The addition of $HCl$ will not suppress the ionisation of $H _2SO _4$ because $H _2SO _4$ is stronger acid than $HCl$.

$H _2O⇌H^{ + }   +       OH^{ - }\quad $

When $NH _4Cl$ is added to $NH _4OH$ solution, concentration of $NH _4^{+}$ ions increases so the equilibrium shift towards left.So the dissociation of ammonium hydroxide is reduced. 

Relative strength of two acids can be compared by their degree of dissociation.

Since, $NaCl$ is soluble to a very significant extent, when $AgCl$ is added to $NaCl$ solution, the common ion $[Cl^-]$ increases in the solution. To have the solubility product or $K _{sp}$ of $AgCl$ constant, $[Ag^+]$ will decrease or $AgCl$ will percipitate out from the solution. This is common ion effect. Hence solubility of $AgCl$ in $NaCl$ solution will be less than that in pure water.

$100ml$ of $20.8$% $BaCl _2$ solution= $20.8g$  $BaCl _2$

NaCl is highly soluble and when it is added to AgCl it decreases the solubility of AgCl because of common ion $Cl^-$ and solution become super saturated.

Relative strength of two bases are measured by their degree of dissociation.

In common ion effect there should be common ions in the two or more chemicals. Here, $HCl+NaCl$ shows common ion effect. Because $Cl^-$ is common in both.

Solution:- (A) $Al{Cl} _{3}$

At higher temperature the value of $kw$ increases.This is in according with le-chatelier principle.

Due to common ion effect, if a salt of weak acid or base is added to a solution of its acid or base respectively, the dissociation of acid or base is diminished.

Common ion effect is normally applied in selective precipitation of ions which involves a technique of separating ions in an aqueous solution by using a reagent that precipitates one or more of the ions. addition of common ions in a solution that is already having that given ion normally leads to a formation of a precipitate. The common ion normally decreases the solubility of a slightly insoluble salt.

Its not (1) because they are both solids.

The remaining are solute solvent pairs. So all the other pairs will show common ion effect if the solute is soluble to some extent and exists in an equilibrium with the solvent. The answer given is (3).

From the law of Mass action, the dissociation of $NH _4OH$ takes place and we have,
$\dfrac{[NH _4^+][OH^-]}{[NH _4OH]} =K$
Ammonium chloride, a strong electrolyte, ionises almost completely as follows:
$NH _4Cl \leftrightarrow NH _4^+ + Cl^-$
So, if excess of $NH _4Cl$ is added before adding $NH _4OH$, the concentration of $NH _4^+$ ions is increased and consequently the concentration of $OH^-$ ions is decreased.

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

$(A)$ is wrong because the addition of $NaCN$ to $HCN$, due to common ion $(CN^{\circleddash})$, the degree of dissociation of $HCN$ is suppressed and hence less $[H^{\oplus}]$ and increase in $pH.$

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Due to common ion effect, the solubility of $HgI _2$ is expected to be less in an aqueous solution of KI than in water as
$HgI _2 + KI \longrightarrow K _2[HgI _4]$.
since $I^{\circleddash}$ ion is large sized and therefore is highly polarisable.
But (R) is not the correct explanation of (A)

$AB \rightarrow A^+ + B^-$
$BC \rightarrow B^+ + C^-$
Since $B^+$ is incapable of forming a complex salt it tends to decrease the solubility by Le-Chatelier's principle.

Fuorine is the most electronegative halogen element, but is the least stable, and the most basic. It turns out that when moving vertically in the periodic table, the size of the atom trump its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodine ion, the negative charge is spread out over a significantly  larger volume therefore, effective nuclear charge decreases. 

${NH} _{4}Cl\longrightarrow {NH} _{4}^{+}+{Cl}^{-}$
${NH} _{4}OH\longrightarrow {NH} _{4}^{+}+{OH}^{-}$
${K} _{b}=\cfrac { \left[ { NH } _{ 4 }^{ + } \right] \left[ OH \right]  }{ \left[ { NH } _{ 4 }OH \right]  } $
$[{NH} _{4}^{+}]=$ is due to salt because ${NH} _{4}OH$ ionise less amount due to common ions effect
$1.8\times {10}^{-5}=\cfrac{0.1\times [{OH}^{-}]}{0.05}$ 
$9\times {10}^{-6}=[{OH}^{-}]$

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Any acid weaker than $HCl$ will be suppressed by $HCl$. Hence, among the given options, only sulphuric acid is an acid with comparable acidity strength to $HCl$. The same can also be verified using $K _a$ values from the data.

$AgCl \rightleftharpoons Ag^+ + Cl^-$
$NaCl \rightarrow Na^+ + Cl^-$
Sodium chloride is a strong electrolyte and is completely dissociated. This increases the chloride ion concentration in solution and suppresses the ionization of $AgCl$. Hence, the solubility of $AgCl$ decreases.

$AgCl \rightleftharpoons Ag^+ + Cl^-$
$KCl \rightarrow K^+ + Cl^-$

$H _3PO _4$$=$$H^+$$+$$H _2PO _4^-$

The solubility of silver chloride decreases in the presence of sodium chloride because of common ion effect.

HCl undergoes complete ionization as it is a strong acid.
On the other hand, acetic acid  $HC _2H _3O _2$ is a weak acid and is ionized to a small extent.

The net effect of the common ion is that it reduces the solubility of the solute in the solution.
The common ion effect is observed when the addition of an ion common to two solutes causes precipitation or reduces ionization.
Thus, if to a solution of weak electrolyte, a solution of strong electrolyte is added which furnishes an ion common to that furnished by the weak electrolyte, the ionization of the weak electrolyte is suppressed.
For example, ammonium hydroxide is a weak electrolyte and ionizes to a small extent to give ammonium ions and hydroxide ions.
$\displaystyle  NH _4OH \rightleftharpoons NH _4^+  +  OH^-$
A strong electrolyte NaOH is added which furnishes hydroxide ions that are common ions. This suppresses the equilibrium for the dissociation of ammonium hydroxide.
$\displaystyle  NaOH \rightarrow Na^+  + OH^-$

When $H _2SO _4$ is added to a $CaSO _4$ solution, due to common ion effect, the solubility of $CaSO _4$ decreases. Further, as the acid concentration increases, the $pH$ of the solution lowers due to addition of $H _2SO _4$.

The common ion effect is observed when the addition of an ion common to two solutes causes precipitation or reduces ionization.
Thus, if to a solution of a weak electrolyte, a solution of strong electrolyte is added which furnishes an ion common to that furnished by the weak electrolyte, the ionization of the weak electrolyte is suppressed.

According to Ostwald's dilution law

When acetic acid is dissolved in water, it partially dissociates (2%).

Those ions will show common ion effect in which one or both them does not dissociate completely.

Common ion effect is observed when a solution of weak electrolyte is mixed with a solution of strong electrolyte, which provides an ion common to that provided by weak electrolyte.

Purification of $NaCl$ by the passage of $HCl$ through brine is based on the common ion effect.
$HCl$ is a strong electrolyte and provides an ion $(Cl^-)$ that is common to that provided by the weak electrolyte. Thus, the ionization of weak electrolytes is suppressed.

$Ksp(Mg(OH) _2)=9.0\times 10^{-12}$

8 gram (0.2 mole) of NaOH (molecular weight 40 g/mol) completely neutralizes 9.8 gram (0.1 mole) of $H _2SO _4$ (molecular weight 98 g/mol).
Since the molar concentration of both the compound are approximately same, the resulting solution will be neutral. Its pH will be 7.

The chloride ion concentration in 0.01M HCl will be 0.01 M.
The chloride ion concentration due to dissociation of AgCl is neglected due to very low value of solubility product of AgCl.
The expression for the solubility product is as shown below.
$K _{sp}=[Ag^+][Cl^-]$
Substitute values in the above expression.
$2.8 \times 10^{-10}=[Ag^+] \times 0.01$
Hence, $[Ag^+]= \frac {2.8 \times 10^{-10}} {0.01}=2.8 \times 10^{-8}$mol/L.

total pressure P total = $16.8 \, bar$

$K _a$ is a measure of the strength of the acid i.e., larger the value of $K _a$, the stronger is the acid.
Thus, the correct order of acidic strength is
$HCOOH > CH _3COOH > HClO$
Stronger the acid, lesser will be the value of pH. Hence, the correct order of pH is $HClO > CH _3COOH > HCOOH$.

The solubility of $AgCl$ or its ion formation will depend inversely on the concentration

$Ag _3 PO _4$ is a weak electrolyte and $AgNO _3$ is a strong electrolyte containing common ion $(NO _3^-ion)$. Thus common ion effect is observed and the solubility of $Ag _3 PO _4$ is suppressed.
Hence, $Ag _3 PO _4$ is least soluble in 0.1M $AgNO _3$.

The common ion presence in the solution decrease the solubility of a given sparingly soluble compound. So, solubility of AgI in NaI solution is less.

$C{H} _{3}COOH\rightleftharpoons C{H} _{3}CO{O}^{-} + {H}^{+}$

Suppose solubility of $AgCNS$ and $AgBr$ in a solution are $a$ and $b$ respectively.

Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Assertion: AgCl will not dissolve in a concentrated NaCl solution.
Reason: The chloride ions from NaCl suppress the solubility of AgCl.
This is an example of a common ion effect. The chloride ions are common ions.
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

$NaCl$ is a strong electrolyte. It completely dissociates to form ions. Hence, it is highly soluble in water.
$AgCl$ dissociates to little extent. Hence, it is sparingly soluble in water.
When $NaCl$ is added to a solution of $AgCl$, due to common ion effect (chloride ion is the common ion), the dissociation of $AgCl$ is suppressed. Also as the concentration of chloride ion (from $NaCl$) increases, the ionic product of $AgCl$ exceeds its solubility product and precipitation occurs.

$HA,  \alpha =\sqrt{\displaystyle\frac{10^{-5}}{0.1}}=0.01$ 
(A) On adding $HCl, \left [ HA \right ]=\frac{0.1}{2},\left [ HCl \right ]=\displaystyle\frac{10^{-3}}{2}$
$HCl \rightarrow H^{+}+Cl^{-}$
$HA \rightleftharpoons H^{+}  +A^{-}$
$\displaystyle\frac{0.1}{2}\left ( 1- \alpha  \right )\left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) \displaystyle\frac{0.1 \alpha }{2} $
$\Rightarrow 10^{-5}=\displaystyle\frac{\displaystyle\frac{0.1\alpha }{2} \times \left (\displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1 \alpha }{2}  \right ) }{\displaystyle\frac{0.1}{2}\left ( 1-\alpha  \right )}$
Neglecting $\alpha $
$10^{-5}=\alpha \times \left ( \displaystyle\frac{10^{-3}}{2}+\displaystyle\frac{0.1\alpha  }{2} \right )$
$\Rightarrow 0.1 \alpha ^{2}+10^{-3} \alpha -2\times 10^{-5}=0$
$\Rightarrow  \alpha ^{2}+10^{-2}\alpha -2\times 10^{-4}=0$
$\Rightarrow \alpha =\displaystyle\frac{-10^{-2}+\sqrt{10^{-4}+8\times 10^{-4}}}{2}=\displaystyle\frac{2\times 10^{-2}}
{2}=2$
Which is similiar to initial.
(B) In case of two weak acids
$\left [\mathrm H^{+}  \right ]=\sqrt{10^{-5}\times \frac{0.1}{2}+2\times 10^{-6}\times \displaystyle\frac{0.5}{2}}$
$=\sqrt{\displaystyle\frac{10^{-6}}{2}+\displaystyle\frac{10^{-6}}{2}}=10^{-3}$
$\alpha =\displaystyle\frac{10^{-5}}{10^{-3}}=10^{-2}$
which is same as earlier
(C) $HNO _{3}\rightarrow H^{+}+NO _{3}^{-}$
$\displaystyle\frac{0.1}{2}       \displaystyle\frac{0.1M}{2}$
$HA \rightleftharpoons H^{+}   +  A^{-}$
$\displaystyle\frac{0.1M}{2}\left ( 1-\alpha  \right )  \displaystyle\frac{0.1}{2}+\displaystyle\frac{0.1}{2}\alpha                
\displaystyle\frac{0.1}{2}\alpha$
$\Rightarrow 10^{-5}=\displaystyle\frac{\frac{0.1\alpha }{2}\times \displaystyle\frac{0.1 }{2} \left (1+\alpha  \right )  }{\displaystyle\frac{0.1}{2}\left (1-\alpha  \right )}$
Neglecting $\alpha $ due to common ion effect.
$\Rightarrow 10^{-5}=\displaystyle\frac{0.1 \alpha }{2}$
$\Rightarrow \alpha =2\times 10^{-4}$
Hence, it decreases from $0.01  to  2\times 10^{-4}$.

The common ion effect describes the changes that occur with the introduction of ions to a solution containing that same ion.
The role that the common ion effect in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium.
$BaCl _2\rightarrow Ba^{2+} + 2Cl^{-}$
$HCl \rightarrow H^{+} + Cl^{-}$
As $Cl^{-} $ is the common ion so, the $ Ba^{2+}$and $ 2Cl^{-}$ combines ( associate ) to give the undissociated $BaCl _2$. 
Hence , there is increase in concentration of undissociated $BaCl _2$.

$Ag^+ + Cl^- \rightarrow AgCl$
$NaCl \rightarrow Na^+ + Cl^-$
As $Cl^- $ is the common ion in both the solution ( i.e. $NaCl $ and $AgCl$ ) , hence $AgCl$ being weaker electrolyte is precipitated or solubility of the silver chloride would decrease.
Common ion Effect : The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions. Some of the salt will be precipitated until the ion product is equal to the solubility product. In short, the common ion effect is the suppression of the degree of dissociation of a weak electrolyte containing a common ion

The correct answer is $(A)$.

While the NH4OH is weak base it does not ionises completely. Thus due to presence of common ion NH4+ in NH4Cl, it supresses the ionisation of weak base NH4OH in order to decrease the OH-concentration so that higher group cations will not get precipitated. 

$\text{On reacting with NaOH(aq) the Mg+2 gives precipitation reaction.}$