Salt hydrolysis Questions in English

(A) The $pH$ of a salt solution depends on the strength of the parent acid.
$NaClO$ is the salt of a weak acid $HClO$.
$NaClO_2$,$NaClO_3$,and $NaClO_4$ are salts of stronger acids $HClO_2$,$HClO_3$,and $HClO_4$ respectively.
Since $HClO$ is the weakest acid among the given oxyacids,its conjugate base $ClO^-$ is the strongest base.
Therefore,the salt $NaClO$ undergoes the highest degree of hydrolysis,resulting in the maximum $pH$.

(B) Sodium borate $(Na_2B_4O_7)$ is a salt of a strong base $(NaOH)$ and a weak acid (boric acid,$H_3BO_3$).
When dissolved in water,it undergoes anionic hydrolysis.
The borate ions $(B_4O_7^{2-})$ react with water to produce $OH^-$ ions,which makes the solution alkaline.
Therefore,the $pH$ of the solution is $> 7$.

(A) $NaCl$ is a salt formed by the reaction of a strong acid $(HCl)$ and a strong base $(NaOH)$.
Since both the acid and the base are strong,the salt does not undergo hydrolysis in water.
Therefore,the resulting solution is neutral,and its $pH$ is $7$ at $25 \ ^{\circ}C$.

(C) Sodium chloride $(NaCl)$ is a salt formed from 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,an aqueous solution of $NaCl$ is neutral,having a $pH$ of approximately $7$ at $25 \ ^{\circ}C$.

(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.
The carbonate ion $(CO_3^{2-})$ reacts with water to produce $OH^-$ ions:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$.
Due to the presence of $OH^-$ ions,the solution becomes basic in nature.
Therefore,the $pH$ of the solution is greater than $7$.

(D) Sodium carbonate $(Na_2CO_3)$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,the carbonate ion $(CO_3^{2-})$ undergoes hydrolysis:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$
This reaction produces hydroxide ions $(OH^-)$,which increases the concentration of $OH^-$ in the solution,making it basic with a $pH > 7$.

(A) The hydrolysis constant $(K_h)$ for a salt of a weak acid and a strong base is given by the formula:
$K_h = \frac{K_w}{K_a}$
Given:
$K_w = 1 \times 10^{-14}$
$K_a = 4.5 \times 10^{-10}$
Substituting the values:
$K_h = \frac{1 \times 10^{-14}}{4.5 \times 10^{-10}} = 0.222 \times 10^{-4} = 2.22 \times 10^{-5}$
Therefore,the correct option is $A$.

(D) Ammonium acetate $(CH_3COONH_4)$ is a salt formed from a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
For a salt of a weak acid and a weak base,the $pH$ of the aqueous solution is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
Since the dissociation constants for acetic acid $(K_a = 1.8 \times 10^{-5})$ and ammonium hydroxide $(K_b = 1.8 \times 10^{-5})$ are approximately equal,$pK_a \approx pK_b$.
Therefore,$pH \approx 7$,making the solution almost neutral.

(A) For a salt of a weak acid and a weak base,the hydrolysis constant $K_h$ is given by the expression: $K_h = \frac{h^2}{(1-h)^2}$,where $h$ is the degree of hydrolysis.
Since $K_h$ is a constant at a given temperature,the degree of hydrolysis $h$ 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 $0.2 \, M$ as well.

(A) The salt $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
For such a salt,the $pH$ is given by the formula: $pH = \frac{1}{2}[pK_w + pK_a + \log c]$.
Given: $pK_w = 14$,$pK_a = 4.74$,and $c = 0.01 \ M = 10^{-2} \ M$.
Substituting the values:
$pH = \frac{1}{2}[14 + 4.74 + \log(10^{-2})]$
$pH = \frac{1}{2}[18.74 - 2]$
$pH = \frac{1}{2}[16.74] = 8.37$.

(B) The salt $NaX$ is formed from a weak acid $HX$ and a strong base $NaOH$. The hydrolysis reaction is: $X^{-} + H_2O ⇌ HX + OH^{-}$.
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 $h$ for a salt of a weak acid and strong base is given by $h = \sqrt{\frac{K_h}{C}}$,where $C = 0.1 \ M$.
$h = \sqrt{\frac{10^{-9}}{0.1}} = \sqrt{10^{-8}} = 10^{-4}$.
To express this as a percentage: $h \times 100 = 10^{-4} \times 100 = 0.01 \%$.

(A) Given: $K_{a} = 1.0 \times 10^{-5}$ and concentration $C = 0.001 \ M = 10^{-3} \ M$.
$K_{w} = 1.0 \times 10^{-14}$.
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 $(h)$ for a salt of a weak acid and strong base is given by $h = \sqrt{\frac{K_{h}}{C}}$.
Substituting the values: $h = \sqrt{\frac{10^{-9}}{10^{-3}}} = \sqrt{10^{-6}} = 10^{-3}$.

(B) Given: $pK_a = 5$,so $K_a = 10^{-5}$.
Concentration $C = 0.1 \ M = 10^{-1} \ M$.
$K_w = 10^{-14}$.
For a salt of a weak acid and a strong base,the degree of hydrolysis $\alpha_h$ is given by:
$\alpha_h = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_a \cdot C}}$
$\alpha_h = \sqrt{\frac{10^{-14}}{10^{-5} \cdot 10^{-1}}} = \sqrt{\frac{10^{-14}}{10^{-6}}} = \sqrt{10^{-8}} = 10^{-4}$.
Percentage hydrolysis = $\alpha_h \times 100 = 10^{-4} \times 100 = 10^{-2} \%$.

(A) $NaHCO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
In water,it dissociates as: $NaHCO_3 \rightarrow Na^{+} + HCO_3^-$.
The bicarbonate ion undergoes hydrolysis: $HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^{-}$.
Since $OH^{-}$ ions are produced in the solution,the resulting aqueous solution is weakly basic.

(A) Salts formed from a strong acid and a strong base do not undergo hydrolysis in water because their constituent ions do not react with $H^+$ or $OH^-$ ions to change the $pH$ of the solution.
$KClO_4$ is a salt of a strong acid $(HClO_4)$ and a strong base $(KOH)$. Therefore,it dissociates completely into $K^+$ and $ClO_4^-$ ions,which do not react with water.
In contrast,$NH_4Cl$ (salt of weak base and strong acid) and $CH_3COONa$ (salt of weak acid and strong base) undergo hydrolysis.

(C) An aqueous solution of $AlCl_3$ undergoes hydrolysis to form $Al(OH)_3$ and $HCl$ as follows:
$AlCl_3 + 3H_2O \rightarrow Al(OH)_3 + 3HCl$
Upon heating the solution to dryness,the volatile $HCl$ gas escapes.
The remaining $Al(OH)_3$ is thermally unstable and decomposes to form aluminium oxide $(Al_2O_3)$ and water vapor:
$2Al(OH)_3 \rightarrow Al_2O_3 + 3H_2O$
Therefore,the final product obtained is $Al_2O_3$.

(C) $(NH_4)_2SO_4$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(H_2SO_4)$.
Upon hydrolysis,it produces a strong acid:
$(NH_4)_2SO_4 + 2H_2O \to 2NH_4OH + H_2SO_4$.
The presence of $H_2SO_4$ leads to an increase in the acidity of the soil upon repeated application.

(D) The $pH$ of an aqueous solution of a salt of a strong base and a weak acid depends on the strength of the acid.
Stronger the acid,weaker is its conjugate base,and lower is the $pH$ of its salt solution.
The order of acidic strength of the corresponding hydrides is $H_2O < H_2S < H_2Se < H_2Te$.
Since the acidic strength increases down the group,the basic strength of the conjugate bases $(O^{2-}, S^{2-}, Se^{2-}, Te^{2-})$ decreases in the order $O^{2-} > S^{2-} > Se^{2-} > Te^{2-}$.
Therefore,the $pH$ of the isomolar solutions follows the order $pH_1 > pH_2 > pH_3 > pH_4$.

(A) Sodium carbonate $(Na_2CO_3)$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
On hydrolysis,it reacts with water as follows:
$Na_2CO_3 + 2H_2O \to 2NaOH + H_2CO_3$
The $NaOH$ produced is a strong electrolyte and undergoes complete ionization:
$2NaOH \to 2Na^+ + 2OH^-$
Thus,one molecule of $Na_2CO_3$ produces $2$ hydroxide ions $(OH^-)$ upon complete hydrolysis.

(A) Anhydrous $CuSO_4$ is a white powder. When it is hydrated,it forms the pentahydrate $CuSO_4 \cdot 5H_2O$,which is blue in color. In aqueous solution,the $Cu^{2+}$ ion undergoes hydrolysis,where it reacts with water to form $[Cu(H_2O)_6]^{2+}$ and releases $H^+$ ions,making the solution acidic.

(A) $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
Upon hydrolysis,it reacts with water as follows: $FeCl_3 + 3H_2O \rightleftharpoons Fe(OH)_3 + 3HCl$.
Since $HCl$ is a strong acid and $Fe(OH)_3$ is a weak base,the resulting aqueous solution contains an excess of $H^+$ ions,making it acidic in nature.

(B) The hydrolysis of anhydrous $AlCl_{3}$ in water is a vigorous reaction that produces aluminum hydroxide and hydrogen chloride gas.
The chemical equation is: $AlCl_{3} + 3H_{2}O \rightarrow Al(OH)_{3} + 3HCl$.
Therefore,the correct product formed is $Al(OH)_{3}$.

(B) Sodium carbonate $(Na_2CO_3)$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$.
When dissolved in water,it undergoes hydrolysis as follows:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$
From the stoichiometry of the reaction,$1 \text{ mole}$ of $CO_3^{2-}$ ions produces $1 \text{ mole}$ of $OH^-$ ions.
Therefore,the hydrolysis of $1 \text{ mole}$ of $Na_2CO_3$ produces $1 \text{ mole}$ of hydroxide ions $(OH^-)$.

(A) For a salt of a weak acid and a weak base,the degree of hydrolysis $(h)$ is independent of the concentration of the salt solution.
The formula for the degree of hydrolysis is given by: $h = \sqrt{\frac{K_w}{K_a \times K_b}}$
Given values: $K_w = 10^{-14}$,$K_a = 10^{-5}$,$K_b = 10^{-5}$.
Substituting the values: $h = \sqrt{\frac{10^{-14}}{10^{-5} \times 10^{-5}}} = \sqrt{\frac{10^{-14}}{10^{-10}}} = \sqrt{10^{-4}} = 10^{-2}$.

(A) $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
In water,the $Fe^{3+}$ ion undergoes hydrolysis:
$Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$
Due to the release of $H^+$ ions,the solution becomes acidic.
This process is known as the hydrolysis of the cation.

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
When dissolved in water,it undergoes cationic hydrolysis:
$NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$
Since $H^+$ ions are produced,the solution becomes acidic.
For a salt of a strong acid and a weak base,the $pH$ is given by the formula:
$pH = \frac{1}{2} [pK_w - pK_b - \log C]$
Given $C = 1 \, M$,$\log(1) = 0$.
$pH = \frac{1}{2} [14 - 4.74 - 0] = \frac{9.26}{2} = 4.63$.
Thus,the $pH$ is between $6$ and $7$ (specifically,it is less than $7$).

(A) Hydrolysis is a chemical process in which a molecule of water is added to a substance. This process is typically observed in salts derived from a weak acid and a weak base,a weak acid and a strong base,or a strong acid and a weak base. In these cases,the ions of the salt react with water to produce an acidic or basic solution. Since salts are $Ionic$ compounds,the process of hydrolysis involves the interaction of $Ionic$ species with water.

(C) $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$.
For the salt of a weak acid and strong base,the $pH$ is given by the formula:
$pH = 7 + \frac{1}{2} pK_a + \frac{1}{2} \log C$
Given $pK_b(CN^-) = 4.7$,we know $pK_a(HCN) = 14 - 4.7 = 9.3$.
Concentration $C = 0.5\,M = 5 \times 10^{-1}\,M$.
Substituting the values:
$pH = 7 + \frac{1}{2}(9.3) + \frac{1}{2} \log(5 \times 10^{-1})$
$pH = 7 + 4.65 + \frac{1}{2}(\log 5 + \log 10^{-1})$
$pH = 11.65 + \frac{1}{2}(0.699 - 1)$
$pH = 11.65 + \frac{1}{2}(-0.301)$
$pH = 11.65 - 0.15 = 11.5$.

(A) solution with $pH < 7$ is acidic in nature.
$FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
Due to the hydrolysis of the $Fe^{3+}$ ion,the solution becomes acidic $(pH < 7)$.
$NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,resulting in a basic solution $(pH > 7)$.
$NaOH$ is a strong base $(pH > 7)$.
$NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,resulting in a neutral solution $(pH = 7)$.

(D) The salt of a weak acid and a strong base undergoes anionic hydrolysis.
For a salt of a weak acid and a strong base,the hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_a}$.
Given $K_w = 1.0 \times 10^{-14}$ and $K_a = 3.5 \times 10^{-3}$,we have $K_h = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-3}} \approx 2.857 \times 10^{-12}$.
The degree of hydrolysis $h$ is given by the formula $h = \sqrt{\frac{K_h}{C}}$,where $C$ is the concentration of the salt.
Given $C = 0.05 \ M$,we have $h = \sqrt{\frac{2.857 \times 10^{-12}}{0.05}} = \sqrt{5.714 \times 10^{-11}} = \sqrt{57.14 \times 10^{-12}} \approx 7.559 \times 10^{-6}$.

(B) The salts $Na_3BO_3$,$Na_3PO_4$,$NaCN$,and $CH_3COONa$ are formed from the reaction of a weak acid ($H_3BO_3$,$H_3PO_4$,$HCN$,and $CH_3COOH$ respectively) and a strong base $(NaOH)$.
Since these salts are derived from a strong base and a weak acid,they undergo anionic hydrolysis in water.
This results in an increase in the concentration of $OH^-$ ions in the solution,making the aqueous solution alkaline (basic).

(A) Ferrous sulfate $(FeSO_4)$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Fe(OH)_2)$.
In water,$Fe^{+2}$ ions undergo hydrolysis:
$Fe^{+2} + 2H_2O \rightleftharpoons Fe(OH)_2 + 2H^+$
Due to the production of $H^+$ ions,the solution becomes acidic.

(D) $Al_2(SO_4)_3$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Al(OH)_3)$.
When dissolved in water,it undergoes hydrolysis.
The $Al^{3+}$ ion reacts with water to form $Al(OH)_3$ and $H^+$ ions.
$Al^{3+} + 3H_2O \rightleftharpoons Al(OH)_3 + 3H^+$.
Due to the presence of excess $H^+$ ions,the resulting aqueous solution is acidic in nature.

(C) For a salt of a weak acid and a weak base,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{\frac{K_w}{K_a \times K_b}}$
Given: $K_w = 10^{-14}$,$K_a = 1.8 \times 10^{-5}$,$K_b = 1.8 \times 10^{-5}$
Substituting the values: $h = \sqrt{\frac{10^{-14}}{1.8 \times 10^{-5} \times 1.8 \times 10^{-5}}}$
$h = \sqrt{\frac{10^{-14}}{3.24 \times 10^{-10}}} = \sqrt{\frac{1}{3.24} \times 10^{-4}}$
$h = \frac{1}{1.8} \times 10^{-2} \approx 0.555 \times 10^{-2}$
Thus,the degree of hydrolysis is $0.55 \times 10^{-2}$.

(B) $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
The hydrolysis of $NH_4^+$ ion is given by: $NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$.
The concentration of $H^+$ ions is given by the formula: $[H^+] = \sqrt{\frac{K_w \cdot K_h}{C}} = \sqrt{\frac{K_w}{K_b} \cdot C}$.
As the solution is diluted,the concentration $C$ decreases.
Since $[H^+] \propto \sqrt{C}$,the concentration of $H^+$ ions decreases upon dilution.
As $[H^+]$ decreases,the $pH$ $(pH = -\log[H^+])$ increases.

(B) $KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
The degree of hydrolysis $h$ is given as $3.7\% = 0.037$.
The formula for the hydrolysis constant $K_h$ for a salt of a weak acid and strong base is $K_h = \frac{K_w}{K_a}$.
Also,$K_h = c \times h^2$ (where $c$ is concentration,$c = \frac{1}{V} = \frac{1}{100} \ M$).
Equating the two: $\frac{K_w}{K_a} = \frac{h^2}{V}$.
$K_w = \frac{h^2 \times K_a}{V} = \frac{(0.037)^2 \times 7.2 \times 10^{-10}}{100}$.
$K_w = \frac{0.001369 \times 7.2 \times 10^{-10}}{100} = 0.000098568 \times 10^{-10} = 0.98568 \times 10^{-14} \approx 0.986 \times 10^{-14}$.

(B) $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so its solution is neutral $(pH \approx 7)$.
$NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$,so its solution is acidic $(pH < 7)$.
$NaHCO_3$ is an acidic salt,making its solution weakly basic.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,which undergoes anionic hydrolysis to produce $OH^-$ ions,making the solution strongly basic $(pH > 7)$.
Therefore,$Na_2CO_3$ has the highest $pH$.

(B) $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
When dissolved in water,it undergoes anionic hydrolysis:
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
Due to the production of $OH^-$ ions,the solution becomes alkaline (basic) in nature.

(C) $CH_3COONa$ is a salt of a weak acid and a strong base.
The formula for the $pH$ of a salt of a weak acid and strong base is:
$pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C$
Given $pK_a = 4.74$ and $C = 0.1 \ M = 10^{-1} \ M$.
Substituting the values:
$pH = 7 + \frac{1}{2}(4.74) + \frac{1}{2}\log(10^{-1})$
$pH = 7 + 2.37 + \frac{1}{2}(-1)$
$pH = 7 + 2.37 - 0.5$
$pH = 8.87$

(D) $CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$.
In an aqueous solution,it undergoes hydrolysis to form an acidic solution $(pH < 7)$.
Since it is an ionic compound,it dissociates into ions ($Cu^{2+}$ and $SO_4^{2-}$) in water.
Therefore,it acts as a conductor of electricity.

(C) The hydrolysis reaction for $CH_3COONa$ is: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
Given concentration $C = 0.1 \, M$ and degree of hydrolysis $h = 1 \% = 0.01$.
The concentration of $[OH^-]$ is given by $C \times h = 0.1 \times 0.01 = 10^{-3} \, M$.
Using the ionic product of water $K_w = 10^{-14}$,we find $[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{10^{-3}} = 10^{-11} \, M$.
Finally,$pH = -\log[H^+] = -\log(10^{-11}) = 11$.
Thus,the values are $[OH^-] = 10^{-3} \, M, [H^+] = 10^{-11} \, M, pH = 11$.

(D) The degree of hydrolysis $(h)$ for a salt of a weak acid and a weak base is given by the formula: $h = \sqrt{K_h} = \sqrt{\frac{K_w}{K_a \times K_b}}$.
Since $K_w$,$K_a$,and $K_b$ are constants at a given temperature,the degree of hydrolysis $(h)$ is independent of the concentration of the salt solution.
Therefore,the degree of hydrolysis for all the given concentrations of $NH_4CN$ will be the same.

(A) Anionic hydrolysis occurs when a salt of a weak acid and a strong base is dissolved in water.
For such a salt (e.g.,$CH_3COONa$),the hydrolysis reaction is: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
The $pH$ of the resulting solution is given by the formula: $pH = \frac{1}{2}pK_W + \frac{1}{2}pK_a + \frac{1}{2}\log c$,where $pK_W$ is the ionic product of water,$pK_a$ is the dissociation constant of the weak acid,and $c$ is the concentration of the salt.

(A) The hydrolysis reaction $A^{-} + H_2O \rightleftharpoons HA + OH^{-}$ shows the production of $OH^{-}$ ions in the solution.
This indicates that the anion $A^{-}$ undergoes hydrolysis to form a weak acid $HA$ and releases $OH^{-}$ ions,making the solution basic.
This type of hydrolysis is characteristic of a salt formed from a $Strong \text{ } Base$ and a $Weak \text{ } Acid$.

(C) The reaction between acetic acid and sodium hydroxide is:
$CH_3COOH + NaOH \rightleftharpoons CH_3COONa + H_2O$
At the equivalence point,the total volume doubles,so the concentration of the salt $(CH_3COONa)$ becomes $C = \frac{0.1 \times V}{2V} = 0.05 \, M$.
The $pH$ of a salt of a weak acid and a strong base is given by:
$pH = \frac{1}{2} [pK_w + pK_a + \log C]$
Given $K_w = 10^{-14}$,$K_a = 1.9 \times 10^{-5}$,and $C = 0.05 \, M$:
$pK_w = 14$
$pK_a = -\log(1.9 \times 10^{-5}) = 5 - 0.2788 = 4.7212$
$pH = \frac{1}{2} [14 + 4.7212 + \log(0.05)]$
$pH = \frac{1}{2} [18.7212 + (-1.301)]$
$pH = \frac{1}{2} [17.4202] = 8.7101 \approx 8.71$

(A) For a salt of a weak acid and strong base,the degree of hydrolysis $h$ is given by $h = \sqrt{\frac{K_w}{K_a \times C}}$.
Given: $K_w = 10^{-14}$,$K_a = 1.3 \times 10^{-9}$,$C = \frac{1}{100} = 0.01 \ M$.
$h = \sqrt{\frac{10^{-14}}{1.3 \times 10^{-9} \times 0.01}} = \sqrt{\frac{10^{-14}}{1.3 \times 10^{-11}}} = \sqrt{0.769 \times 10^{-3}} = \sqrt{7.69 \times 10^{-4}} \approx 0.0277$.
Percentage of hydrolysis $= h \times 100 = 0.0277 \times 100 = 2.77\%$.
The $pH$ of the salt solution is given by $pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C$.
$pK_a = -\log(1.3 \times 10^{-9}) = 9 - \log(1.3) = 9 - 0.1139 = 8.886$.
$pH = 7 + \frac{1}{2}(8.886) + \frac{1}{2}\log(0.01) = 7 + 4.443 - 1 = 10.443 \approx 10.4$.

(D) The $pH$ of water is $7$ (neutral). When substance $Y$ is added,the $pH$ increases to $13$,which indicates that the resulting solution is strongly basic. $A$ salt that produces a basic solution upon hydrolysis is formed from a weak acid and a strong base.

(B) Both $CH_3COOK$ and $HCOOK$ undergo hydrolysis in water to form their respective weak acids,$CH_3COOH$ and $HCOOH$.
The degree of hydrolysis $(h)$ is given by $h = \sqrt{\frac{K_h}{C}}$.
Since $K_h = \frac{K_w}{K_a}$,we have $h = \sqrt{\frac{K_w}{K_a \times C}}$.
The concentration of the acid formed is $[Acid] = C \times h = \sqrt{\frac{K_w \times C}{K_a}}$.
Since $K_a$ of $HCOOH$ $(1.8 \times 10^{-4})$ is greater than $K_a$ of $CH_3COOH$ $(1.8 \times 10^{-5})$,the concentration of $HCOOH$ formed will be less than the concentration of $CH_3COOH$ formed.
Therefore,$[CH_3COOH] > [HCOOH]$.

(A) Ammonium formate $(NH_4HCOO)$ is a salt of a weak acid $(HCOOH)$ and a weak base $(NH_4OH)$.
The $pH$ of a salt of a weak acid and a weak base is given by the formula:
$pH = 7 + \frac{1}{2} pK_a - \frac{1}{2} pK_b$
Given:
$K_a = 1 \times 10^{-4} \implies pK_a = -\log(1 \times 10^{-4}) = 4$
$K_b = 1 \times 10^{-5} \implies pK_b = -\log(1 \times 10^{-5}) = 5$
Substituting the values into the formula:
$pH = 7 + \frac{1}{2}(4) - \frac{1}{2}(5)$
$pH = 7 + 2 - 2.5$
$pH = 6.5$