Salt hydrolysis Questions in English

(B) An acidic salt is a salt that contains at least one replaceable hydrogen atom in its structure.
$NaHSO_3$ (Sodium bisulfite) is formed by the partial neutralization of $H_2SO_3$ by $NaOH$.
It contains a replaceable $H^+$ ion,which makes it an acidic salt.

(C) $NaCl$ is a salt formed from a strong acid $(HCl)$ and a strong base $(NaOH)$.
In aqueous solution,$Na^{+}$ acts as an extremely weak conjugate acid and $Cl^{-}$ acts as an extremely weak conjugate base.
Neither $Na^{+}$ reacts with $OH^{-}$ nor $Cl^{-}$ reacts with $H^{+}$ to form undissociated acid or base.
Therefore,$NaCl$ does not undergo hydrolysis.

(C) $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$.
Since it is a salt of a strong base and a weak acid,it undergoes anionic hydrolysis in water to produce $OH^-$ ions,making the solution alkaline.

(A) $MgCl_2$ is a salt of a strong acid $(HCl)$ and a weak base $(Mg(OH)_2)$.
In an aqueous solution,$MgCl_2$ undergoes hydrolysis to produce $HCl$,which is a strong acid.
Therefore,the resulting solution is acidic in nature.
Thus,the $pH$ of the solution will be $< 7$.

(A) Salt hydrolysis occurs for $CH_3COONa$ (salt of a weak acid and strong base) in water to form $CH_3COOH$ and $NaOH$. However,$C_6H_5CH_3$ (toluene) does not undergo hydrolysis under normal conditions. $CH_3COCH_3$ (acetone) also does not undergo hydrolysis. Therefore,only $CH_3COONa$ undergoes hydrolysis.

(A) For the hydrolysis of a salt of a weak acid and a strong base,the hydrolysis constant is given by $K_h = \frac{K_w}{K_a}$.
Given $K_w = 1 \times 10^{-14}$ and $K_a = 1 \times 10^{-5}$,we have $K_h = \frac{10^{-14}}{1 \times 10^{-5}} = 10^{-9}$.
The degree of hydrolysis $\alpha$ is related to $K_h$ and concentration $C$ by the formula $\alpha = \sqrt{\frac{K_h}{C}}$ (assuming $\alpha \ll 1$).
Given $C = 0.001 \ M = 10^{-3} \ M$,we have $\alpha = \sqrt{\frac{10^{-9}}{10^{-3}}} = \sqrt{10^{-6}} = 1 \times 10^{-3}$.

(C) The hydrolysis reaction for the cation $B^{+}$ is given by: $B^{+} + H_2O \rightleftharpoons BOH + H^{+}$.
For the hydrolysis of a salt of a weak base and a strong acid,the hydrolysis constant $K_H$ is related to the dissociation constant of the base $K_b$ and the ionic product of water $K_w$ by the formula: $K_H = \frac{K_w}{K_b}$.
Given $K_w = 1.0 \times 10^{-14}$ and $K_b = 1.0 \times 10^{-6}$.
Substituting the values: $K_H = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-6}} = 1.0 \times 10^{-8}$.

(A) $NH_4Cl$ is a salt formed from a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
When $NH_4Cl$ undergoes hydrolysis in water,the reaction is: $NH_4Cl + H_2O \rightleftharpoons NH_4OH + HCl$.
Since $HCl$ is a strong acid and $NH_4OH$ is a weak base,the resulting solution contains a higher concentration of $H^+$ ions,making the solution acidic.

(B) Sodium sulphate $(Na_2SO_4)$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$.
Salts formed from the neutralization of a strong acid and a strong base do not undergo hydrolysis in water; they only dissociate into their respective ions.
$Na_2SO_4(aq) \to 2Na^+(aq) + SO_4^{2-}(aq)$

(C) $Na_2CO_3$ 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 with a $pH > 7$.
$KCl$ and $NaCl$ are salts of strong acids and strong bases,resulting in a neutral solution $(pH \approx 7)$.
$CuSO_4$ is a salt of a weak base $(Cu(OH)_2)$ and a strong acid $(H_2SO_4)$,resulting in an acidic solution $(pH < 7)$.

(A) $BaCl_2$ is a salt of a strong acid $(HCl)$ and a strong base $(Ba(OH)_2)$. Therefore,its aqueous solution is neutral with a $pH$ of $7$.
$AlCl_3$,$LiCl$,and $BeCl_2$ undergo cationic hydrolysis in water to produce acidic solutions because they involve weak bases or small,highly charged cations.
Consequently,their $pH$ values are less than $7$.
Thus,the $pH$ value is highest for the solution of $BaCl_2$.

(D) The ionization constant of the weak base $NH_{4}OH$ is given as $K_{b} = 1.77 \times 10^{-5}$.
Ammonium chloride $(NH_{4}Cl)$ is a salt of a weak base $(NH_{4}OH)$ and a strong acid $(HCl)$.
The hydrolysis constant $(K_{h})$ for such a salt is given by the formula:
$K_{h} = \frac{K_{w}}{K_{b}}$
Where $K_{w}$ is the ionic product of water,which is $1.0 \times 10^{-14}$ at $298 \ K$.
Substituting the values:
$K_{h} = \frac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}}$
$K_{h} \approx 0.56497 \times 10^{-9} = 5.65 \times 10^{-10}$.

(C) The salt $BA$ is formed from a weak acid $(HA)$ and a weak base $(BOH)$.
In aqueous solution,the salt undergoes hydrolysis as follows:
$BA + H_2O \rightleftharpoons BOH + HA$
The $pH$ of a salt of a weak acid and a weak base is given by the formula:
$pH = \frac{1}{2} pK_w + \frac{1}{2} pK_a - \frac{1}{2} pK_b$
Given $pK_w = 14.00$,$pK_a = 4.80$,and $pK_b = 4.78$.
Substituting these values into the formula:
$pH = \frac{1}{2} (14.00 + 4.80 - 4.78)$
$pH = \frac{1}{2} (14.02)$
$pH = 7.01$

(A) The hydrolysis of ammonium sulphate,$(NH_4)_2SO_4$,in the soil produces sulphuric acid as follows:
$(NH_4)_2SO_4 + 2H_2O \longrightarrow H_2SO_4 + 2NH_4OH$
Since $H_2SO_4$ is a strong acid,its accumulation in the soil leads to an increase in soil acidity.

(B) For the salt of a weak acid and a weak base,the $pH$ is calculated using the formula:
$pH = 7 + \frac{1}{2} pK_a - \frac{1}{2} pK_b$
Given values are $pK_a = 3.2$ and $pK_b = 3.4$.
Substituting these values into the formula:
$pH = 7 + \frac{1}{2}(3.2) - \frac{1}{2}(3.4)$
$pH = 7 + 1.6 - 1.7$
$pH = 6.9$

(D) The $pH$ of a solution is inversely related to the concentration of $H^+$ ions.
$HNO_3$ and $HCl$ are strong acids,so they fully dissociate to provide a high concentration of $H^+$ ions,resulting in a very low $pH$ (typically $< 1$).
$CH_3COOH$ is a weak acid,which partially dissociates,resulting in a higher $pH$ than strong acids but still acidic $(pH < 7)$.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$. It undergoes anionic hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
This hydrolysis produces $OH^-$ ions,making the solution basic,which results in a $pH > 7$.
Therefore,$0.2 \ M \ CH_3COONa$ has the maximum $pH$ value.

(C) At the equivalence point,the salt sodium formate $(HCOONa)$ is formed. It is a salt of a weak acid and a strong base.
The $pH$ of such a salt solution is given by the formula: $pH = 7 + 0.5 \ pK_a + 0.5 \ \log \ C$.
First,calculate the volume of $NaOH$ required: $M_1V_1 = M_2V_2 \implies 0.1 \ M \times 25 \ mL = 0.1 \ M \times V_2 \implies V_2 = 25 \ mL$.
Total volume of the solution $= 25 \ mL + 25 \ mL = 50 \ mL$.
The concentration of the salt $(C)$ is: $C = \frac{0.1 \ M \times 25 \ mL}{50 \ mL} = 0.05 \ M$.
Now,substitute the values into the $pH$ formula:
$pH = 7 + 0.5 \times 3.74 + 0.5 \ \log \ (0.05)$
$pH = 7 + 1.87 + 0.5 \ \log \ (5 \times 10^{-2})$
$pH = 8.87 + 0.5 \times (\log \ 5 - 2)$
Since $\log \ 5 = \log \ (10/2) = 1 - 0.30 = 0.70$,
$pH = 8.87 + 0.5 \times (0.70 - 2) = 8.87 + 0.5 \times (-1.30) = 8.87 - 0.65 = 8.22$.

(D) The salt $A^{+}B^{-}$ is a salt of a strong base and a weak acid $HB$.
Hydrolysis reaction: $B^{-} + H_2O \leftrightarrow HB + OH^{-}$.
Hydrolysis constant $K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-12}} = 10^{-2}$.
For a salt of a weak acid and strong base,$K_h = \frac{C\alpha^2}{1-\alpha}$.
Given $C = 0.01 \ M$,we have $\frac{0.01 \alpha^2}{1-\alpha} = 10^{-2}$,which simplifies to $\alpha^2 = 1 - \alpha$ or $\alpha^2 + \alpha - 1 = 0$.
Solving for $\alpha$ using the quadratic formula: $\alpha = \frac{-1 + \sqrt{1 + 4}}{2} = \frac{\sqrt{5}-1}{2} \approx 0.618$.
$[OH^{-}] = C\alpha = 0.01 \times 0.618 = 6.18 \times 10^{-3} \ M$.
$[H^{+}] = \frac{K_w}{[OH^{-}]} = \frac{10^{-14}}{6.18 \times 10^{-3}} \approx 1.6 \times 10^{-12} \ M$.

(B) $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.
For the salt of a strong base and a weak acid,the hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_a}$.
Given $K_b$ for $OCN^- = 10^{-10}$,we know $K_a \times K_b = K_w = 10^{-14}$.
Thus,$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.
$K_h = \frac{10^{-14}}{10^{-4}} = 10^{-10}$.
The degree of hydrolysis $h$ is given by $h = \sqrt{\frac{K_h}{C}}$,where $C = 0.01 \ M = 10^{-2} \ M$.
$h = \sqrt{\frac{10^{-10}}{10^{-2}}} = \sqrt{10^{-8}} = 10^{-4}$.
Percentage of hydrolysis $= h \times 100 = 10^{-4} \times 100 = 0.01 \%$.

(C) For the salt of a weak base and a strong acid,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_b \cdot C}}$.
From this,we can see that $h \propto \frac{1}{\sqrt{C}}$,which implies $\frac{h_1}{h_2} = \sqrt{\frac{C_2}{C_1}}$.
Given: $C_1 = 0.1 \ M$,$h_1 = 0.01$,and $C_2 = 0.4 \ M$.
Substituting the values: $\frac{0.01}{h_2} = \sqrt{\frac{0.4}{0.1}} = \sqrt{4} = 2$.
Therefore,$h_2 = \frac{0.01}{2} = 0.005$.

(D) For a salt of a weak acid $(WA)$ and a weak base $(WB)$,the $pH$ of the solution is given by the formula: $pH = 7 + \frac{1}{2} pK_a - \frac{1}{2} pK_b$.
Since the $pH$ depends on the relative values of $pK_a$ and $pK_b$,it can be equal to,greater than,or less than $7$ depending on whether $K_a = K_b$,$K_a < K_b$,or $K_a > K_b$ respectively.
Therefore,the $pH$ depends upon the $K_a$ and $K_b$ values.

(D) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
Upon hydrolysis,the $NH_4^+$ ion reacts with water to produce $H_3O^+$ ions,making the solution acidic with a $pH < 7$ at $298 \ K$.
$KCN$ and $CH_3COONa$ are salts of a weak acid and a strong base,resulting in a basic solution $(pH > 7)$.
$NaBr$ is a salt of a strong acid and a strong base,resulting in a neutral solution $(pH = 7)$.

(A) $(i)$ Borax $(Na_2B_4O_7 \cdot 10H_2O)$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_3BO_3)$,so it undergoes anionic hydrolysis to produce an alkaline solution.
$(ii)$ Potash alum $(K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O)$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Al(OH)_3)$,resulting in an acidic solution.
$(iii)$ $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,so it undergoes anionic hydrolysis to produce an alkaline solution.
$(iv)$ $NH_4NO_3$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HNO_3)$,resulting in an acidic solution.
Therefore,$(i)$ and $(iii)$ produce an alkaline solution.

(D) For a salt of a weak acid and a weak base,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
First,calculate $pK_a$ and $pK_b$:
$pK_a = -\log(6.2 \times 10^{-10}) \approx 9.21$
$pK_b = -\log(1.8 \times 10^{-5}) \approx 4.74$
Now,substitute these values into the formula:
$pH = 7 + \frac{1}{2}(9.21 - 4.74)$
$pH = 7 + \frac{1}{2}(4.47)$
$pH = 7 + 2.235 = 9.235$
Since the $pH$ is greater than $7$ but close to neutral,the solution is weakly basic.

(B) The degree of hydrolysis $h$ for a salt of a weak acid and strong base is given by $h = \sqrt{\frac{K_h}{C}}$.
Here,$K_h = \frac{K_w}{K_a}$,so $h = \sqrt{\frac{K_w}{K_a \times C}}$.
Given $h = 1 \% = 0.01 = 10^{-2}$,$C = 0.1 \ M$,and $K_w = 10^{-14}$.
Substituting the values: $10^{-2} = \sqrt{\frac{10^{-14}}{K_a \times 0.1}}$.
Squaring both sides: $10^{-4} = \frac{10^{-14}}{K_a \times 0.1}$.
$K_a = \frac{10^{-14}}{10^{-4} \times 0.1} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$.

(C) $CH_3COONa$ is a salt of a weak acid $(WA)$ and a strong base $(SB)$,therefore it is alkaline with $pH > 7$ $(A \to iii)$.
$NH_4Cl$ is a salt of a strong acid $(SA)$ and a weak base $(WB)$,therefore it is acidic with $pH < 7$ $(B \to ii)$.
$NaNO_3$ is a salt of a strong acid $(SA)$ and a strong base $(SB)$,therefore it is neutral with $pH = 7$ $(C \to iv)$.
$CH_3COONH_4$ is a salt of a weak acid $(WA)$ and a weak base $(WB)$,so its $pH$ depends on the relative values of $pK_a$ and $pK_b$,making it almost neutral $(D \to i)$.
Thus,the correct match is $A \to iii, B \to ii, C \to iv, D \to i$.

(C) The salt $C_6H_5NH_3^+ Cl^-$ is a salt of a weak base $(C_6H_5NH_2)$ and a strong acid $(HCl)$.
For the hydrolysis of such a salt,the hydrolysis constant $K_h$ is given by the formula:
$K_h = \frac{K_w}{K_b}$
Given that $K_w = 10^{-14}$ and $K_b = 10^{-4}$,
$K_h = \frac{10^{-14}}{10^{-4}} = 10^{-10}$

(B) The $pH$ of a salt solution depends on the strength of the parent acid and base.
For a salt of a strong base and a weak acid,the $pH$ is greater than $7$.
As the $pH$ of the salt solution increases,the strength of the corresponding acid decreases.
Given $pH$ values: $NaA = 7$,$NaB = 9$,$NaC = 10$,$NaD = 11$.
Since $NaA$ has a $pH$ of $7$,it is a salt of a strong acid and a strong base.
Since $NaB$,$NaC$,and $NaD$ have $pH > 7$,they are salts of a strong base and a weak acid.
Among the given acids,$HA$ is the strongest because its salt $NaA$ is neutral $(pH = 7)$,whereas the others form basic solutions.

(B) For a salt of a weak acid and a weak base,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{K_h} = \sqrt{\frac{K_w}{K_a \times K_b}}$.
Since the expression for the degree of hydrolysis $(h)$ does not contain any concentration term,it is independent of the concentration of the salt solution.
Therefore,if the degree of hydrolysis is $50 \ \%$ at $0.1 \ M$,it will remain $50 \ \%$ at $0.5 \ M$ as well.

(C) The degree of hydrolysis $(h)$ for a salt of a weak acid and a strong base is given by the formula $h = \sqrt{\frac{K_h}{C}}$.
Given $K_h = 3.175 \times 10^{-5}$ and concentration $C = 0.01 \ M$.
Substituting the values: $h = \sqrt{\frac{3.175 \times 10^{-5}}{0.01}} = \sqrt{3.175 \times 10^{-3}} = \sqrt{31.75 \times 10^{-4}} = 5.63 \times 10^{-2}$.

(A) The $pH$ of a salt solution formed by a weak base $(NH_4OH)$ and a weak acid $(CH_3COOH)$ is given by the formula: $pH = \frac{1}{2}pK_w + \frac{1}{2}pK_a - \frac{1}{2}pK_b$.
This formula shows that the $pH$ is independent of the concentration of the salt.
For other options like $NH_4Cl$,$NH_4NO_3$,and $(NH_4)_2SO_4$,the salts are formed from a weak base and a strong acid.
The $pH$ of such salts is given by $pH = \frac{1}{2}(pK_w - pK_b - \log C)$,which depends on the concentration $C$.
Therefore,the correct anion is $CH_3COO^{\ominus}$.

(C) $NaHA$ is an amphiprotic salt derived from a weak acid $H_2A$ and a strong base $NaOH$.
The $pH$ of an amphiprotic salt solution is given by the formula:
$pH = \frac{pK_{a_1} + pK_{a_2}}{2}$
This formula is independent of the concentration $C$ of the salt,provided the concentration is not extremely dilute.

(D) Sodium acetate is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$.
The formula for the $pH$ of a salt of a strong base and a weak acid is given by:
$pH = 7 + \frac{1}{2}pK_a + \frac{1}{2} \log c$
Given:
$c = 0.1 \ M = 10^{-1} \ M$
$K_a = 1.0 \times 10^{-5}$
Step $1$: Calculate $pK_a$
$pK_a = - \log K_a = - \log (10^{-5}) = 5$
Step $2$: Calculate $\log c$
$\log c = \log (10^{-1}) = -1$
Step $3$: Calculate $pH$
$pH = 7 + \frac{1}{2}(5) + \frac{1}{2}(-1)$
$pH = 7 + 2.5 - 0.5$
$pH = 9.0$

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
The formula for the $pH$ of such a salt solution is:
$pH = \frac{1}{2} [pK_w - pK_b - \log C]$
Given:
$C = 0.02 \ M = 2 \times 10^{-2} \ M$
$K_b = 10^{-5} \implies pK_b = 5$
$K_w = 10^{-14} \implies pK_w = 14$
Substituting the values:
$pH = \frac{1}{2} [14 - 5 - \log(2 \times 10^{-2})]$
$pH = \frac{1}{2} [9 - (\log 2 + \log 10^{-2})]$
$pH = \frac{1}{2} [9 - (0.301 - 2)]$
$pH = \frac{1}{2} [9 - (-1.699)]$
$pH = \frac{1}{2} [10.699]$
$pH = 5.3495 \approx 5.35$

(B) For the salt $NaX$,which is a salt of a weak acid and a strong base,the hydrolysis reaction is:
$X^{-} + H_2O \rightleftharpoons HX + OH^{-}$
The hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$.
The degree of hydrolysis $h$ is given by the formula $h = \sqrt{\frac{K_h}{C}}$,where $C = 0.1 \ M$.
Substituting the values:
$h = \sqrt{\frac{10^{-9}}{0.1}} = \sqrt{10^{-8}} = 10^{-4}$.
The percentage of hydrolysis is calculated as:
$\% h = h \times 100 = 10^{-4} \times 100 = 10^{-2} = 0.01 \ \%$.

(C) For a salt of a weak acid and a weak base $(WAWB)$,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{\frac{K_w}{K_a \times K_b}}$.
Since this expression does not contain the concentration term $(C)$,the degree of hydrolysis for $CH_3COONH_4$ is independent of the concentration of the salt solution.

(A) The salt $NaOCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCN)$.
For such a salt,the degree of hydrolysis $h$ is given by the formula: $h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_a \times C}}$.
Given: $K_w = 10^{-14}$,$K_a = 3.33 \times 10^{-4}$,and $C = 0.003 \ M = 3 \times 10^{-3} \ M$.
Substituting the values: $h = \sqrt{\frac{10^{-14}}{3.33 \times 10^{-4} \times 3 \times 10^{-3}}} = \sqrt{\frac{10^{-14}}{10^{-6}}} = \sqrt{10^{-8}} = 10^{-4}$.
The percentage hydrolysis is $h \times 100 = 10^{-4} \times 100 = 0.01 \%$.

(B) $NH_4NO_3$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HNO_3)$.
In an aqueous solution,the $NH_4^+$ ion undergoes hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$.
Since the solution is acidic,we calculate the $H^+$ concentration using $[H^+] = \sqrt{\frac{K_w \times C}{K_b}}$.
To find the hydroxyl ion concentration $[OH^-]$,we use the relation $[H^+][OH^-] = K_w$,which gives $[OH^-] = \frac{K_w}{[H^+]}$.
Substituting the expression for $[H^+]$,we get $[OH^-] = \frac{K_w}{\sqrt{\frac{K_w \times C}{K_b}}} = \sqrt{\frac{K_w^2 \times K_b}{K_w \times C}} = \sqrt{\frac{K_w \times K_b}{C}}$.

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it is acidic.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,so it is basic.
$Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,so it is neutral.
$CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$,so its nature depends on the relative values of $K_a$ and $K_b$.
In $CH_3COONa$,the $CH_3COO^{-}$ ion undergoes hydrolysis: $CH_3COO^{-} + H_2O \rightleftharpoons CH_3COOH + OH^{-}$.
Since $OH^{-}$ ions are produced,the solution is basic.

(C) When an aqueous solution of copper $(II)$ sulphate is allowed to stand or is heated,it undergoes slow hydrolysis.
This process results in the formation of a green-colored precipitate known as basic copper sulphate.
The chemical composition of this basic copper sulphate is represented as $CuSO_4 \cdot 3Cu(OH)_2$.

(A) The salts $NaClO$,$NaClO_2$,$NaClO_3$,and $NaClO_4$ are derived from the corresponding acids $HClO$,$HClO_2$,$HClO_3$,and $HClO_4$ and the strong base $NaOH$.
Among these acids,$HClO$ is the weakest acid.
Since the salt of a weak acid and a strong base is the most basic,$NaClO$ will be the most basic aqueous solution.

(B) The salt $NaX$ is formed from a weak acid $(HX)$ and a strong base $(NaOH)$.
For such a salt,the hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_a}$.
Given $K_w = 10^{-14}$ and $K_a = 10^{-5}$,we have $K_h = \frac{10^{-14}}{10^{-5}} = 10^{-9}$.
The formula for the degree of hydrolysis $(h)$ for a salt of a weak acid and strong base is $h = \sqrt{\frac{K_h}{C}}$.
Here,$C = 0.1 \ M = 10^{-1} \ M$.
$h = \sqrt{\frac{10^{-9}}{10^{-1}}} = \sqrt{10^{-8}} = 10^{-4}$.
The percentage of hydrolysis is $h \times 100 = 10^{-4} \times 100 = 0.01 \%$.

(C) $pH > 7$ indicates that the solution is basic in nature.
This occurs due to the hydrolysis of the anion of a weak acid,which happens when the salt is formed from a $weak \ acid$ and a $strong \ base$.

(A) $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$.
Since it is a salt of a strong acid and a strong base,it does not undergo hydrolysis in water.
Therefore,the aqueous solution of $NaCl$ remains neutral at $25 \ ^oC$.
The $pH$ of a neutral solution at $25 \ ^oC$ is $7$.

(B) At the equivalence point,the moles of $HA$ equal the moles of $NaOH$ added.
Moles of $HA = 0.1 \ M \times 0.025 \ L = 0.0025 \ mol$.
Volume of $NaOH$ required $= \frac{0.0025 \ mol}{0.05 \ M} = 0.05 \ L = 50 \ mL$.
Total volume of the solution $= 25 \ mL + 50 \ mL = 75 \ mL = 0.075 \ L$.
Concentration of salt $NaA$ formed $= \frac{0.0025 \ mol}{0.075 \ L} = \frac{1}{30} \ M$.
For the salt of a weak acid and strong base,$[OH^{-}] = \sqrt{K_h \times C} = \sqrt{\frac{K_w}{K_a} \times C}$.
$[H^{+}] = \frac{K_w}{[OH^{-}]} = \sqrt{\frac{K_w \times K_a}{C}}$.
Substituting the values: $[H^{+}] = \sqrt{\frac{10^{-14} \times 10^{-5}}{1/30}} = \sqrt{30 \times 10^{-19}} = \sqrt{3 \times 10^{-18}} = 1.732 \times 10^{-9} \ M$.

(C) When equal volumes of $0.2 \ M \ NaOH$ and $0.2 \ M \ CH_3COOH$ are mixed,the number of moles of each reactant is equal.
$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
Since the reaction is stoichiometric,both reactants are completely consumed,resulting in the formation of a salt,$CH_3COONa$.
The concentration of the salt formed is $0.1 \ M$ (since volume doubles).
$CH_3COONa$ is a salt of a weak acid and a strong base,which undergoes anionic hydrolysis: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
The $pH$ of such a salt solution is given by: $pH = 7 + \frac{1}{2}(pK_a + \log C)$.
Given $K_a = 10^{-5}$,so $pK_a = 5$.
$C = 0.1 \ M$.
$pH = 7 + \frac{1}{2}(5 + \log 0.1) = 7 + \frac{1}{2}(5 - 1) = 7 + \frac{1}{2}(4) = 7 + 2 = 9$.

(C) The given reaction is the hydrolysis of the formate ion $(HCOO^-)$,which is the conjugate base of a weak acid $(HCOOH)$.
The equilibrium constant for this hydrolysis reaction is denoted as $K_h$.
The relationship between $K_h$,$K_w$ (ionic product of water),and $K_a$ (acid dissociation constant) is given by:
$K_h = \frac{K_w}{K_a}$
Given $K_w = 1.0 \times 10^{-14}$ at $25^{\circ}C$ and $K_a = 1.8 \times 10^{-4}$.
$K_h = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}}$
$K_h = 0.5556 \times 10^{-10} = 5.56 \times 10^{-11}$
Thus,the correct option is $C$.

(D) For a salt of a weak acid $(wA)$ and a weak base $(wB)$,the $pH$ of the aqueous solution is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
Given: $K_a = 1.0 \times 10^{-5}$,so $pK_a = -\log(1.0 \times 10^{-5}) = 5.0$.
Given: $pK_b = 4.4$.
Substituting these values into the formula: $pH = 7 + \frac{1}{2}(5.0 - 4.4) = 7 + \frac{1}{2}(0.6) = 7 + 0.3 = 7.3$.
However,the question states the $pH$ is $6.7$. Since the formula for the $pH$ of a salt of a weak acid and a weak base is independent of the concentration of the salt,the $pH$ remains $7.3$ regardless of the concentration.
Therefore,the given $pH$ value of $6.7$ is inconsistent with the provided $pK_a$ and $pK_b$ values,and the concentration cannot be determined from the $pH$.

(B) The degree of hydrolysis $(h)$ for a salt depends on the strength of the acid and base from which it is derived.
$NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so it does not undergo hydrolysis.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,which undergoes anionic hydrolysis.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,which undergoes cationic hydrolysis.
$CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
For a salt of a weak acid and a weak base,the degree of hydrolysis is given by $h = \sqrt{K_h} = \sqrt{\frac{K_w}{K_a \times K_b}}$.
Since both $K_a$ and $K_b$ are small,the value of $h$ is significantly higher compared to salts derived from a strong acid/weak base or weak acid/strong base.
Therefore,$CH_3COONH_4$ undergoes the highest degree of hydrolysis.

(B) For a salt of a weak acid and a weak base,the degree of hydrolysis $(h)$ is given by: $h = \sqrt{\frac{K_w}{K_a K_b}}$.
Given $K_w = 10^{-14}$,$K_a = 10^{-6}$,and $K_b = 10^{-6}$.
$h = \sqrt{\frac{10^{-14}}{10^{-6} \times 10^{-6}}} = \sqrt{\frac{10^{-14}}{10^{-12}}} = \sqrt{10^{-2}} = 0.1$.
Therefore,$\alpha = 0.1$ or $10\%$.
The $pH$ of a salt of a weak acid and a weak base is given by: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
Since $K_a = K_b = 10^{-6}$,$pK_a = pK_b = 6$.
$pH = 7 + \frac{1}{2}(6 - 6) = 7 + 0 = 7$.
Thus,the $pH$ is $7$ and $\alpha$ is $10\%$.