Acids and Bases Questions in English

(N/A) Auto-protolysis refers to the self-ionization of water,where water molecules react with each other to form ions.
It is represented by the equation:
$H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O^{+}_{(aq)} + OH^{-}_{(aq)}$
Significance:
Due to auto-protolysis,water exhibits amphoteric nature,meaning it can act as both an acid and a base.
It acts as a base when reacting with acids stronger than itself and as an acid when reacting with bases stronger than itself.
Examples:
$1$. As an acid: $H_2O_{(l)} + NH_{3(aq)} \longrightarrow NH_{4(aq)}^{+} + OH^{-}_{(aq)}$
$2$. As a base: $H_2O_{(l)} + H_2S_{(aq)} \longrightarrow H_3O^{+}_{(aq)} + HS^{-}_{(aq)}$

(B) Ammonia $(NH_3)$ reacts with water to form ammonium hydroxide $(NH_4OH)$,which dissociates to provide hydroxide ions $(OH^-)$ in the solution.
$NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$
Due to the presence of $OH^-$ ions,the aqueous solution of ammonia is basic in nature.

(N/A) The strength of an acid is inversely proportional to its $pK_{a}$ value,meaning a smaller $pK_{a}$ indicates a stronger acid.
$1$. Strong acids: $pK_{a} < 1$.
$2$. Moderately strong acids: $1 < pK_{a} < 5$.
$3$. Weak acids: $5 < pK_{a} < 15$.
$4$. Extremely weak acids: $pK_{a} > 15$.

(N/A) Magnesium hydroxide,$Mg(OH)_{2}$,is a better antacid than sodium hydrogencarbonate,$NaHCO_{3}$.
Magnesium hydroxide is insoluble in water,so it does not cause the $pH$ of the stomach to rise above neutrality,preventing the stimulation of further acid production.
In contrast,sodium hydrogencarbonate is soluble in water. An excess of it can make the stomach environment alkaline,which triggers the stomach to produce even more acid,potentially worsening the condition.

(A) The aqueous solution of hydrogen chloride $(HCl)$ gas is known as $Hydrochloric \ acid$.

(B) The reaction between $H_{2}SO_{4}$ and $HNO_{3}$ is an acid-base reaction.
In this process,$H_{2}SO_{4}$ acts as a strong acid (proton donor) and $HNO_{3}$ acts as a base (proton acceptor).
The reaction is: $HNO_{3} + H_{2}SO_{4} \rightleftharpoons H_{2}NO_{3}^{+} + HSO_{4}^{-} \rightleftharpoons NO_{2}^{+} + H_{3}O^{+} + HSO_{4}^{-}$.

(N/A) Acids from nature: $(i)$ Hydrochloric acid $(HCl)$ is present in gastric juice,which is secreted by the stomach lining and is essential for digestion. $(ii)$ Citric acid is found in lemon and orange juices. $(iii)$ Tartaric acid is found in tamarind paste.
Bases from nature: $A$ common example is magnesium hydroxide $(Mg(OH)_2)$,often used as an antacid. Bases generally turn red litmus paper blue,have a bitter taste,and feel soapy to the touch.
Salts from nature: $A$ common example is sodium chloride $(NaCl)$,also known as common salt. It is an essential component of our diet and is formed by the neutralization reaction between hydrochloric acid $(HCl)$ and sodium hydroxide $(NaOH)$.

(N/A) Definition: According to the Arrhenius theory,acids are substances that dissociate in water to give hydrogen ions $(H_{(aq)}^{+})$. According to the Arrhenius theory,bases are substances that produce hydroxyl ions $(OH_{(aq)}^{-})$.
General formula for an acid is $HX$ and for a base is $MOH$.
Ionization of acid $HX$ in aqueous solution:
$HX_{(aq)} \longrightarrow H_{(aq)}^{+} + X_{(aq)}^{-}$
$OR$
$HX_{(aq)} + H_{2}O_{(l)} \longrightarrow H_{3}O_{(aq)}^{+} + X_{(aq)}^{-}$
Note: $A$ bare proton $H^{+}$ is very reactive and cannot exist freely in aqueous solutions. Thus,it bonds to the oxygen atom of a water molecule to form a hydronium ion $(H_{3}O^{+})$.
$H_{(aq)}^{+} + H_{2}O_{(l)} \longrightarrow H_{3}O_{(aq)}^{+}$
Ionization of base $MOH$ in aqueous solution:
$MOH_{(aq)} \longrightarrow M_{(aq)}^{+} + OH_{(aq)}^{-}$
Limitations of the Arrhenius concept:
$(i)$ It is applicable only to aqueous solutions.
$(ii)$ It does not account for the basicity of substances like ammonia $(NH_{3})$ which do not possess a hydroxyl group.
$(iii)$ It does not explain the stability of the $H^{+}$ ion.

(N/A) hydrogen ion $(H^{+})$ by itself is a bare proton with a very small size (radius $\approx 10^{-15} \ m$) and an intense electric field.
Therefore,the proton binds itself to a water molecule at one of the two available lone pairs on the oxygen atom via a coordinate covalent bond,forming the hydronium ion $(H_{3}O^{+})$.
This hydronium ion $(H_{3}O^{+})$ species has a trigonal pyramidal geometry. The existence of $H_{3}O^{+}$ has been detected in many compounds,such as $H_{3}O^{+}Cl^{-}$ in the solid state.
Furthermore,the hydronium ion can be further hydrated to form various ionic species such as $H_{5}O_{2}^{+}$,$H_{7}O_{3}^{+}$,$H_{9}O_{4}^{+}$,etc.
Thus,a proton does not exist independently in an aqueous solution; instead,it exists as a hydronium ion or oxonium ion.

(N/A) The Danish chemist,Johannes Bronsted and the English chemist,Thomas $M$. Lowry provided a more general definition of acids and bases.
Definition: According to the Bronsted-Lowry theory,an acid is a substance capable of donating a hydrogen ion $(H^{+})$ and a base is a substance capable of accepting a hydrogen ion $(H^{+})$. In short,acids are proton donors and bases are proton acceptors.
Example $1$: Consider the dissolution of $NH_{3}$ in $H_{2}O$ represented by the following equation:
$NH_{3(aq)} + H_{2}O_{(l)} \rightleftharpoons NH_{4(aq)}^{+} + OH_{(aq)}^{-}$
In this reaction,the water molecule acts as a proton donor (acid) and the ammonia molecule acts as a proton acceptor (base). In the reverse reaction,$H^{+}$ is transferred from $NH_{4}^{+}$ to $OH^{-}$. Here,$NH_{4}^{+}$ acts as a Bronsted acid and $OH^{-}$ acts as a Bronsted base.
Conjugate Acid-Base Pair: An acid-base pair that differs only by one proton is called a 'conjugate acid-base pair'.
$(i)$ $NH_{4}^{+}$ and $NH_{3}$
$(ii)$ $H_{2}O$ and $OH^{-}$
Here,$OH^{-}$ is the conjugate base of the acid $H_{2}O$,and $H_{2}O$ is the conjugate acid of $OH^{-}$. $NH_{4}^{+}$ is the conjugate acid of the base $NH_{3}$,and $NH_{3}$ is the conjugate base of $NH_{4}^{+}$.
Note: $A$ conjugate acid has one extra proton,and each conjugate base has one less proton. If a Bronsted acid is strong,its conjugate base is weak,and vice-versa.
Example $2$: Consider the ionization of hydrochloric acid in water: $HCl_{(aq)} + H_{2}O_{(l)} \rightarrow H_{3}O_{(aq)}^{+} + Cl_{(aq)}^{-}$. Here,$HCl$ acts as an acid by donating a proton to the $H_{2}O$ molecule,which acts as a base.

(N/A) Definition: $A$ conjugate acid-base pair is a pair of chemical species that differ by only one proton $(H^+)$. When an acid loses a proton,it forms its conjugate base,and when a base gains a proton,it forms its conjugate acid.
For the given species:
$1$. $HNO_2$: Conjugate base is $NO_2^-$.
$2$. $CN^{-}$: Conjugate acid is $HCN$.
$3$. $HClO_4$: Conjugate base is $ClO_4^-$.
$4$. $F^{-}$: Conjugate acid is $HF$.
$5$. $OH^{-}$: Conjugate acid is $H_2O$.
$6$. $CO_3^{2-}$: Conjugate acid is $HCO_3^-$.
$7$. $S^{2-}$: Conjugate acid is $HS^{-}$.

(N/A) Definition: $G$.$N$. Lewis in $1923$ defined an acid as a species that accepts an electron pair and a base as a species that donates an electron pair.
Comparison of Bronsted-Lowry and Lewis Base: Regarding bases,there is little difference between the Bronsted-Lowry and Lewis concepts,as the base provides a lone pair in both cases. For example,the Lewis base $NH_{3}$ donates an electron pair to $H^{+}$ to form $NH_{4}^{+}$. Thus,$NH_{3}$ acts as a base in both principles. All Lewis bases are Bronsted-Lowry bases.
Comparison of Bronsted-Lowry and Lewis Acid: In the Bronsted-Lowry concept,an acid must contain a proton $(H^{+})$. However,in the Lewis concept,many acids do not contain a proton. Electron-deficient species like $BF_{3}$,$AlCl_{3}$,$FeCl_{3}$,$NO_{2}^{+}$,$Mg^{2+}$,and $Co^{3+}$ can act as Lewis acids. Thus,not all Lewis acids act as Bronsted-Lowry acids.
Example reaction: $BF_{3} + :NH_{3} \longrightarrow F_{3}B:NH_{3}$
Here,$BF_{3}$ is the Lewis acid and $:NH_{3}$ is the Lewis base.
Species like $H_{2}O$,$NH_{3}$,and $OH^{-}$ which can donate a pair of electrons act as Lewis bases.

(B) conjugate acid-base pair differs by a single proton $(H^{+})$.
In the reaction $HCl_{(aq)} + H_2O_{(aq)} \rightleftharpoons H_3O^{+}_{(aq)} + Cl^{-}_{(aq)}$:
$1$. $HCl$ acts as an acid and loses a proton to form its conjugate base,$Cl^{-}$. Thus,$(HCl, Cl^{-})$ is a conjugate acid-base pair.
$2$. $H_2O$ acts as a base and accepts a proton to form its conjugate acid,$H_3O^{+}$. Thus,$(H_3O^{+}, H_2O)$ is a conjugate acid-base pair.

(A) conjugate acid-base pair differs by a single proton $(H^+)$.
In the reaction $NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH_{4(aq)}^+ + OH_{(aq)}^-$:
$1$. $NH_3$ acts as a base and accepts a proton to form its conjugate acid $NH_4^+$. Thus,$(NH_3, NH_4^+)$ is a conjugate acid-base pair.
$2$. $H_2O$ acts as an acid and donates a proton to form its conjugate base $OH^-$. Thus,$(H_2O, OH^-)$ is a conjugate acid-base pair.

(A) According to the $Br\o nsted-Lowry$ theory,an acid is a proton $(H^{+})$ donor.
In the forward reaction,$CH_{3}COOH$ donates a proton to $H_{2}O$,so $CH_{3}COOH$ is the acid.
In the reverse reaction,$H_{3}O^{+}$ donates a proton to $CH_{3}COO^{-}$,so $H_{3}O^{+}$ is the conjugate acid.
Therefore,the species acting as acids are $CH_{3}COOH$ and $H_{3}O^{+}$.

(A) According to the Bronsted-Lowry theory,the conjugate base of a weak acid is relatively strong,and the conjugate base of a strong acid is weak.
$CH_3COOH$ is a weak acid,so its conjugate base,$CH_3COO^{-}$,is a relatively strong base.
$H_3O^{+}$ is a strong acid,so its conjugate base,$H_2O$,is a weak base.
Therefore,$H_2O$ is the weaker base compared to $CH_3COO^{-}$.

(B) Lewis acids are electron-pair acceptors,while Lewis bases are electron-pair donors.
$1$. $CO_2$: Carbon is electron-deficient in the resonance structure,acting as a Lewis acid.
$2$. $BCl_3$: Boron has an incomplete octet ($6$ electrons),acting as a Lewis acid.
$3$. $NH_3$: Nitrogen has a lone pair of electrons,acting as a Lewis base.
$4$. $CH_3NH_2$: Nitrogen has a lone pair of electrons,acting as a Lewis base.
$5$. $NO_2^+$: The positive charge makes it an electron-pair acceptor,acting as a Lewis acid.
$6$. $C_6H_5NH_3^+$: The positive charge on the nitrogen atom makes it an electron-pair acceptor,acting as a Lewis acid.
Therefore,the classification is:
Acids: $CO_2, BCl_3, NO_2^+, C_6H_5NH_3^+$
Bases: $NH_3, CH_3NH_2$

(A) Lewis acid is an electron pair acceptor,while a Bronsted acid is a proton $(H^{+})$ donor.
$1$. $CO_{2}$: It has no $H$ atom to donate,so it is not a Bronsted acid. It acts as a Lewis acid by accepting electron pairs from bases like $OH^{-}$.
$2$. $BCl_{3}$: It has an incomplete octet (electron deficient),acting as a Lewis acid. It has no $H$ atom,so it is not a Bronsted acid.
$3$. $NO_{2}^{+}$: It is electron deficient (positive charge),acting as a Lewis acid. It has no $H$ atom,so it is not a Bronsted acid.
$4$. $NH_{3}$ and $CH_{3}NH_{2}$: These are Bronsted bases (proton acceptors) and Lewis bases (electron pair donors).
$5$. $C_{6}H_{5}NH_{3}^{+}$: This is a Bronsted acid as it can donate a proton.
Therefore,the species that are only Lewis acids but not Bronsted acids are $CO_{2}, BCl_{3}$,and $NO_{2}^{+}$.

(N/A) The conjugate base is formed by removing a proton $(H^{+})$ from the acid.
$HF - H^{+} \rightarrow F^{-}$
$CH_{3}NH_{3}^{+} - H^{+} \rightarrow CH_{3}NH_{2}$
$H_{3}PO_{4} - H^{+} \rightarrow H_{2}PO_{4}^{-}$
$HPO_{4}^{2-} - H^{+} \rightarrow PO_{4}^{3-}$

The conjugate acid is formed by adding a proton $(H^{+})$ to the given base.
$1$. For $HS^{-}$,the conjugate acid is $HS^{-} + H^{+} \rightarrow H_{2}S$.
$2$. For $NH_{3}$,the conjugate acid is $NH_{3} + H^{+} \rightarrow NH_{4}^{+}$.
$3$. For $C_{6}H_{5}COO^{-}$,the conjugate acid is $C_{6}H_{5}COO^{-} + H^{+} \rightarrow C_{6}H_{5}COOH$.
$4$. For $OH^{-}$,the conjugate acid is $OH^{-} + H^{+} \rightarrow H_{2}O$.
Therefore,the conjugate acids are $H_{2}S, NH_{4}^{+}, C_{6}H_{5}COOH, H_{2}O$.

(N/A) The conjugate acid is formed by adding $H^+$ and the conjugate base is formed by removing $H^+$.
$(i)$ For $(CH_3)_2NH$: Conjugate acid is $(CH_3)_2NH_2^+$ and conjugate base is $(CH_3)_2N^-$.
$(ii)$ For $HPO_4^{2-}$: Conjugate acid is $H_2PO_4^-$ and conjugate base is $PO_4^{3-}$.
$(iii)$ For $HS^-$: Conjugate acid is $H_2S$ and conjugate base is $S^{2-}$.

(N/A) $(i)$ $CH_3NH_2$ (Base) $+ H_2O$ (Acid) $\rightleftharpoons$ $CH_3NH_3^+ + OH^-$
$(ii)$ $CO_2$ (Acid) $+ H_2O$ (Base) $\rightleftharpoons$ $H_2CO_3$
$(iii)$ $H_2PO_4^-$ (Acid) $+ CO_3^{2-}$ (Base) $\rightleftharpoons$ $HPO_4^{2-} + HCO_3^-$
$(iv)$ $NH_2NH_2$ (Base) $+ H_2O$ (Acid) $\rightleftharpoons$ $NH_2NH_3^+ + OH^-$
$(v)$ $C_6H_6$ (Base) $+ NO_2^+$ (Acid) $\rightleftharpoons$ $C_6H_5NO_2 + H^+$
$(vi)$ $C_6H_6$ (Acid) $+ NH_2^-$ (Base) $\rightleftharpoons$ $C_6H_5^- + NH_3$

(N/A) The conjugate bases are formed by removing a proton $(H^+)$ from the acid.
$1$. For $HClO_4$,the conjugate base is $ClO_4^-$.
$2$. For $H_2SO_4$,the conjugate base is $HSO_4^-$.
$3$. For $HNO_3$,the conjugate base is $NO_3^-$.
$4$. For $HPO_4^{2-}$,the conjugate base is $PO_4^{3-}$.
Since $HClO_4$,$H_2SO_4$,and $HNO_3$ are strong acids,their conjugate bases ($ClO_4^-$,$HSO_4^-$,$NO_3^-$) are very weak. $HPO_4^{2-}$ is a weak acid,so its conjugate base $PO_4^{3-}$ is relatively stronger.

(A) In the given reaction,$HCl$ acts as a Bronsted-Lowry acid by donating a proton to $H_2O$. The resulting $H_3O^{+}$ ion acts as the conjugate acid of $H_2O$. Between $HCl$ and $H_3O^{+}$,$HCl$ is a strong acid because it undergoes complete dissociation in aqueous solution.

(A) In the reaction $HCl_{(aq)} + H_2O_{(aq)} \rightleftharpoons H_3O^{+}_{(aq)} + Cl^{-}_{(aq)}$,$H_2O$ acts as a base by accepting a proton $(H^+)$ to form $H_3O^+$.
$Cl^-$ is the conjugate base of the strong acid $HCl$.
According to the Brønsted-Lowry theory,the conjugate base of a strong acid is always a very weak base.
Therefore,$H_2O$ and $Cl^-$ are the bases,and $Cl^-$ is the weak base.

(A-D) Arrhenius Acid-Base Theory: Strength is determined by the degree of ionization in aqueous solution.
- Strong Acids: Acids that undergo complete ionization in aqueous solution. Examples: Perchloric acid $(HClO_4)$,Hydrochloric acid $(HCl)$,Hydrobromic acid $(HBr)$,Hydroiodic acid $(HI)$,Nitric acid $(HNO_3)$,and Sulphuric acid $(H_2SO_4)$.
- Strong Bases: Bases that undergo complete ionization in aqueous solution. Examples: Lithium hydroxide $(LiOH)$,Sodium hydroxide $(NaOH)$,Potassium hydroxide $(KOH)$,Caesium hydroxide $(CsOH)$,and Barium hydroxide $(Ba(OH)_2)$.
- Weak Acid-Base: These undergo partial ionization in aqueous solution and exist primarily as unionized molecules. Examples: $Mg(OH)_2$ and $Ca(OH)_2$ are weak bases; $HCN, H_2S, H_3PO_4$ are weak acids.
$(B)$ Bronsted-Lowry Acid-Base Theory: $A$ strong acid is a good proton donor,and a strong base is a good proton acceptor.
- Strong Acid: Strong acids donate protons readily. Their conjugate bases are very weak. Examples: $(HClO_4, HCl, HBr, HI, HNO_3, H_2SO_4)$ are strong acids because they are excellent proton donors.
- Conjugate Bases: The conjugate bases of these strong acids,such as $(ClO_4^-, Cl^-, Br^-, I^-, NO_3^-, HSO_4^-)$,are much weaker bases than water.

(N/A) The characteristics and uses of the $K_{a}$ value are as follows:
$(i)$ The larger the value of $K_{a}$,the stronger the acid.
$(ii)$ $K_{a}$ is a dimensionless quantity.
$(iii)$ The $[H^{+}]$ concentration of a weak acid and its $pH$ can be calculated using the value of $K_{a}$.
$(iv)$ The degree of ionization $\alpha$ can be calculated with the help of the $K_{a}$ value.
$(v)$ $pK_{a}$ is calculated using the value of $K_{a}$ as $pK_{a} = -\log(K_{a})$.
If the $pK_{a}$ value is higher,the acid becomes less strong.

$K_{a} = 1 \times 10^{-1}$	$K_{a} = 1 \times 10^{-2}$	$K_{a} = 1 \times 10^{-3}$
$pK_{a} = 1$	$pK_{a} = 2$	$pK_{a} = 3$

(N/A) $(i)$ $A$ higher value of $K_{b}$ indicates a stronger base.
$(ii)$ $K_{b}$ is a dimensionless quantity.
$(iii)$ $K_{b}$ is used to calculate the concentration of $[OH^{-}]$ ions for a weak base,which is then used to determine the $pOH$.
$(iv)$ The degree of ionization $(\alpha)$ of a base can be calculated using the value of $K_{b}$.
$(v)$ $pK_{b}$ is calculated using the value of $K_{b}$ as $pK_{b} = -\log(K_{b})$. $A$ higher $pK_{b}$ value indicates a weaker base.

$K_{b}$	$1 \times 10^{-1}$	$1 \times 10^{-2}$	$1 \times 10^{-3}$
$pK_{b}$	$+1$	$+2$	$+3$

(A) Weak acids and weak bases partially dissociate in aqueous solution,establishing an equilibrium between the undissociated molecules and the ions.

$A$. Weak Acids	Ionic Equilibriums
$1$. Acetic acid $(CH_{3}COOH)$	$CH_{3}COOH_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + CH_{3}COO^{-}_{(aq)}$
$2$. Benzoic acid $(C_{6}H_{5}COOH)$	$C_{6}H_{5}COOH_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + C_{6}H_{5}COO^{-}_{(aq)}$
$3$. Hydrocyanic acid $(HCN)$	$HCN_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + CN^{-}_{(aq)}$
$4$. Formic acid $(HCOOH)$	$HCOOH_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O^{+}_{(aq)} + HCOO^{-}_{(aq)}$

$B$. Weak Bases	Ionic Equilibriums
$1$. Ammonia $(NH_{3})$	$NH_{3(aq)} + H_{2}O_{(l)} \rightleftharpoons N{H_{4}}^{+}_{(aq)} + OH^{-}_{(aq)}$
$2$. Aniline $(C_{6}H_{5}NH_{2})$	$C_{6}H_{5}NH_{2(aq)} + H_{2}O_{(l)} \rightleftharpoons C_{6}H_{5}N{H_{3}}^{+}_{(aq)} + OH^{-}_{(aq)}$
$3$. Methylamine $(CH_{3}NH_{2})$	$CH_{3}NH_{2(aq)} + H_{2}O_{(l)} \rightleftharpoons CH_{3}N{H_{3}}^{+}_{(aq)} + OH^{-}_{(aq)}$
$4$. Dimethylamine $(CH_{3})_{2}NH$	$(CH_{3})_{2}NH_{(aq)} + H_{2}O_{(l)} \rightleftharpoons (CH_{3})_{2}N{H_{2}}^{+}_{(aq)} + OH^{-}_{(aq)}$

(N/A) $NH_3$ (ammonia) is a weak base and its conjugate acid is $NH_4^+$. The following equilibrium is established for $NH_3$ in water:
$(i) NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH_{4(aq)}^{+} + OH_{(aq)}^{-}$
$K_b = \frac{[NH_4^+][OH^{-}]}{[NH_3]}$
$NH_4^+$ acts as a conjugate acid. Its equilibrium in aqueous solution is:
$(ii) NH_{4(aq)}^{+} + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^{+} + NH_{3(aq)}$
The ionization constant for the weak acid $NH_4^+$ is $K_a$:
$K_a = \frac{[H_3O^{+}][NH_3]}{[NH_4^+]}$
Adding reaction $(i)$ and $(ii)$ gives the net reaction:
$2H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^{+} + OH_{(aq)}^{-}$
This is the auto-ionization of water,where $K_w = [H_3O^{+}][OH^{-}]$.
Multiplying the equilibrium constants $K_a$ and $K_b$:
$K_a \times K_b = \frac{[H_3O^{+}][NH_3]}{[NH_4^+]} \times \frac{[NH_4^+][OH^{-}]}{[NH_3]}$
$K_a \times K_b = [H_3O^{+}][OH^{-}] = K_w$
Thus,for a conjugate acid-base pair,$K_a \times K_b = K_w$.

(N/A) $NH_3$ (ammonia) is a weak base and its conjugate acid is $NH_4^+$. The following equilibrium is established for $NH_3$:
$(i) NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH_{4(aq)}^+ + OH_{(aq)}^-$
$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$
$NH_4^+$ is the conjugate acid of $NH_3$. Its equilibrium in aqueous solution is:
$(ii) NH_{4(aq)}^+ + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^+ + NH_{3(aq)}$
The ionization constant of the weak acid $NH_4^+$ is $K_a$:
$K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]}$
Adding reaction $(i)$ and $(ii)$ gives the net reaction:
$(i) + (ii) = 2H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^+ + OH_{(aq)}^-$
This is the self-ionization of water,where $K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$.
Multiplying the equilibrium constants $K_a$ and $K_b$:
$K_a \times K_b = \frac{[H_3O^+][NH_3]}{[NH_4^+]} \times \frac{[NH_4^+][OH^-]}{[NH_3]}$
$K_a \times K_b = [H_3O^+][OH^-] = K_w$
Thus,$K_a \times K_b = K_w$.

(N/A) For a weak base $B$,the equilibrium in aqueous solution is:
$B_{(aq)} + H_2O_{(l)} \rightleftharpoons BH^{+}_{(aq)} + OH^{-}_{(aq)} \quad \dots (i)$
The base dissociation constant $K_b$ is given by:
$K_b = \frac{[BH^{+}][OH^{-}]}{[B]}$
For the conjugate acid $BH^{+}$,the dissociation equilibrium is:
$BH^{+}_{(aq)} + H_2O_{(l)} \rightleftharpoons B_{(aq)} + H_3O^{+}_{(aq)} \quad \dots (ii)$
The acid dissociation constant $K_a$ is given by:
$K_a = \frac{[B][H_3O^{+}]}{[BH^{+}]}$
Multiplying $K_a$ and $K_b$:
$K_a \times K_b = \left( \frac{[B][H_3O^{+}]}{[BH^{+}]} \right) \times \left( \frac{[BH^{+}][OH^{-}]}{[B]} \right)$
$K_a \times K_b = [H_3O^{+}][OH^{-}] = K_w$
Taking the negative logarithm on both sides:
$-\log(K_a \times K_b) = -\log(K_w)$
$-\log K_a - \log K_b = -\log K_w$
$pK_a + pK_b = pK_w$

As an example,the ionization of a dibasic acid $H_2X$ in aqueous solution is represented in two steps:
$(i) \ H_2X_{(aq)} + H_2O_{(l)} \rightleftharpoons H^{+}_{(aq)} + HX^{-}_{(aq)}$
$(ii) \ HX^{-}_{(aq)} + H_2O_{(l)} \rightleftharpoons H^{+}_{(aq)} + X^{2-}_{(aq)}$
If the equilibrium constants for these steps are $K_{a1}$ and $K_{a2}$ respectively,then:
$K_{a1} = \frac{[H^{+}][HX^{-}]}{[H_2X]}$ and $K_{a2} = \frac{[H^{+}][X^{2-}]}{[HX^{-}]}$
The overall reaction is the sum of $(i)$ and $(ii)$:
$H_2X_{(aq)} + 2H_2O_{(l)} \rightleftharpoons 2H^{+}_{(aq)} + X^{2-}_{(aq)}$
The overall equilibrium constant $K_a$ is given by:
$K_a = \frac{[H^{+}]^2 [X^{2-}]}{[H_2X]} = K_{a1} \times K_{a2}$
For any polyprotic acid,the overall dissociation constant is the product of the individual ionization constants:
$K_a = K_{a1} \times K_{a2} \times K_{a3} \dots$
Generally,$K_{a1} > K_{a2} > K_{a3} \dots$ because it is progressively more difficult to remove a positively charged proton from a negatively charged ion.

Polyprotic acids are acids that have more than one ionizable proton per molecule. These are also known as polybasic acids.
$1$. Diprotic acid: An acid that has two ionizable protons per molecule. It ionizes in two steps:
$(i)$ $H_2X + H_2O \rightleftharpoons H_3O^{+} + HX^{-}, K_{a(1)}$
$(ii)$ $HX^{-} + H_2O \rightleftharpoons H_3O^{+} + X^{2-}, K_{a(2)}$
Here,$K_{a(1)} > K_{a(2)}$. Examples: $H_2SO_4$,$H_2CO_3$,$H_2C_2O_4$.
$2$. Triprotic acid: An acid that has three ionizable protons per molecule. It ionizes in three steps:
$(i)$ $H_3A + H_2O \rightleftharpoons H_3O^{+} + H_2A^{-}, K_{a(1)}$
$(ii)$ $H_2A^{-} + H_2O \rightleftharpoons H_3O^{+} + HA^{2-}, K_{a(2)}$
$(iii)$ $HA^{2-} + H_2O \rightleftharpoons H_3O^{+} + A^{3-}, K_{a(3)}$
Here,$K_{a(1)} > K_{a(2)} > K_{a(3)}$. Example: $H_3PO_4$.

Polyprotic acid: Acids that have more than one ionizable proton per molecule. These are also known as polybasic acids.
Diprotic acid: An acid that has two ionizable protons per molecule. These are also known as dibasic acids.
General ionization for a diprotic acid $(H_2X)$:
$H_2X_{(aq)} \rightleftharpoons 2H^+_{(aq)} + X^{2-}_{(aq)}$
The ionization occurs in two steps:
$(i) H_2X + H_2O \rightleftharpoons H_3O^+ + HX^- \quad K_{a1}$
$(ii) HX^- + H_2O \rightleftharpoons H_3O^+ + X^{2-} \quad K_{a2}$
Here,$K_{a1} > K_{a2}$ and the overall dissociation constant $K_a = K_{a1} \times K_{a2}$.
Examples of diprotic acids: Oxalic acid $(H_2C_2O_4)$,Sulphuric acid $(H_2SO_4)$,Carbonic acid $(H_2CO_3)$,and Sulphurous acid $(H_2SO_3)$.
Examples of triprotic acids: Phosphoric acid $(H_3PO_4)$ and Citric acid.
In a polyprotic acid solution,a mixture of various acid species like $H_2A$,$HA^-$,and $A^{2-}$ exists in equilibrium.

(N/A) The strength of an acid is determined by its ability to donate $H^+$ ions.
$(i)$ Bond Strength: As the strength of the $H-A$ bond decreases,the energy required to break the bond decreases,making $HA$ a stronger acid.
$(ii)$ Bond Polarity: As the electronegativity difference between $H$ and $A$ increases,the bond becomes more polar,facilitating the release of $H^+$. Thus,$\text{Polarity} \propto \text{Acidity}$.
$(iii)$ Trends in a Period: Across a period,electronegativity of $A$ increases,which increases bond polarity. For example: $CH_4 < NH_3 < H_2O < HF$. Here,acid strength increases as electronegativity of $A$ increases.
$(iv)$ Trends in a Group: Down a group,the size of $A$ increases,which significantly decreases the $H-A$ bond strength. This is the dominant factor. For example: $HF < HCl < HBr < HI$. Here,acid strength increases as the size of $A$ increases.

(A) For a conjugate acid-base pair,the relationship between the dissociation constant of the acid $(K_a)$ and the dissociation constant of its conjugate base $(K_b)$ is given by the equation: $K_a \times K_b = K_w$.
Given $K_a = 1.8 \times 10^{-4}$ and $K_w = 1.0 \times 10^{-14}$ at $298 \ K$.
Substituting the values: $K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}}$.
$K_b = 0.555 \times 10^{-10} = 5.55 \times 10^{-11}$.

(A) For a conjugate acid-base pair,the relationship between the dissociation constants is given by: $K_a \times K_b = K_w$.
Given: $K_a = 1.76 \times 10^{-5}$ and $K_w = 1.00 \times 10^{-14}$ at $298 \ K$.
$K_b = \frac{K_w}{K_a} = \frac{1.00 \times 10^{-14}}{1.76 \times 10^{-5}}$.
$K_b \approx 5.68 \times 10^{-10}$.

(A) The conjugate base of an acid is formed by removing one proton $({H^+})$ from the species.
For ${H_2}PO_3^-$,removing one ${H^+}$ gives ${HPO_3^{2-}}$.
The conjugate acid of a base is formed by adding one proton $({H^+})$ to the species.
For ${HCO_3^-}$,adding one ${H^+}$ gives ${H_2CO_3}$.
Therefore,the conjugate base is ${HPO_3^{2-}}$ and the conjugate acid is ${H_2CO_3}$.

(N/A) $1$. $A$ $Brønsted-Lowry$ base is a proton $(H^+)$ acceptor. $NH_3$ has a lone pair of electrons on the nitrogen atom,which allows it to accept a proton to form the ammonium ion $(NH_4^+)$. Thus,$NH_3 + H^+ \rightarrow NH_4^+$.
$2$. $A$ $Lewis$ base is an electron pair donor. Since $NH_3$ has a lone pair,it can donate this pair to an electron-deficient species,acting as a $Lewis$ base.
$3$. $A$ $Lewis$ acid is an electron pair acceptor. $BF_3$ has an incomplete octet (only $6$ electrons around $B$),making it electron-deficient and capable of accepting an electron pair.
$4$. $A$ $Brønsted-Lowry$ acid is a proton $(H^+)$ donor. $BF_3$ does not contain any hydrogen atoms,so it cannot donate a proton,and therefore,it cannot act as a $Brønsted-Lowry$ acid.

(C) Toilet cleaning liquid typically contains $HCl$ (hydrochloric acid).
To neutralize the acid on the skin,a mild base like aqueous $NaHCO_{3}$ (sodium bicarbonate) is used.
$NaOH$ is avoided because it is highly corrosive and can cause severe chemical burns.

(B) Compounds that are more acidic than $H_{2}CO_{3}$ (carbonic acid) will react with $NaHCO_{3}$ to liberate $CO_{2}$.
$(CH_{3})_{3}NH^{+}Cl^{-}$ is a salt of a weak base and a strong acid,making the cation $(CH_{3})_{3}NH^{+}$ acidic enough to react with $HCO_{3}^{-}$:
$(CH_{3})_{3}NH^{+}Cl^{-} + NaHCO_{3} \longrightarrow (CH_{3})_{3}N + NaCl + H_{2}CO_{3}$
$H_{2}CO_{3} \longrightarrow H_{2}O + CO_{2} \uparrow$
Therefore,option $B$ is the correct answer.

(D) The amphoteric nature of water is explained by the Brønsted-Lowry acid/base concept,not the Lewis concept.
In the reaction $H_{2}O + NH_{3} \rightleftharpoons NH_{4}^{+} + OH^{-}$,water acts as a Brønsted-Lowry acid (proton donor).
In the reaction $H_{2}S + H_{2}O \rightleftharpoons H_{3}O^{+} + HS^{-}$,water acts as a Brønsted-Lowry base (proton acceptor).
Since Assertion $A$ claims the Lewis concept explains this,$A$ is false.
Reason $R$ correctly describes the behavior of water in these reactions,so $R$ is true.
Therefore,$A$ is false but $R$ is true.

(D) Bronsted acid is a species that can donate $H^{+}$ ions,and a Bronsted base is a species that can accept $H^{+}$ ions.
$HCO_3^{-}$ can act as both:
$HCO_3^{-} + H^{+} \rightleftharpoons H_2CO_3$ (Acting as a Bronsted base)
$HCO_3^{-} \rightleftharpoons H^{+} + CO_3^{2-}$ (Acting as a Bronsted acid)
Such species are known as amphoteric species.

(B) conjugate base is formed by the removal of a proton $(H^+)$ from an acid.
For $HCO_3^-$,removing a proton yields $CO_3^{2-}$.
For $NH_3$,removing a proton yields $NH_2^-$.
Thus,the conjugate bases are $CO_3^{2-}$ and $NH_2^-$.
$HCO_3^- \rightleftharpoons CO_3^{2-} + H^+$
$NH_3 \rightleftharpoons NH_2^- + H^+$

(B) The basic strength of a conjugate base is inversely proportional to the acidic strength of its corresponding acid.
The order of acidic strength of the corresponding acids is:
$HCl > CH_3COOH > H_2O > ROH$
Since $HCl$ is the strongest acid,its conjugate base $Cl^{-}$ is the weakest base.
Since $ROH$ (alcohol) is a weaker acid than $H_2O$,its conjugate base $RO^{-}$ is a stronger base than $OH^{-}$.
Therefore,the decreasing order of basic strength is:
$RO^{-} > OH^{-} > CH_3COO^{-} > Cl^{-}$

(D) diprotic acid is an acid that can donate two protons ($H^+$ ions) per molecule.
$1$. $H_3PO_4$: Triprotic acid.
$2$. $H_2SO_4$: Diprotic acid.
$3$. $H_3PO_3$: Diprotic acid (contains two $P-OH$ bonds).
$4$. $H_2CO_3$: Diprotic acid.
$5$. $H_2S_2O_7$: Diprotic acid.
$6$. $H_3BO_3$: Monoprotic acid (Lewis acid).
$7$. $H_3PO_2$: Monoprotic acid (contains one $P-OH$ bond).
$8$. $H_2CrO_4$: Diprotic acid.
$9$. $H_2SO_3$: Diprotic acid.
The diprotic acids are: $H_2SO_4, H_3PO_3, H_2CO_3, H_2S_2O_7, H_2CrO_4, H_2SO_3$.
Total count = $6$.

(A) The increase in temperature is directly proportional to the amount of heat released during the neutralization reaction. \\ The heat of neutralization is maximum for the reaction between a strong acid and a strong base. \\ In option $(A)$,$30 \ mmol$ of $HCl$ (strong acid) reacts with $30 \ mmol$ of $NaOH$ (strong base),resulting in complete neutralization. \\ In option $(B)$,$CH_3COOH$ is a weak acid,so some heat is absorbed during its dissociation,leading to a lower temperature rise compared to $(A)$. \\ Options $(C)$ and $(D)$ involve fewer moles of reaction compared to $(A)$. \\ Therefore,the largest increase in temperature occurs in option $(A)$.

(D) $H_3PO_4 \rightleftharpoons H^{+} + H_2PO_4^{-}; K_{a_1}$
$H_2PO_4^{-} \rightleftharpoons H^{+} + HPO_4^{2-}; K_{a_2}$
$HPO_4^{2-} \rightleftharpoons H^{+} + PO_4^{3-}; K_{a_3}$
Adding these steps,the overall reaction is: $H_3PO_4 \rightleftharpoons 3H^{+} + PO_4^{3-}$.
The overall ionization constant is $K = K_{a_1} \times K_{a_2} \times K_{a_3}$.
Taking $\log$ on both sides: $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$. Thus,statement $A$ is true.
For polybasic acids,the order of ionization constants is $K_{a_1} > K_{a_2} > K_{a_3}$ because as the negative charge on the anion increases,the electrostatic attraction for the $H^{+}$ ion increases,making it harder to remove subsequent protons. Thus,statement $C$ is true.
Since $K_{a_1} > K_{a_2} > K_{a_3}$,$H_3PO_4$ is the strongest acid among the species,making statement $B$ true.
Statement $D$ is incorrect as there is no such relationship.

(C) To determine the basic strength,we analyze the nature of the compounds:
$(I)$ $KOH$ is a strong base (alkali metal hydroxide).
$(II)$ $Zn(OH)_2$ is an amphoteric hydroxide.
$(III)$ $HNO_3$ is a strong acid.
$(IV)$ $Ca(OH)_2$ is a strong base (alkaline earth metal hydroxide).
Comparing the basicity: $KOH$ (Group $1$) is a stronger base than $Ca(OH)_2$ (Group $2$). $Zn(OH)_2$ is amphoteric,meaning it acts as a weak base compared to strong bases but is more basic than the strong acid $HNO_3$.
Therefore,the order of basic strength is $KOH > Ca(OH)_2 > Zn(OH)_2 > HNO_3$,which corresponds to $I > IV > II > III$.