Salt hydrolysis Questions in English

(A) $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
When dissolved in water,the $Fe^{3+}$ ion undergoes hydrolysis:
$Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$.
The production of $H^+$ ions makes the solution acidic.

(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 hydrolysis to form a basic solution:
$Na_2CO_3 + 2H_2O \rightleftharpoons 2NaOH + H_2CO_3$
Since $NaOH$ is a strong base,the resulting solution is alkaline in nature.

(C) $NaCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HCN)$.
When dissolved in water,the $CN^-$ ion undergoes hydrolysis:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$
Due to the production of $OH^-$ ions,the aqueous solution becomes basic.

(B) $Ba(NO_3)_2$ is a salt of a strong acid $(HNO_3)$ and a strong base $(Ba(OH)_2)$.
Since both the acid and the base are strong,the salt does not undergo hydrolysis in water.
Therefore,the resulting solution is neutral with a $pH$ of $7$.
In contrast,$(NH_4)_2SO_4$ is a salt of a weak base and a strong acid (acidic),
$CrCl_3$ and $CuSO_4$ are salts of weak bases and strong acids (acidic).

(C) . The aqueous solution of $CuSO_4 \cdot 5H_2O$ is acidic because the $Cu^{2+}$ ion undergoes hydrolysis in water to produce $H^{+}$ ions.
The reaction is: $Cu^{2+} + 2H_2O \to Cu(OH)_2 + 2H^{+}$.
The presence of $H^{+}$ ions makes the solution acidic,which turns blue litmus paper red.

(C) $Na_2HPO_4$ dissociates in water to give $2Na^+$ and $HPO_4^{2-}$ ions.
The $HPO_4^{2-}$ ion undergoes hydrolysis: $HPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^- + OH^-$.
Since the hydrolysis produces $OH^-$ ions,the resulting aqueous solution is basic in nature.

(A) $FeCl_3$ is a salt derived from a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
In water,it undergoes cationic hydrolysis:
$Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$.
Due to the production of $H^+$ ions,the concentration of hydronium ions increases,making the solution acidic $(pH < 7)$.

(D) When a salt of a weak acid and a strong base is dissolved in water,the anion of the weak acid undergoes hydrolysis to produce $OH^-$ ions.
This increases the concentration of $OH^-$ ions in the solution,making it alkaline $(pH > 7)$.
Therefore,the salt is formed from a weak acid and a strong base.

(B) The correct answer is $(B)$.
An aqueous solution of $Na_2CO_3$ is alkaline because it undergoes salt hydrolysis.
$Na_2CO_3$ dissociates into $2Na^+$ and $CO_3^{2-}$.
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 alkaline.

(B) Ammonium carbonate $(NH_4)_2CO_3$ is a salt of a weak base $(NH_4OH)$ and a weak acid $(H_2CO_3)$.
In aqueous solution,it undergoes hydrolysis:
$NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$
The dissociation constant of $NH_4OH$ $(K_b = 1.8 \times 10^{-5})$ is slightly greater than the dissociation constant of $HCO_3^-$ $(K_a = 5.6 \times 10^{-11})$,but the overall hydrolysis of the carbonate ion is more significant in determining the pH.
Since the hydrolysis of the carbonate ion produces $OH^-$ ions,the resulting solution is weakly basic.

(A) $CH_3COONa$ 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 basic in nature.
Therefore,it will turn red litmus blue.

(A) $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
When dissolved in water,it undergoes cationic hydrolysis:
$FeCl_3 + 3H_2O ⇌ Fe(OH)_3 + 3HCl$.
Since $HCl$ is a strong acid,the resulting solution contains an excess of $H^+$ ions,making the solution acidic with a $pH < 7$.

(B) $NaCl$ is a salt formed by the reaction of a strong acid $(HCl)$ and a strong base $(NaOH)$.
Since both the acid and base are strong,the resulting salt does not undergo hydrolysis,and the solution remains neutral with a $pH$ of $7$.
$HCl_{(aq)} + NaOH_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$

(D) $NaOCl$ is a salt of a strong base $(NaOH)$ and a weak acid $(HOCl)$.
When dissolved in water,the $OCl^-$ ion undergoes hydrolysis to form $OH^-$ ions,making the solution basic.
$OCl^- + H_2O \rightleftharpoons HOCl + OH^-$

(A) $Al_2(SO_4)_3$ is a salt of a weak base,$Al(OH)_3$,and a strong acid,$H_2SO_4$.
Due to the hydrolysis of the $Al^{3+}$ ion,the solution produces $H^+$ ions,making the aqueous solution acidic.

(A) The aqueous solution of $AlCl_3$ is acidic because the $Al^{3+}$ ion undergoes hydrolysis in water.
The reaction is: $Al^{3+} (aq) + 3H_2O (l) \rightleftharpoons Al(OH)_3 (s) + 3H^+ (aq)$.
Since $H^+$ ions are produced,the solution becomes acidic.
This process is known as cation hydrolysis.

(A) $AlCl_3$ is a salt of a weak base $(Al(OH)_3)$ and a strong acid $(HCl)$.
When dissolved in water,it undergoes hydrolysis:
$AlCl_3 + 3H_2O \rightleftharpoons Al(OH)_3 + 3HCl$.
Since $HCl$ is a strong acid,it dissociates completely to provide $H^+$ ions,making the solution acidic.

(A) $AlCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Al(OH)_3)$.
When $AlCl_3$ is dissolved in water,the $Al^{3+}$ ion undergoes cationic hydrolysis because it is a small,highly charged cation.
The reaction is: $Al^{3+} + 6H_2O \rightleftharpoons [Al(H_2O)_6]^{3+}$.
This hydrated complex further releases $H^+$ ions: $[Al(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H_3O^+$.
The presence of $H_3O^+$ ions makes the aqueous solution acidic.

(C) Sodium carbonate $(Na_2CO_3)$ is a salt formed from the neutralization of a strong base,sodium hydroxide $(NaOH)$,and a weak acid,carbonic acid $(H_2CO_3)$.
When dissolved in water,it undergoes hydrolysis:
$Na_2CO_3 + 2H_2O ⇌ 2NaOH + H_2CO_3$.
Since $NaOH$ is a strong base and $H_2CO_3$ is a weak acid,the resulting solution contains an excess of $OH^-$ ions,making the solution alkaline (basic).

(B) Aluminium sulphate $(Al_2(SO_4)_3)$ is a salt of a weak base $(Al(OH)_3)$ and a strong acid $(H_2SO_4)$.
When dissolved in water,$Al^{3+}$ ions undergo hydrolysis:
$Al^{3+} + 3H_2O ⇌ Al(OH)_3 + 3H^+$
Due to the production of $H^+$ ions,the solution becomes acidic in nature.

(C) $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
Due to the hydrolysis of the $NH_4^+$ ion,the solution produces $H^+$ ions,making the aqueous solution acidic.

(B) $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
When dissolved in water,the $Fe^{3+}$ ion undergoes hydrolysis:
$Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$
Due to the production of $H^+$ ions,the resulting solution becomes acidic.

(A) $CuSO_4$ is a salt of a weak base,$Cu(OH)_2$,and a strong acid,$H_2SO_4$.
Due to the hydrolysis of the $Cu^{2+}$ ion,the solution produces $H^+$ ions,making the aqueous solution acidic.

(B) $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
When dissolved in water,it undergoes hydrolysis to form an acidic solution.
$NH_4Cl + H_2O \rightleftharpoons NH_4OH + HCl$
$NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$
$NaCl$,$KCl$,and $Na_2SO_4$ are salts of strong acids and strong bases,so they do not undergo hydrolysis.

(C) $FeCl_3$ is a salt of a weak base $(Fe(OH)_3)$ and a strong acid $(HCl)$.
When dissolved in water,the $Fe^{3+}$ ion undergoes hydrolysis:
$Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$.
The production of $H^+$ ions makes the solution acidic.

(D) Mohr's salt,with the chemical formula $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$,is a classic example of a double salt.
It is formed by the combination of two simple salts,ferrous sulfate and ammonium sulfate,in a $1:1$ molar ratio.

(D) Sodium acetate $(CH_3COONa)$ is a salt formed from a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$.
When dissolved in water,it undergoes anionic hydrolysis:
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
Since $OH^-$ ions are produced in the solution,the resulting aqueous solution is alkaline $(pH > 7)$.

(B) $NH_4CN$ is a salt formed from a weak acid $(HCN)$ and a weak base $(NH_4OH)$.
For a salt of a weak acid and a weak base,the hydrolysis constant $(K_h)$ is given by the formula:
$K_h = \frac{K_w}{K_a \times K_b}$
Where $K_w$ is the ionic product of water,$K_a$ is the dissociation constant of the weak acid,and $K_b$ is the dissociation constant of the weak base.
Therefore,the correct option is $(B)$.

(A) The correct answer is $A$.
$CH_3COONa$ is a salt formed from a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
Salts of weak acids and strong bases undergo anionic hydrolysis in water to form a basic solution.
$KNO_3$,$NaCl$,and $K_2SO_4$ are salts of strong acids and strong bases,which do not undergo hydrolysis.

(B) mixed salt is a salt that contains more than one type of cation or anion.
$Ca(OCl)Cl$ (bleaching powder) is a mixed salt because it contains two different anions,$Cl^{-}$ and $OCl^{-}$,associated with the $Ca^{2+}$ cation.
$NaHCO_3$ is an acid salt.
$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$ is a double salt.
$Mg(OH)Br$ is a basic salt.

(A) 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.
$A$ $(KNO_3)$ is a salt of 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 be hydrolysed.

(B) $pH$ of $9$ indicates that the solution is basic.
$NH_4NO_3$ is a salt of a strong acid $(HNO_3)$ and a weak base $(NH_4OH)$,which results in an acidic solution $(pH < 7)$.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,which undergoes anionic hydrolysis to produce a basic solution $(pH > 7)$.
$CH_3COONH_4$ is a salt of a weak acid and a weak base,which results in a nearly neutral solution $(pH \approx 7)$.
$CaCO_3$ is insoluble in water.
Therefore,the correct salt is $CH_3COONa$.

(B) $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
When it undergoes hydrolysis,the acetate ion $(CH_3COO^-)$ reacts with water to produce $OH^-$ ions.
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.
Due to the presence of $OH^-$ ions,the resulting aqueous solution is basic in nature.

(D) salt of strong acid and strong base cannot be hydrolysed.
In this case,the ions of the salt do not react with water to produce $H^+$ or $OH^-$ ions,as they are the conjugate bases/acids of strong acids/bases,which are extremely weak.
Therefore,the equilibrium cannot be shifted,and the solution remains neutral.

(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:
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
Since $OH^-$ ions are produced in the solution,the aqueous solution of $CH_3COONa$ is alkaline (basic) in nature.

(C) Salt hydrolysis occurs when a salt is derived from a weak acid and a strong base,a weak base and a strong acid,or a weak acid and a weak base.
$CH_3COOK$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(KOH)$.
When dissolved in water,the acetate ion $(CH_3COO^-)$ undergoes hydrolysis to form $CH_3COOH$ and $OH^-$,making the solution basic.
$KCl$,$NaNO_3$,and $K_2SO_4$ are salts of strong acids and strong bases,which do not undergo hydrolysis.

(C) The acidity of a salt solution depends on the extent of hydrolysis,which is determined by the charge density of the cation.
$Fe^{3+}$ and $Al^{3+}$ are highly acidic due to high charge density.
However,$FeCl_3$ undergoes hydrolysis to form $Fe(OH)_3$ and $HCl$.
Comparing the acidity,$FeCl_3$ is generally considered more acidic than $AlCl_3$ because $Fe^{3+}$ has a higher charge-to-size ratio and forms a more acidic solution upon hydrolysis in aqueous media.
Therefore,the correct answer is $C$.

(C) $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
In aqueous solution,$Fe^{3+}$ ions undergo hydrolysis:
$Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$.
The production of $H^+$ ions makes the solution acidic,resulting in a $pH < 7.0$.

(C) For a salt derived from a weak acid $(HA)$ and a weak base $(BOH)$,the hydrolysis reaction is: $B^+ + A^- + H_2O \rightleftharpoons BOH + HA$.
The equilibrium constant for this reaction is the hydrolysis constant $(K_h)$.
It is related to the ionic product of water $(K_w)$,the dissociation constant of the acid $(K_a)$,and the dissociation constant of the base $(K_b)$ by the expression: $K_h = \frac{K_w}{K_a \times K_b}$.

(A) The hydrolysis of a salt formed from a strong base and a weak acid results in a basic solution.
$KCN$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
When $KCN$ undergoes hydrolysis,it produces $OH^-$ ions,making the solution basic:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$
Therefore,the correct option is $A$.

(C) For the salt hydrolysis of ammonium benzoate $(C_6H_5COONH_4)$,the degree of hydrolysis is given as $\alpha = 0.25 \% = 0.0025$.
Concentration $C = 0.5 \, M$.
The formula for the hydrolysis constant $(K_h)$ for a salt of a weak acid and a weak base is $K_h = \frac{\alpha^2 C}{1 - \alpha}$.
Since $\alpha$ is very small $(0.0025 \ll 1)$,we can approximate $1 - \alpha \approx 1$.
Thus,$K_h = \alpha^2 C$.
Substituting the values: $K_h = (0.0025)^2 \times 0.5$.
$K_h = (6.25 \times 10^{-6}) \times 0.5 = 3.125 \times 10^{-6}$.

(A) For the hydrolysis of a salt of a weak acid and a strong base,the equilibrium reaction is $A^{-} + H_{2}O \rightleftharpoons HA + OH^{-}$.
The equilibrium constant for this reaction is the hydrolysis constant,$K_{h}$.
$K_{h} = \frac{[HA][OH^{-}]}{[A^{-}]}$.
Multiplying the numerator and denominator by $[H^{+}]$,we get $K_{h} = \frac{[HA][OH^{-}][H^{+}]}{[A^{-}][H^{+}]}$.
Since $K_{w} = [H^{+}][OH^{-}]$ and $K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$,it follows that $K_{h} = \frac{K_{w}}{K_{a}}$.

(A) Potassium cyanide $(KCN)$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
When dissolved in water,it undergoes anionic hydrolysis:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$.
The production of $OH^-$ ions increases the concentration of hydroxide ions in the solution.
Therefore,the $pH$ of the solution will increase,making it basic.

(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 anionic hydrolysis:
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$
This reaction produces hydroxide ions $(OH^-)$,which increases the concentration of $OH^-$ ions in the solution.
Consequently,the solution becomes basic (alkaline),leading to an increase in alkalinity and a $pH$ value greater than $7$.

(D) The $pH$ of an aqueous salt solution depends on the strength of the parent acid and base.
$NaClO$,$NaClO_2$,$NaClO_3$,and $NaClO_4$ are salts of a strong base $(NaOH)$ and their respective oxyacids: $HClO$,$HClO_2$,$HClO_3$,and $HClO_4$.
As the oxidation state of chlorine increases,the strength of the corresponding oxyacid increases $(HClO < HClO_2 < HClO_3 < HClO_4)$.
Since $HClO_4$ is the strongest acid among these,its conjugate base $ClO_4^-$ is the weakest,resulting in the least hydrolysis and the lowest $pH$ for the salt $NaClO_4$.

(A) Potassium cyanide $(KCN)$ is a salt of a strong base $(KOH)$ and a weak acid $(HCN)$.
When dissolved in water,it undergoes anionic hydrolysis:
$CN^{-} + H_2O \rightleftharpoons HCN + OH^{-}$
Due to the production of $OH^{-}$ ions,the concentration of hydroxide ions in the solution increases.
Consequently,the $pH$ of the solution increases,making it basic.

(D) $CH_3COONa$ is a salt of a weak acid,$CH_3COOH$,and a strong base,$NaOH$.
Since it is a salt of a weak acid and a strong base,it undergoes anionic hydrolysis to produce $OH^-$ ions,making the solution basic.
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$.

(A) $MgCl_2$ is a salt of a strong acid $(HCl)$ and a weak base $(Mg(OH)_2)$.
When dissolved in water,it undergoes cationic hydrolysis: $Mg^{2+} + 2H_2O ⇌ Mg(OH)_2 + 2H^+$.
Due to the production of $H^+$ ions,the solution becomes acidic.
Therefore,the $pH$ of the solution is $< 7$.

(A) $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$.
Due to salt hydrolysis,the aqueous solution of $Na_2CO_3$ is basic in nature.
Therefore,the $pH$ of the solution is greater than $7$.

(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 of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,making its solution weakly basic.
$Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$. It undergoes hydrolysis to form $OH^-$ ions: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$. Since it produces a strong base,it has the highest $pH$ among the given options.