl(s) \rightleftharpoons Na^+(aq) + Cl^-(aq)$,the ionic product $Q = [Na^+][Cl^-]$ increases.
When the ionic product $Q$ exceeds the solubility product $K_{sp}$,the equilibrium shifts to the left,causing the precipitation of $NaCl$.

(B) The solubility of a sparingly soluble salt like $AgI$ decreases in the presence of a common ion due to the common ion effect.
When $NaI$ is added to a saturated solution of $AgI$,the concentration of $I^-$ ions increases.
According to Le Chatelier's principle,the equilibrium $AgI(s) \rightleftharpoons Ag^+(aq) + I^-(aq)$ shifts to the left,resulting in the precipitation of $AgI$ and a decrease in its solubility.

(D) The presence of $NH_4Cl$ (a strong electrolyte) suppresses the dissociation of $NH_4OH$ (a weak base) due to the common ion effect ($NH_4^+$ ions).
This significantly reduces the concentration of $OH^-$ ions in the solution.
For the precipitation of $Ba(OH)_2$ to occur,the ionic product $[Ba^{2+}][OH^-]^2$ must exceed the solubility product constant $(K_{sp})$ of $Ba(OH)_2$.
Since the concentration of $OH^-$ is kept very low,the ionic product does not exceed $K_{sp}$,and no precipitation occurs.
Furthermore,$Ba(OH)_2$ is actually a strong base and is soluble in water.
Therefore,the Assertion is incorrect and the Reason is incorrect.

(B) The common table salt $(NaCl)$ is purified by passing $HCl$ gas through a saturated solution of salt.
This process is based on the common ion effect,where the increase in the concentration of $Cl^-$ ions forces the equilibrium to shift towards the precipitation of $NaCl$.
The reaction is: $NaCl(aq) + HCl(g) \to NaCl(s) + H^+(aq) + Cl^-(aq)$.

(C) The Assertion is true because $NH_4Cl$ is added to $NH_4OH$ to provide a common ion $(NH_4^+)$,which suppresses the dissociation of $NH_4OH$ due to the common ion effect.
This reduction in the concentration of $OH^-$ ions ensures that only the hydroxides of the third group cations (like $Fe^{3+}$,$Al^{3+}$,$Cr^{3+}$) are precipitated,preventing the precipitation of hydroxides of subsequent groups.
The Reason is false because the purpose is not to convert ions into chlorides,but to control the $OH^-$ concentration for selective precipitation.

(D) $Sb(III)$ belongs to the $II$ group of qualitative analysis and is precipitated as $Sb_2S_3$ by passing $H_2S$ in an acidic medium (presence of $HCl$).
The $HCl$ provides $H^+$ ions,which suppress the dissociation of $H_2S$ due to the common ion effect,keeping the concentration of $S^{2-}$ low,which is sufficient for $II$ group precipitation but not for higher groups.
In an alkaline medium,$OH^-$ ions react with $H^+$ ions from $H_2S$ to form water $(H^+ + OH^- \rightleftharpoons H_2O)$.
This shifts the equilibrium $H_2S \rightleftharpoons 2H^+ + S^{2-}$ to the right,significantly increasing the concentration of $S^{2-}$ ions.
Since the concentration of $S^{2-}$ is very high in an alkaline medium,$Sb(III)$ will definitely precipitate as a sulphide.
Therefore,the Assertion is false because $Sb(III)$ is indeed precipitated in an alkaline medium,and the Reason is false because the concentration of $S^{2-}$ is actually very high,not inadequate.

(D) Antimony sulphide $(Sb_2S_3)$ is soluble in yellow ammonium sulphide $( (NH_4)_2S_x )$ to form soluble ammonium thioantimonate $( (NH_4)_3SbS_4 )$.
The chemical reaction is: $Sb_2S_3 + 3(NH_4)_2S + 2S \longrightarrow 2(NH_4)_3SbS_4$.
Since $Sb_2S_3$ is soluble,the Assertion is false.
The common ion effect due to $S^{2-}$ ions would generally decrease solubility,but it is not the reason for the behavior of $Sb_2S_3$ in this context,and the premise of the Assertion itself is incorrect. Thus,both the Assertion and the Reason are incorrect.

(N/A) Ionization of phenol: $C_6H_5OH + H_2O \leftrightarrow C_6H_5O^{-} + H_3O^{+}$
Initial concentration: $0.05 \ M$,$0$,$0$
At equilibrium: $0.05 - x$,$x$,$x$
$K_a = \frac{[C_6H_5O^{-}][H_3O^{+}]}{[C_6H_5OH]} = \frac{x^2}{0.05 - x}$
Since $K_a$ is very small,$x$ is negligible compared to $0.05$. So,$0.05 - x \approx 0.05$.
$x = \sqrt{K_a \times C} = \sqrt{1.0 \times 10^{-10} \times 0.05} = \sqrt{5.0 \times 10^{-12}} = 2.236 \times 10^{-6} \ M$.
Thus,$[C_6H_5O^{-}] = 2.236 \times 10^{-6} \ M$.
Now,in the presence of $0.01 \ M$ sodium phenolate $(C_6H_5ONa)$:
$C_6H_5ONa \rightarrow C_6H_5O^{-} + Na^{+}$
$[C_6H_5O^{-}] = 0.01 \ M$ (due to common ion effect).
Let $\alpha$ be the degree of ionization of phenol.
$[C_6H_5OH] = 0.05 \ M$,$[H_3O^{+}] = 0.05 \alpha$,$[C_6H_5O^{-}] = 0.01 \ M$.
$K_a = \frac{[C_6H_5O^{-}][H_3O^{+}]}{[C_6H_5OH]} \Rightarrow 1.0 \times 10^{-10} = \frac{0.01 \times 0.05 \alpha}{0.05}$.
$1.0 \times 10^{-10} = 0.01 \alpha \Rightarrow \alpha = 1.0 \times 10^{-8}$.

$c = 0.05 \ M$
$p K_{ a } = 4.74$
$p K_{ a } = -\log(K_{ a }) \implies K_{ a } = 10^{-4.74} = 1.82 \times 10^{-5}$
For weak acid,$\alpha = \sqrt{\frac{ K_{ a } }{ c }} = \sqrt{\frac{ 1.82 \times 10^{-5} }{ 0.05 }} = 1.908 \times 10^{-2}$
When $HCl$ is added,the concentration of $H^{+}$ ions increases,shifting the equilibrium backward (common ion effect),thus decreasing the degree of dissociation.
Case $(a)$: $0.01 \ M$ $HCl$ added.
$K_{ a } = \frac{ [CH_{3}COO^{-}][H^{+}] }{ [CH_{3}COOH] } \approx \frac{ \alpha c \times 0.01 }{ c }$
$1.82 \times 10^{-5} = \alpha \times 0.01 \implies \alpha = 1.82 \times 10^{-3}$
Case $(b)$: $0.1 \ M$ $HCl$ added.
$1.82 \times 10^{-5} = \alpha \times 0.1 \implies \alpha = 1.82 \times 10^{-4}$

(N/A) The common ion effect is a phenomenon based on $Le \text{ } Chatelier's$ principle.
It is defined as the shift in equilibrium upon adding a substance that provides more of an ionic species already present in the dissociation equilibrium.
Example: Consider the dissociation equilibrium of acetic acid:
$CH_3COOH_{(aq)} \rightleftharpoons H^{+}_{(aq)} + CH_3COO^{-}_{(aq)}$
$\therefore K_a = \frac{[H^{+}][CH_3COO^{-}]}{[CH_3COOH]}$
If we add sodium acetate $(CH_3COONa)$ to this solution,the concentration of acetate ions $[CH_3COO^{-}]$ increases.
According to $Le \text{ } Chatelier's$ principle,the equilibrium shifts to the left to consume the excess acetate ions,thereby decreasing the concentration of $H^{+}$ ions and suppressing the dissociation of acetic acid.

(N/A) Definition: The common ion effect is defined as the shift in equilibrium upon adding a substance that provides more of an ionic species already present in the dissociation equilibrium. This phenomenon is based on $Le \ Chatelier's$ principle.
Example: Consider the dissociation equilibrium of acetic acid $(CH_{3}COOH)$:
$CH_{3}COOH_{(aq)} \rightleftharpoons H_{(aq)}^{+} + CH_{3}COO_{(aq)}^{-}$
The ionization constant is given by $K_{a} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}$.
If sodium acetate $(CH_{3}COONa)$ is added to this solution,it provides a high concentration of acetate ions $(CH_{3}COO^{-})$. According to $Le \ Chatelier's$ principle,the increase in the concentration of the product $(CH_{3}COO^{-})$ causes the equilibrium to shift to the left (towards the reactants) to minimize the effect of the added ions.
Consequently,the concentration of $H^{+}$ ions decreases,and the degree of dissociation of acetic acid is suppressed. This is the common ion effect.

(N/A) The dissociation of acetic acid is given by the equilibrium: $CH_{3}COOH_{(aq)} \rightleftharpoons H^{+}_{(aq)} + CH_{3}COO^{-}_{(aq)}$.
When $0.05 \ M$ acetate ions $(CH_{3}COO^{-})$ are added to the $0.05 \ M$ acetic acid $(CH_{3}COOH)$ solution,the concentration of the common ion,$CH_{3}COO^{-}$,increases.
According to Le Chatelier's principle,the increase in the concentration of the product $(CH_{3}COO^{-})$ causes the equilibrium to shift to the left.
As a result,the dissociation of $CH_{3}COOH$ is suppressed,leading to a decrease in the concentration of $H^{+}$ ions ($[H^{+}]$ decreases).
Consequently,the degree of dissociation of acetic acid decreases,and the $pH$ of the resulting solution increases.

(C) In the presence of $HCl$,the dissociation of $H_2S$ is suppressed due to the common ion effect of $H^+$ ions: $H_2S \rightleftharpoons 2H^+ + S^{2-}$.
This leads to a very low concentration of $S^{2-}$ ions.
The solubility product $(K_{sp})$ of second group sulfides is very low,so they precipitate even at low $S^{2-}$ concentrations.
However,the $K_{sp}$ of fourth group sulfides is relatively high,so they do not precipitate under these conditions.

(N/A) The dissociation of $H_2S$ is suppressed by the presence of $0.3 \ M \ HCl$ (a strong acid).
$1$. For the first dissociation: $H_2S \rightleftharpoons H^+ + HS^-$
$K_{a1} = \frac{[H^+][HS^-]}{[H_2S]} = 1.0 \times 10^{-7}$
Since $[H^+] \approx 0.3 \ M$ and $[H_2S] \approx 0.1 \ M$,we have:
$[HS^-] = \frac{K_{a1} \times [H_2S]}{[H^+]} = \frac{1.0 \times 10^{-7} \times 0.1}{0.3} = 3.33 \times 10^{-8} \ M$
$2$. For the second dissociation: $HS^- \rightleftharpoons H^+ + S^{2-}$
$K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]} = 1.3 \times 10^{-13}$
$[S^{2-}] = \frac{K_{a2} \times [HS^-]}{[H^+]} = \frac{1.3 \times 10^{-13} \times 3.33 \times 10^{-8}}{0.3} = 1.44 \times 10^{-20} \ M$

(N/A) The solubility of a sparingly soluble salt is decreased in the presence of a common ion. $A$ common ion effect occurs when the concentration of either the cation or the anion of the salt increases in the solution.
$1$. In a saturated solution of $AgCl$,if $NaCl$ is added,the $Cl^{-}$ ion is common. If $AgNO_{3}$ is added,the $Ag^{+}$ ion is common. According to Le Chatelier's principle,the increase in the concentration of these ions shifts the equilibrium $AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$ to the left,causing precipitation and decreasing the solubility of the salt.
$2$. Similarly,in a saturated solution of $NaCl$,if $HCl$ gas is passed,the concentration of $Cl^{-}$ increases,which forces the equilibrium $NaCl_{(s)} \rightleftharpoons Na^{+}_{(aq)} + Cl^{-}_{(aq)}$ to shift to the left,resulting in the precipitation of solid $NaCl$ and a decrease in its solubility.

(N/A) The solubility of a sparingly soluble salt decreases in the presence of a common ion. This is known as the common ion effect.
According to Le Chatelier's principle,when the concentration of one of the ions (either the cation or the anion) of a sparingly soluble salt is increased by adding a strong electrolyte containing a common ion,the equilibrium shifts to the left to consume the excess ions,resulting in the precipitation of the salt and a decrease in its solubility.
Example $1$: In a saturated solution of $AgCl$,if $NaCl$ is added,the $Cl^{-}$ ion concentration increases. If $AgNO_{3}$ is added,the $Ag^{+}$ ion concentration increases. In both cases,the equilibrium $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$ shifts to the left,causing more $AgCl$ to precipitate and decreasing its solubility.
Example $2$: When $HCl$ gas is passed through a saturated solution of $NaCl$,the concentration of $Cl^{-}$ ions increases. This causes the equilibrium $NaCl(s) \rightleftharpoons Na^{+}(aq) + Cl^{-}(aq)$ to shift to the left,leading to the precipitation of solid $NaCl$ and a decrease in its solubility.

(A) The common ion effect is widely used in various chemical processes:
$1$. Purification of $NaCl$: To remove impurities like $Na_{2}SO_{4}$ and $MgSO_{4}$ from a saturated $NaCl$ solution, $HCl$ gas is passed through it. The increase in $[Cl^{-}]$ concentration causes the precipitation of $NaCl$ due to the common ion effect.
$2$. Salting out of soap: Soap is a sodium salt of a fatty acid $(RCOONa)$. When $NaCl$ is added to the soap solution, $Na^{+}$ acts as a common ion. The equilibrium $RCOONa_{(aq)} \rightleftharpoons RCOO^{-}_{(aq)} + Na^{+}_{(aq)}$ shifts to the left, causing the soap to precipitate as a solid.
$3$. Quantitative analysis: To ensure complete precipitation of an ion, an excess of a reagent containing a common ion is added. For example, excess $H_{2}SO_{4}$ is added for $BaSO_{4}$ precipitation, and excess $NaCl$ is used for $AgCl$ precipitation.
$4$. Qualitative analysis:
$(i)$ In Group-$II$ analysis, $HCl$ is added to $H_{2}S$ to suppress the dissociation of $H_{2}S$ via the common ion effect of $H^{+}$, ensuring only $CuS$ and $CdS$ precipitate.
$(ii)$ In Group-$III$ $A$ analysis, $NH_{4}Cl$ is added before $NH_{4}OH$ to suppress the dissociation of $NH_{4}OH$ via the common ion effect of $NH_{4}^{+}$, limiting $[OH^{-}]$ to precipitate only $Fe^{3+}$, $Al^{3+}$, and $Cr^{3+}$.
$5$. Buffer solutions: The common ion effect is essential in maintaining the $pH$ of buffer solutions used in drugs, cosmetics, and fertilizers.

(N/A) The solubility product expression for $AgCl$ is $K_{sp} = [Ag^+][Cl^-]$.
In a $0.1 \ M \ NaCl$ solution,the concentration of $Cl^-$ ions is $0.1 \ M$ due to the common ion effect.
Let $s$ be the solubility of $AgCl$ in the presence of $NaCl$.
Then,$[Ag^+] = s$ and $[Cl^-] = 0.1 + s \approx 0.1 \ M$ (since $s$ is very small).
Substituting these values into the $K_{sp}$ expression:
$1.6 \times 10^{-10} = s \times 0.1$
$s = \frac{1.6 \times 10^{-10}}{0.1} = 1.6 \times 10^{-9} \ M$.

(C) When $HCl$ gas is passed through the solution,the concentration of $Cl^{-}$ ions increases significantly due to the dissociation of $HCl$.
According to the common ion effect,the ionic product of $NaCl$ $([Na^{+}][Cl^{-}])$ exceeds its solubility product $(K_{sp})$.
Since $NaCl$ has a lower solubility compared to $MgCl_{2}$ and $CaCl_{2}$ in the presence of high $Cl^{-}$ concentration,$NaCl$ crystallizes out of the solution.
$MgCl_{2}$ and $CaCl_{2}$ remain in the solution as they are more soluble.

(C) Dimethylamine is a weak base: $(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-$
Given: $C = 0.05 \ M$,$K_b = 5 \times 10^{-4}$,and $[OH^-]_{NaOH} = 0.1 \ M$.
Due to the common ion effect,the concentration of $OH^-$ is dominated by $NaOH$,so $[OH^-] \approx 0.1 \ M$.
The expression for $K_b$ is: $K_b = \frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]} = \frac{C\alpha \times 0.1}{C(1-\alpha)} \approx \frac{0.05 \times \alpha \times 0.1}{0.05} = 0.1 \alpha$.
$5 \times 10^{-4} = 0.1 \alpha \implies \alpha = 5 \times 10^{-3}$.
Percentage dissociation $= \alpha \times 100 = 5 \times 10^{-3} \times 100 = 0.5 \% = 5 \times 10^{-1} \%$.

(C) The dissociation equilibrium of the weak acid $HA$ is given by: $HA \rightleftharpoons H^{+} + A^{-}$.
Initial concentrations: $[HA] = 0.01 \ M$,$[H^{+}] = 0.1 \ M$ (from $HCl$),$[A^{-}] = 0$.
At equilibrium,let the concentration of $A^{-}$ be $x \ M$. Then $[HA] = (0.01 - x) \approx 0.01 \ M$ and $[H^{+}] = (0.1 + x) \approx 0.1 \ M$ (since $\alpha << 1$).
The acid dissociation constant expression is $K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$.
Substituting the values: $2.0 \times 10^{-6} = \frac{0.1 \times x}{0.01}$.
Solving for $x$: $x = \frac{2.0 \times 10^{-6} \times 0.01}{0.1} = 2.0 \times 10^{-7} \ M$.
The degree of dissociation $\alpha = \frac{x}{C} = \frac{2.0 \times 10^{-7}}{0.01} = 2.0 \times 10^{-5}$.
Thus,the value is $2 \times 10^{-5}$.

(D) The solubility equilibrium for $AgBr$ is: $AgBr_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Br^{-}_{(aq)}$.
First,calculate the solubility product constant $(K_{sp})$ using the given concentrations in $KBr$ solution:
$K_{sp} = [Ag^{+}][Br^{-}] = (5 \times 10^{-10} \ M) \times (10^{-3} \ M) = 5 \times 10^{-13}$.
When $AgBr_{(s)}$ is added to $10^{-2} \ M$ $AgNO_3$ solution,the common ion effect occurs,and the concentration of $Ag^{+}$ becomes $10^{-2} \ M$.
Using the $K_{sp}$ value:
$K_{sp} = [Ag^{+}][Br^{-}]$
$5 \times 10^{-13} = (10^{-2} \ M) \times [Br^{-}]$
$[Br^{-}] = \frac{5 \times 10^{-13}}{10^{-2}} = 5 \times 10^{-11} \ M$.

(A) . $A$ dilute aqueous solution of $NH_3$ exists as $NH_4OH$,which dissociates as: $NH_4OH \rightleftharpoons NH_4^+ + OH^-$.
On adding solid ammonium chloride $(NH_4Cl)$,it dissociates completely: $NH_4Cl \rightarrow NH_4^+ + Cl^-$.
Due to the common ion effect of $NH_4^+$,the equilibrium shifts to the left,decreasing the concentration of $OH^-$.
Since $pOH = -\log[OH^-]$,a decrease in $[OH^-]$ increases $pOH$,which in turn decreases the $pH$ $(pH = 14 - pOH)$.

(C) The reaction is: $AgNO_3 + KI \rightarrow AgI(s) + KNO_3$.
Initial moles of $AgNO_3 = 25 \ mL \times 1 \ M = 25 \ mmol$.
Initial moles of $KI = 25 \ mL \times 1.05 \ M = 26.25 \ mmol$.
Since $AgNO_3$ is the limiting reagent,all $Ag^{+}$ ions react with $I^{-}$ to form $AgI$ precipitate.
After the reaction,$25 \ mmol$ of $AgI$ is formed,$25 \ mmol$ of $K^{+}$ and $NO_3^{-}$ remain as spectator ions,and $1.25 \ mmol$ of $I^{-}$ remains in excess.
$AgI$ is a sparingly soluble salt,so it exists in equilibrium: $AgI(s) \rightleftharpoons Ag^{+}(aq) + I^{-}(aq)$.
Due to the common ion effect from the excess $I^{-}$,the concentration of $Ag^{+}$ becomes extremely small.
Since $I^{-}$ is present in excess $(1.25 \ mmol)$,the ion present in the smallest quantity is $Ag^{+}$.

(B) The dissociation of ammonium hydroxide is given by: $NH_{4}OH \rightleftharpoons NH_{4}^{+} + OH^{-}$.
Ammonium chloride is a strong electrolyte and dissociates completely: $NH_{4}Cl \rightarrow NH_{4}^{+} + Cl^{-}$.
Due to the common ion effect of $NH_{4}^{+}$,the equilibrium of $NH_{4}OH$ shifts to the left.
This decreases the concentration of $OH^{-}$ ions to such an extent that only the hydroxides of group-$III$ cations (like $Fe^{3+}$,$Al^{3+}$,$Cr^{3+}$) are precipitated,as they have very low $K_{sp}$ values (in the range of $10^{-18}$ to $10^{-38}$),while other group cations remain in solution.

(C) Statement $I$ is incorrect. $BaCl_2$ is highly soluble in water and does not precipitate upon the addition of $HCl$ gas at room temperature because the common ion effect is not sufficient to exceed the solubility product $(K_{sp})$ of $BaCl_2$.
Statement $II$ is correct. When $HCl_{(g)}$ is passed through a saturated solution of $NaCl$,the concentration of $Cl^-$ ions increases significantly. According to the common ion effect,the ionic product $[Na^+][Cl^-]$ exceeds the solubility product $(K_{sp})$ of $NaCl$,leading to the precipitation of $NaCl$.

(B) The solubility $s$ of $AgBr$ is $\sqrt{K_{sp}} = \sqrt{12 \times 10^{-14}} \approx 3.46 \times 10^{-7} \ M$.
Upon adding $10^{-7} \ mol$ of $AgNO_3$ to $1 \ L$,the common ion effect suppresses the solubility of $AgBr$. Let $s'$ be the new solubility of $AgBr$.
$[Ag^{+}] = s' + 10^{-7}$,$[Br^{-}] = s'$.
$K_{sp} = (s' + 10^{-7})s' = 12 \times 10^{-14}$.
Solving the quadratic $s'^2 + 10^{-7}s' - 12 \times 10^{-14} = 0$,we get $s' \approx 2.7 \times 10^{-7} \ M$.
Converting to $mol \ m^{-3}$ $(1 \ M = 10^3 \ mol \ m^{-3})$: $[Br^{-}] = 2.7 \times 10^{-4} \ mol \ m^{-3}$,$[Ag^{+}] = 3.7 \times 10^{-4} \ mol \ m^{-3}$,$[NO_3^-] = 10^{-4} \ mol \ m^{-3}$.
Conductivity $\kappa = \sum \lambda_i C_i = (8 \times 10^{-3} \times 2.7 \times 10^{-4}) + (6 \times 10^{-3} \times 3.7 \times 10^{-4}) + (7 \times 10^{-3} \times 10^{-4}) = 21.6 \times 10^{-7} + 22.2 \times 10^{-7} + 7 \times 10^{-7} = 50.8 \times 10^{-7} \ S \ m^{-1}$.
Rounding to the nearest integer provided in options,the answer is $55$.

(B) The solubility of $BaSO_4$ in pure water is $S = 10^{-5} \ M$.
$K_{sp} = S^2 = (10^{-5})^2 = 10^{-10}$.
In the presence of $0.1 \ M \ K_2SO_4$,the common ion effect occurs due to $SO_4^{2-}$ ions.
Let the new solubility be $s'$.
$BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)$
$[Ba^{2+}] = s'$,$[SO_4^{2-}] = 0.1 + s' \approx 0.1 \ M$.
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = s' \times 0.1 = 10^{-10}$.
$s' = 10^{-9} \ mol \ L^{-1}$.
To convert to $g \ L^{-1}$,multiply by molar mass $(233 \ g \ mol^{-1})$:
$s' = 10^{-9} \times 233 = 2.33 \times 10^{-7} \ g \ L^{-1}$.

(B) Assertion $(A)$ is correct because adding a strong acid increases $[H^+]$ and adding a strong base increases $[OH^-]$,which changes the $pH$ of pure water.
Reason $(R)$ is incorrect because the addition of an acid or base actually suppresses the ionisation of water due to the common ion effect,rather than increasing it.
Therefore,$(A)$ is correct but $(R)$ is not correct.

(A) In qualitative analysis,group $III$ cations (like $Fe^{3+}$,$Al^{3+}$,$Cr^{3+}$) are precipitated as their hydroxides using $NH_4OH$ in the presence of $NH_4Cl$.
$NH_4Cl$ is a strong electrolyte that dissociates completely to provide a high concentration of $NH_4^{+}$ ions.
According to the common ion effect,the presence of excess $NH_4^{+}$ ions suppresses the dissociation of the weak base $NH_4OH$ $(NH_4OH \rightleftharpoons NH_4^{+} + OH^{\ominus})$.
This significantly decreases the concentration of $OH^{\ominus}$ ions,ensuring that only the hydroxides of group $III$ cations precipitate,while preventing the precipitation of group $IV$ cations.

(C) The common ion effect is the suppression of the degree of dissociation of a weak electrolyte in the presence of a strong electrolyte that shares a common ion.
In the mixture $CH_{3}COOH + NaOH$,the reaction $CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$ occurs,which is a neutralization reaction,not an example of the common ion effect.
In the other options,$AgCl$ in $NaCl$,$H_{2}S$ in $HCl$,and $NH_{4}OH$ in $NH_{4}Cl$ all involve the presence of a common ion that suppresses dissociation.

(C) Ammonia solution contains $NH_4OH$ which is a weak base and ionizes as $NH_4OH \rightleftharpoons NH_4^{+} + OH^{-}$.
When $NH_4Cl$ is added,it provides $NH_4^{+}$ ions,which is a common ion.
Due to the common ion effect,the equilibrium shifts to the left,decreasing the concentration of $OH^{-}$ ions.
Since $pH = 14 - pOH$ and $pOH = -\log[OH^{-}]$,a decrease in $[OH^{-}]$ leads to an increase in $pOH$,which consequently results in a decrease in the $pH$ value.

(C) The dissociation of $H_2S$ is given by the equilibrium: $H_2S \rightleftharpoons 2H^+ + S^{2-}$.
In the presence of $HCl$,the concentration of $H^+$ ions increases significantly.
According to the common ion effect,this shift in equilibrium suppresses the dissociation of $H_2S$,resulting in a lower concentration of $S^{2-}$ ions.
Group-$2$ metal sulfides have very low solubility products $(K_{sp})$,so they precipitate even at this low $S^{2-}$ concentration.
Group-$4$ metal sulfides have higher $K_{sp}$ values and require a higher $S^{2-}$ concentration to precipitate,which is not achieved in the presence of $HCl$.

(C) In the analysis of $III$ group basic radicals,$NH_4Cl$ is added to $NH_4OH$ to suppress the dissociation of $NH_4OH$ due to the common ion effect ($NH_4^+$ ions).
This ensures that the concentration of $OH^-$ ions is low enough to precipitate only the $III$ group hydroxides (like $Fe(OH)_3$,$Al(OH)_3$,$Cr(OH)_3$) while preventing the precipitation of hydroxides of subsequent groups.

(B) The solubility of a sparingly soluble salt like $AgCl$ is governed by the common ion effect.
According to the common ion effect,the presence of a common ion $(Cl^-)$ decreases the solubility of the salt.
The concentration of $Cl^-$ ions provided by the electrolytes are:
$0.1 \ M \ BaCl_2$ provides $0.2 \ M \ Cl^-$ ions.
$0.1 \ M \ AlCl_3$ provides $0.3 \ M \ Cl^-$ ions.
$0.1 \ M \ NaCl$ provides $0.1 \ M \ Cl^-$ ions.
Since $AlCl_3$ provides the highest concentration of common $Cl^-$ ions $(0.3 \ M)$,the equilibrium $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$ shifts furthest to the left.
Therefore,the solubility of $AgCl$ is least in $0.1 \ M \ AlCl_3$.

(B) The solubility of $AgCl$ is governed by the common ion effect.
In the presence of $0.1 \ M \ KCl$,the concentration of $Cl^{-}$ ions increases significantly.
According to Le Chatelier's principle,the equilibrium $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$ shifts to the left,decreasing the solubility of $AgCl$.
$KNO_3$ does not provide a common ion,and its effect on solubility is negligible compared to the common ion effect.
Therefore,the solubility of $AgCl$ is minimum in $0.1 \ M \ KCl$.

(B) $KOH \rightarrow K^{+} + OH^{-}$
$Cd(OH)_2 \rightleftharpoons Cd^{2+} + 2OH^{-}$
Solubility product $K_{sp} = [Cd^{2+}][OH^{-}]^2$
Given $[OH^{-}] = 0.1 \ M$ due to common ion effect.
$K_{sp} = [S] \times [0.1]^2$
$2.5 \times 10^{-14} = [S] \times 0.01$
$[S] = \frac{2.5 \times 10^{-14}}{10^{-2}} = 2.5 \times 10^{-12}$
$[S] = 25 \times 10^{-13} = x \times 10^{-y}$
Thus,$x = 25$ and $y = 13$.

(B) For $BaSO_4$,the solubility product constant $K_{sp} = s^2 = (4 \times 10^{-5})^2 = 1.6 \times 10^{-9}$.
In $0.1 \ M \ Na_2SO_4$,the concentration of the common ion $[SO_4^{2-}] = 0.1 \ M$.
Let the new solubility be $s'$.
$K_{sp} = [Ba^{2+}][SO_4^{2-}] = s' \times 0.1$.
$1.6 \times 10^{-9} = s' \times 0.1$.
$s' = 1.6 \times 10^{-8} \ M$.