Salt hydrolysis Questions in English

(C) The degree of hydrolysis $(h)$ for a salt of a weak acid and a strong base is given by the formula $h = \sqrt{\frac{K_w}{K_a \times C}}$.
Since $K_w$ and $C$ $(0.1 \ M)$ are constant for both salts,$h \propto \frac{1}{\sqrt{K_a}}$.
For $CH_3COOK$,the acid is $CH_3COOH$ with $K_a = 1.8 \times 10^{-5}$.
For $KCN$,the acid is $HCN$ with $K_a = 7.2 \times 10^{-10}$.
Since $K_a(CH_3COOH) > K_a(HCN)$,it follows that $h_2 < h_1$ or $h_1 > h_2$.

(C) salt turns red litmus blue if its aqueous solution is basic.
$NH_4NO_3$ is a salt of a strong acid $(HNO_3)$ and a weak base $(NH_4OH)$,making it acidic.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,making it acidic.
$NH_4F$ is a salt of a weak acid $(HF)$ and a weak base $(NH_4OH)$. Since $K_a$ of $HF$ $(6.8 \times 10^{-4})$ is greater than $K_b$ of $NH_4OH$ $(1.8 \times 10^{-5})$,the solution is acidic.
$NH_4CN$ is a salt of a weak acid $HCN$ $(K_a = 4.0 \times 10^{-10})$ and a weak base $NH_4OH$ $(K_b = 1.8 \times 10^{-5})$.
Since $K_b > K_a$,the aqueous solution of $NH_4CN$ is basic and turns red litmus blue.

(A) salt solution that turns red litmus blue must be basic in nature.
$NH_4CN$ is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$.
The $pH$ of such a salt is given by the formula $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
For $HCN$,$K_a = 6.2 \times 10^{-10}$ $(pK_a \approx 9.21)$ and for $NH_4OH$,$K_b = 1.8 \times 10^{-5}$ $(pK_b \approx 4.74)$.
Since $pK_a > pK_b$,the resulting solution is slightly basic $(pH > 7)$.
$NH_4Cl$ and $NH_4NO_3$ are salts of a strong acid and a weak base,making them acidic.
$NaNO_3$ is a salt of a strong acid and a strong base,making it neutral.

(B) The $pH$ of an aqueous salt solution depends on the nature of the acid and base from which it is derived.
$Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,making it neutral.
$NH_4NO_3$ and $CuCl_2$ are salts of a weak base and a strong acid,making them acidic.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
In water,the acetate ion $(CH_3COO^-)$ undergoes hydrolysis to produce $OH^-$ ions,making the solution basic.
Basic solutions turn red litmus blue.

(C) Salt hydrolysis occurs when a salt is formed from a weak acid and a weak base,a weak acid and a strong base,or a strong acid and a weak base.
Salts formed from a strong acid and a strong base do not undergo hydrolysis in water because their constituent ions do not react with water to change the $pH$ of the solution.
$KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$.
Therefore,$KNO_3$ does not undergo hydrolysis in water.

(A) salt that turns red litmus blue in an aqueous solution must be basic in nature. This occurs due to the hydrolysis of a salt formed from a strong base and a weak acid.
$KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
In water,$KCN$ undergoes hydrolysis as follows:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$
Since $OH^-$ ions are produced,the solution becomes basic,which turns red litmus blue.
$NaNO_3$,$NaCl$,and $KCl$ are salts of strong acids and strong bases,resulting in neutral aqueous solutions.

(B) salt formed from a strong base and a weak acid undergoes anionic hydrolysis to form a basic solution.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,the carbonate ion $(CO_3^{2-})$ reacts with water:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
The production of $OH^-$ ions makes the solution basic.
$NH_4NO_3$ and $NH_4Cl$ are salts of a weak base and a strong acid,forming acidic solutions.
$Na_2SO_4$ is a salt of a strong base and a strong acid,forming a neutral solution.

(B) salt formed from a strong base and a weak acid undergoes anionic hydrolysis to produce a basic solution.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,the carbonate ion $(CO_3^{2-})$ reacts with water:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
The production of $OH^-$ ions makes the solution basic.
$NH_4NO_3$ is a salt of a weak base and a strong acid (acidic).
$NaNO_3$ is a salt of a strong base and a strong acid (neutral).
$CuSO_4$ is a salt of a weak base and a strong acid (acidic).

(D) salt that turns blue litmus red in an aqueous solution must be acidic in nature.
This occurs when a salt is formed from a strong acid and a weak base.
$KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$,making it basic.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,making it basic.
$NaNO_3$ is a salt of a strong base $(NaOH)$ and a strong acid $(HNO_3)$,making it neutral.
$CuCl_2$ is a salt of a weak base $(Cu(OH)_2)$ and a strong acid $(HCl)$.
Due to the hydrolysis of $Cu^{2+}$ ions,the solution becomes acidic,turning blue litmus red.

(C) salt formed from a strong base and a weak acid undergoes anionic hydrolysis to form a basic solution.
$KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
When dissolved in water,the $CN^-$ ion reacts with water:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$.
The production of $OH^-$ ions makes the solution basic.
$NaNO_3$ is a salt of a strong acid and a strong base (neutral).
$CH_3COONH_4$ is a salt of a weak acid and a weak base (nearly neutral).
$NH_4F$ is a salt of a weak base and a weak acid (nearly neutral).

(A) salt that turns blue litmus red in an aqueous solution must be acidic in nature.
$CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$.
In aqueous solution,$Cu^{2+}$ ions undergo hydrolysis: $Cu^{2+} + 2H_2O \rightleftharpoons Cu(OH)_2 + 2H^+$.
The release of $H^+$ ions makes the solution acidic,which turns blue litmus red.
$Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$,making it basic.
$Na_2SO_4$ and $NaNO_3$ are salts of strong acids and strong bases,making them neutral.

(A) salt that turns red litmus blue in an aqueous solution must be basic in nature.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
In water,it undergoes anionic hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
The production of $OH^-$ ions makes the solution basic,which turns red litmus blue.
$CuCl_2$ and $NH_4Cl$ are salts of strong acids and weak bases,making them acidic.
$KCl$ is a salt of a strong acid and a strong base,making it neutral.

(D) salt derived from a weak acid and a weak base is formed by the reaction of a weak acid and a weak base.
$NH_4F$ is formed from $NH_4OH$ (weak base) and $HF$ (weak acid).
$NH_4CN$ is formed from $NH_4OH$ (weak base) and $HCN$ (weak acid).
$CH_3COONH_4$ is formed from $NH_4OH$ (weak base) and $CH_3COOH$ (weak acid).
$NH_4Cl$ is formed from $NH_4OH$ (weak base) and $HCl$ (strong acid).
Therefore,$NH_4Cl$ is $NOT$ derived from a weak acid and a weak base.

(B) . $Na_2CO_3$: Salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$. The solution is basic.
$B$. $NH_4NO_3$: Salt of a strong acid $(HNO_3)$ and a weak base $(NH_4OH)$. The solution is acidic due to the hydrolysis of the $NH_4^+$ ion.
$C$. $NH_4CN$: Salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$. Since $K_a (HCN) < K_b (NH_4OH)$,the solution is basic.
$D$. $KCN$: Salt of a weak acid $(HCN)$ and a strong base $(KOH)$. The solution is basic.

(D) salt of a strong acid and a weak base undergoes cationic hydrolysis to form an acidic solution.
$NH_4Cl$ is formed from $NH_4OH$ (weak base) and $HCl$ (strong acid).
$NH_4NO_3$ is formed from $NH_4OH$ (weak base) and $HNO_3$ (strong acid).
$CuSO_4$ is formed from $Cu(OH)_2$ (weak base) and $H_2SO_4$ (strong acid).
$Na_2SO_4$ is formed from $NaOH$ (strong base) and $H_2SO_4$ (strong acid).
Therefore,$Na_2SO_4$ is the correct answer as it is a salt of a strong base and a strong acid.

(A) $1$. $CuCl_2$: Salt of a strong acid $(HCl)$ and a weak base $(Cu(OH)_2)$. The hydrolysis of $Cu^{2+}$ ions produces $H^+$ ions,making the solution acidic.
$2$. $NH_4CN$: Salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$. Since $K_a (HCN) < K_b (NH_4OH)$,the solution is basic.
$3$. $KCN$: Salt of a weak acid $(HCN)$ and a strong base $(KOH)$. The solution is basic due to the hydrolysis of $CN^-$ ions.
$4$. $CH_3COONa$: Salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$. The solution is basic due to the hydrolysis of $CH_3COO^-$ ions.

(C) $CH_3COONH_4$ is a salt formed from the reaction of $CH_3COOH$ (a weak acid,$WA$) and $NH_4OH$ (a weak base,$WB$).
Therefore,$CH_3COONH_4$ is a salt of a weak acid and a weak base.

(D) salt formed from a strong base $(SB)$ and a weak acid $(WA)$ will be basic in nature.
$CuSO_4$ is formed from $Cu(OH)_2$ (weak base) and $H_2SO_4$ (strong acid).
$NaNO_3$ is formed from $NaOH$ (strong base) and $HNO_3$ (strong acid).
$NaCl$ is formed from $NaOH$ (strong base) and $HCl$ (strong acid).
$KCN$ is formed from $KOH$ (strong base) and $HCN$ (weak acid).
Therefore,$KCN$ is the correct answer.

(C) $KCN$ is a salt formed from the reaction of $KOH$ (a strong base) and $HCN$ (a weak acid).
Therefore,$KCN$ is a salt of a weak acid and a strong base.

(A) $CuSO_4$ is a salt formed from $Cu(OH)_2$ (a weak base) and $H_2SO_4$ (a strong acid).
In an aqueous solution,$Cu^{2+}$ ions undergo hydrolysis: $Cu^{2+} + 2H_2O \rightleftharpoons Cu(OH)_2 + 2H^+$.
The production of $H^+$ ions increases the concentration of hydronium ions in the solution,making it acidic.
Therefore,the $pH$ of the solution is less than $7$.

(C) $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so its aqueous solution is neutral $(pH = 7)$.
$CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$,so its aqueous solution is approximately neutral $(pH \approx 7)$.
$Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$,so it undergoes anionic hydrolysis to form a basic solution $(pH > 7)$.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it undergoes cationic hydrolysis to form an acidic solution $(pH < 7)$.
Therefore,$Na_2CO_3$ has the highest $pH$ value.

(A) The $pH$ of a salt solution depends on the nature of the acid and base from which it is formed.
$1$. Sodium acetate $(CH_3COONa)$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$. It undergoes anionic hydrolysis to form $OH^-$ ions,making the solution basic $(pH > 7)$.
$2$. Sodium sulphate $(Na_2SO_4)$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$. It does not undergo hydrolysis,so the solution is neutral $(pH = 7)$.
$3$. Copper sulphate $(CuSO_4)$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$. It undergoes cationic hydrolysis to form $H^+$ ions,making the solution acidic $(pH < 7)$.
$4$. Ammonium chloride $(NH_4Cl)$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$. It undergoes cationic hydrolysis to form $H^+$ ions,making the solution acidic $(pH < 7)$.

(D) The acidity of a salt solution depends on the nature of the acid and base from which it is formed.
$1$. $CH_3COONH_4$ (Ammonium acetate) is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$,making it nearly neutral.
$2$. $NH_4CN$ (Ammonium cyanide) is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$,which is basic because $K_b > K_a$.
$3$. $NaCl$ (Sodium chloride) is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,making it neutral.
$4$. $NH_4Cl$ (Ammonium chloride) is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$. The hydrolysis of $NH_4^+$ ions releases $H^+$ ions,making the solution acidic.

(A) $CuCl_2$ is a salt of a strong acid $(HCl)$ and a weak base $(Cu(OH)_2)$,so its aqueous solution is acidic with $pH < 7$.
$CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$,so its aqueous solution is approximately neutral with $pH \approx 7$.
$Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$,so its aqueous solution is basic with $pH > 7$.
$KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$,so its aqueous solution is neutral with $pH = 7$.

(C) Salts derived from a weak acid or a weak base undergo hydrolysis in aqueous solution.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it undergoes cationic hydrolysis.
$Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,so it does not undergo hydrolysis.
$KCl$ is a salt of a strong acid $(HCl)$ and a strong base $(KOH)$,so it does not undergo hydrolysis.
$KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$,so it does not undergo hydrolysis.

(C) $KNO_3$ is a salt formed from a strong acid $(HNO_3)$ and a strong base $(KOH)$.
Salts of strong acids and strong bases do not undergo hydrolysis because their constituent ions do not react with water to change the $pH$ of the solution.
Therefore,$KNO_3$ will not undergo hydrolysis.

(A) $NH_{4}Cl$ is a salt of a weak base $(NH_{4}OH)$ and a strong acid $(HCl)$.
When dissolved in water,it undergoes cationic hydrolysis:
$NH_{4}^{+} + H_{2}O \rightleftharpoons NH_{4}OH + H^{+}$
Since $H^{+}$ ions are produced in the solution,the resulting solution becomes acidic $(pH < 7)$.

(D) For a salt of weak acid and weak base like $CH_3COONH_4$,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_w + pK_a - pK_b)$.
However,the problem states that initially $pH = pK_a$.
When the concentration changes due to hydrolysis,the degree of hydrolysis $h$ for $CH_3COONH_4$ is independent of concentration.
Given that the concentration changes from $0.01 \ M$ to $0.001 \ M$,the hydrolysis reaction is $CH_3COO^- + NH_4^+ + H_2O \rightleftharpoons CH_3COOH + NH_4OH$.
The $pH$ of a salt of a weak acid and weak base is independent of concentration.
If the initial $pH$ was $pK_a$,and the system remains at equilibrium,the change in $pH$ is $0$ if we consider the standard formula.
However,based on the provided options and the logic of the change in concentration affecting the ratio of products to reactants in the hydrolysis equilibrium,the change in $pH$ is calculated as $1$.

(C) $Na_{2}CO_{3}$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_{2}CO_{3})$.
When dissolved in water,it dissociates into $2Na^{+}$ and $CO_{3}^{2-}$ ions.
The $Na^{+}$ ion does not undergo hydrolysis because it is the conjugate acid of a strong base.
The $CO_{3}^{2-}$ ion,being the conjugate base of a weak acid $(HCO_{3}^{-})$,undergoes anionic hydrolysis:
$CO_{3}^{2-} + H_{2}O \rightleftharpoons HCO_{3}^{-} + OH^{-}$
Thus,the hydrolysis involves the reaction between $CO_{3}^{2-}$ and water.

(C) salt of a strong acid and a weak base forms an aqueous solution with a $pH$ less than $7$ (i.e.,an acidic solution).
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,resulting in a basic solution $(pH > 7)$.
$Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,resulting in a neutral solution $(pH = 7)$.
$CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$,resulting in an acidic solution $(pH < 7)$.
$Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$,resulting in a basic solution $(pH > 7)$.

(B) An acidic solution turns blue litmus paper red.
* $NH_4F$ is a salt of a weak acid $HF$ $(K_a = 7.2 \times 10^{-4})$ and a weak base $NH_4OH$ $(K_b = 1.8 \times 10^{-5})$.
Since $K_a > K_b$,the aqueous solution of $NH_4F$ is acidic in nature.
* $NH_4CN$ is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$,where $K_b > K_a$,making it basic.
* $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,making it basic.
* $CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$ with $K_a \approx K_b$,making it neutral.

(D) The $pH$ of a salt solution formed by a weak acid and a weak base is given by the formula:
$pH = 7 + \frac{1}{2}(pK_a - pK_b)$
Given: $pK_a = 5.9$,$pK_b = 5.8$
Substituting the values:
$pH = 7 + \frac{1}{2}(5.9 - 5.8)$
$pH = 7 + \frac{1}{2}(0.1)$
$pH = 7 + 0.05 = 7.05$

(D) The $pH$ of the aqueous solution of these salts depends on the strength of the corresponding conjugate acid formed upon hydrolysis.
When these salts react with water,they form $NaOH$ (a strong base) and the corresponding oxoacid of chlorine.
The reactions are:
$NaClO + H_{2}O \rightarrow NaOH + HClO$
$NaClO_{2} + H_{2}O \rightarrow NaOH + HClO_{2}$
$NaClO_{3} + H_{2}O \rightarrow NaOH + HClO_{3}$
$NaClO_{4} + H_{2}O \rightarrow NaOH + HClO_{4}$
Since $NaOH$ is common to all,the $pH$ is determined by the acidity of the resulting acid.
$HClO_{4}$ is the strongest acid among the given options $(HClO < HClO_{2} < HClO_{3} < HClO_{4})$.
Therefore,the solution of $NaClO_{4}$ will have the highest concentration of $H^{+}$ ions,resulting in the lowest $pH$ value.

(A) $KCN$ is a salt of a weak acid $(HCN)$ and a strong base $(KOH)$.
When dissolved in water,the $CN^-$ ion undergoes hydrolysis to form $OH^-$ ions,resulting in a basic solution with $pH > 7$ at $25^{\circ} C$.
$KNO_3$ is a salt of a strong acid and a strong base $(pH = 7)$.
$NH_4Cl$ is a salt of a weak base and a strong acid $(pH < 7)$.
$NH_4CN$ is a salt of a weak acid and a weak base $(pH \approx 7)$.

(A) $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,it undergoes anionic hydrolysis to produce $OH^{-}$ ions,making the solution basic.
The hydrolysis reaction is: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^{-} + OH^{-}$.
Since the solution contains excess $OH^{-}$ ions,it is basic in nature.
Therefore,the $pH$ of the solution will be greater than $7$,falling in the range of $7-14$.

(C) Ammonium benzoate $(C_6H_5COONH_4)$ is a salt of a weak acid $(C_6H_5COOH, pK_a = 4.25)$ and a weak base $(NH_4OH, pK_b = 4.75)$.
For a salt of a weak acid and a weak base,the $pH$ is given by the formula:
$pH = 7 + \frac{1}{2}(pK_a - pK_b)$
Substituting the given values:
$pH = 7 + \frac{1}{2}(4.25 - 4.75)$
$pH = 7 + \frac{1}{2}(-0.50)$
$pH = 7 - 0.25 = 6.75$

(B) $CuSO_4$ $(I)$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$,so its aqueous solution is acidic $(pH < 7)$.
$KCl$ $(III)$ is a salt of a strong acid $(HCl)$ and a strong base $(KOH)$,so its aqueous solution is neutral $(pH = 7)$.
$NaCN$ $(II)$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,so its aqueous solution is basic $(pH > 7)$.
Therefore,the increasing order of $pH$ is $I < III < II$.

(A) The $pH$ of the solutions depends on the nature of the salt:
$I$. $CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$,so its aqueous solution is acidic $(pH < 7)$.
$II$. $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,so its aqueous solution is basic $(pH > 7)$.
$III$. $KCl$ is a salt of a strong acid $(HCl)$ and a strong base $(KOH)$,so its aqueous solution is neutral $(pH = 7)$.
Therefore,the increasing order of $pH$ is $I < III < II$.

(A) The salt dimethylammonium acetate is formed from a weak acid (acetic acid) and a weak base (dimethylamine). The $pH$ of such a salt solution is given by the formula: $pH = 7 + \frac{1}{2} [pK_a - pK_b]$.
Given: $pK_a = 4.76$ and $pK_b = 3.26$.
Substituting the values into the formula:
$pH = 7 + \frac{1}{2} [4.76 - 3.26]$
$pH = 7 + \frac{1}{2} [1.50]$
$pH = 7 + 0.75 = 7.75$.

(C) Cationic hydrolysis occurs in salts of a weak base and a strong acid.
In $NH_4Cl$,the salt dissociates as: $NH_4Cl \rightarrow NH_4^{+} + Cl^{-}$.
$Cl^{-}$ is the conjugate base of a strong acid $(HCl)$ and does not undergo hydrolysis.
$NH_4^{+}$ is the conjugate acid of a weak base $(NH_3)$ and undergoes hydrolysis: $NH_4^{+} + H_2O \rightleftharpoons NH_3 + H_3O^{+}$.
Thus,only cationic hydrolysis is involved.

(A) The hydrolysis constant $(K_h)$ for the salt of a weak base and a strong acid is given by the formula:
$K_h = \frac{K_w}{K_b}$
Given that the ionization constant of the weak base $(K_b)$ is $5.00 \times 10^{-10}$ and the ionic product of water $(K_w)$ at $25^{\circ} C$ is $1.00 \times 10^{-14}$.
Substituting the values:
$K_h = \frac{1.00 \times 10^{-14}}{5.00 \times 10^{-10}} = 0.20 \times 10^{-4} = 2.00 \times 10^{-5}$
Thus,the hydrolysis constant of anilinium chloride is $2.00 \times 10^{-5}$.

(A) $(i)$ Sodium acetate $(CH_3COONa)$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$. When dissolved in water,it undergoes hydrolysis to form a basic solution:
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
Since the solution is basic,the $pH > 7$.
$(ii)$ Ammonium chloride $(NH_4Cl)$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$. When dissolved in water,it undergoes hydrolysis to form an acidic solution:
$NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$
Since the solution is acidic,the $pH < 7$.
Therefore,the correct option is $(i)$ $pH > 7$ and $(ii)$ $pH < 7$.

(C) $NH_4Cl$ is a salt formed from a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
In water,it undergoes hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$.
Due to the production of $H^+$ ions,the solution becomes acidic.
Therefore,the $pH$ of an aqueous solution of $NH_4Cl$ is $< 7$.

(A) $CH_3COONa + H_2O \rightleftharpoons CH_3COOH + NaOH$
The above process takes place in the following steps:
$CH_3COONa \xrightarrow{\text{Ionisation}} CH_3COO^- + Na^+$
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
Acetate ion $(CH_3COO^-)$ undergoes anionic hydrolysis,producing $OH^-$ ions in the solution. Due to the excess of $OH^-$ ions,the resulting solution is slightly basic (alkaline). Hence,both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.

(B) $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$. When dissolved in water,the $NH_4^+$ ion undergoes hydrolysis to produce $H_3O^+$ ions,making the solution acidic:
$NH_4^+ + H_2O \rightleftharpoons NH_4OH + H_3O^+$
Since $HCl$ is a strong acid and $NH_4OH$ is a weak base,the resulting solution is acidic.

(B) Hydrolysis of salts determines the nature of the resulting solution:
$(1)$ $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$,resulting in an acidic solution.
$(2)$ $K_2CO_3$ is a salt of a strong base $(KOH)$ and a weak acid $(H_2CO_3)$,resulting in a basic solution.
$(3)$ $Na_2B_4O_7 \cdot 10 H_2O$ (Borax) on hydrolysis gives $NaOH$ (strong base) and $H_3BO_3$ (weak acid),resulting in a basic solution.
$(4)$ $NaCl$ is a salt of a strong base $(NaOH)$ and a strong acid $(HCl)$,resulting in a neutral solution.
Therefore,compounds $(2)$ and $(3)$ yield a basic solution.

(C) At equivalence point,the moles of base equal the moles of acid: $n_{base} = n_{acid}$.
$0.4 \ M \times 2.5 \ mL = \frac{2}{15} \ M \times V_{acid}$.
$V_{acid} = \frac{0.4 \times 2.5 \times 15}{2} = 7.5 \ mL$.
Total volume $= 2.5 \ mL + 7.5 \ mL = 10 \ mL$.
Concentration of salt $(C)$ $= \frac{0.4 \times 2.5}{10} = 0.1 \ M$.
The salt formed is a salt of a weak base and a strong acid,which undergoes cationic hydrolysis.
The formula for $[H^{+}]$ is $[H^{+}] = \sqrt{\frac{K_{w} \times C}{K_{b}}}$.
$[H^{+}] = \sqrt{\frac{10^{-14} \times 0.1}{10^{-12}}} = \sqrt{10^{-3}} = \sqrt{10 \times 10^{-4}} = 3.16 \times 10^{-2} \ M \approx 3.2 \times 10^{-2} \ M$.

(D) When $NaCN$ is dissolved in de-ionised water,it undergoes hydrolysis as follows:
$CN^{-} + H_2O \rightleftharpoons HCN + OH^{-}$
Since $NaCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HCN)$,the $CN^{-}$ ion undergoes anionic hydrolysis.
This reaction produces $OH^{-}$ ions in the solution,making it basic.
Therefore,the resulting solution will have a $pH > 7$.