Acids and Bases Questions in English

(A) Bronsted-Lowry acid is defined as a substance that releases $H^+$ ions,while a Bronsted-Lowry base is defined as a substance that accepts $H^+$ ions.
In the given options,$Na_2CO_3$ acts as a base by accepting $H^+$ ions.
$SO_4^{2-}$ can only accept $H^+$ ions to form $HSO_4^-$,so it acts only as a base.
$HSO_4^-$ can accept $H^+$ to form $H_2SO_4$ (acting as a base) and can release $H^+$ to form $SO_4^{2-}$ (acting as an acid).
Therefore,$HSO_4^-$ can act as both a Bronsted acid and a Bronsted base.

(B) Bronsted-Lowry acid is a proton $(H^+)$ donor,and a Bronsted-Lowry base is a proton $(H^+)$ acceptor.
For $HCO_3^{-}$:
$1$. As a base: $HCO_3^{-} + H^{+} \rightleftharpoons H_2CO_3$
$2$. As an acid: $HCO_3^{-} \rightleftharpoons H^{+} + CO_3^{2-}$
Since $HCO_3^{-}$ can both donate and accept a proton,it acts as an amphoteric species (both acid and base).

(C) salt can furnish $H^{+}$ ions in an aqueous solution if it contains at least one ionizable hydrogen atom directly bonded to an oxygen atom ($-OH$ group).
$NaH_2PO_2$ (Sodium hypophosphite) has no $P-OH$ bond,so it cannot furnish $H^{+}$ ions.
$Na_2HPO_3$ (Sodium phosphite) has no $P-OH$ bond,so it cannot furnish $H^{+}$ ions.
$Na_2HPO_4$ (Disodium hydrogen phosphate) contains a $P-OH$ bond,which allows it to dissociate in water to release $H^{+}$ ions: $Na_2HPO_4 \rightleftharpoons 2Na^{+} + HPO_4^{2-} + H^{+}$.
Therefore,only $Na_2HPO_4$ can furnish $H^{+}$ ions.

(C) For the dissociation of a weak acid $HF$:
$HF + H_2O \rightleftharpoons H_3O^{+} + F^{-}$,$K_a = \frac{[H_3O^{+}][F^{-}]}{[HF]}$
For the hydrolysis of its conjugate base $F^{-}$:
$F^{-} + H_2O \rightleftharpoons HF + OH^{-}$,$K_b = \frac{[HF][OH^{-}]}{[F^{-}]}$
Multiplying $K_a$ and $K_b$:
$K_a \times K_b = \frac{[H_3O^{+}][F^{-}]}{[HF]} \times \frac{[HF][OH^{-}]}{[F^{-}]} = [H_3O^{+}][OH^{-}] = K_w$
Therefore,the correct relation is $K_a \times K_b = K_w$.

(A) According to the Bronsted-Lowry concept,an acid is a proton $(H^+)$ donor.
$1$. In the forward reaction,$[Al(H_2O)_6]^{+3}$ donates a proton to $HCO_3^-$ to form $[Al(H_2O)_5OH]^{+2}$ and $H_2CO_3$. Thus,$[Al(H_2O)_6]^{+3}$ (labeled $A$) acts as an acid.
$2$. In the reverse reaction,$H_2CO_3$ (labeled $D$) donates a proton to $[Al(H_2O)_5OH]^{+2}$ to form $[Al(H_2O)_6]^{+3}$ and $HCO_3^-$. Thus,$H_2CO_3$ acts as an acid.
Therefore,the species behaving as Bronsted-Lowry acids are $(A)$ and $(D)$.

(C) An amphiprotic species is one that can both donate and accept a proton $(H^+)$.
$H_3O^+$ can only donate a proton,so it is not amphiprotic.
$H_2O$ can accept a proton to form $H_3O^+$ and donate a proton to form $OH^-$.
$HPO_4^{2-}$ can accept a proton to form $H_2PO_4^-$ and donate a proton to form $PO_4^{3-}$.
$HCO_3^-$ can accept a proton to form $H_2CO_3$ and donate a proton to form $CO_3^{2-}$.
$H_2PO_4^-$ can accept a proton to form $H_3PO_4$ and donate a proton to form $HPO_4^{2-}$.
$H_2PO_3^-$ can accept a proton to form $H_3PO_3$ and donate a proton to form $HPO_3^{2-}$.
Since $H_3O^+$ is not amphiprotic,option $A$ is incorrect.
However,in many contexts,the question implies identifying sets where all members are amphiprotic. Given the options,$C$ contains species that are all amphiprotic.

(C) According to the Brønsted-Lowry theory,an acid is a proton $(H^+)$ donor and a base is a proton acceptor.
In the reaction $NH_4^+ + S^{2-} \rightleftharpoons NH_3 + HS^-$,$NH_4^+$ acts as an acid by donating a proton to $S^{2-}$.
$NH_3$ is the conjugate base of the acid $NH_4^+$.
$HS^-$ is the conjugate acid of the base $S^{2-}$.
Therefore,$NH_3$ and $HS^-$ represent a conjugate acid-base pair.

(A) conjugate acid is formed when a base accepts a proton $(H^+)$.
To find the conjugate acid of a species,we add one $H^+$ ion to it.
For the ion $H_2PO_4^-$,the addition of a proton is represented as:
$H_2PO_4^- + H^+ \rightarrow H_3PO_4$.
Therefore,the conjugate acid of $H_2PO_4^-$ is $H_3PO_4$.

(C) The strength of an acid is directly proportional to its $K_a$ value and inversely proportional to its $pK_a$ value.
$pK_a = -\log(K_a)$.
We convert all values to $pK_a$ for comparison:
$HA_1: pK_a = 10$.
$HA_2: K_a = 10^{-4} \implies pK_a = -\log(10^{-4}) = 4$.
$HA_3: pK_a = 2$.
$HA_4: K_a = 10^{-7} \implies pK_a = -\log(10^{-7}) = 7$.
Comparing the $pK_a$ values: $2 < 4 < 7 < 10$.
Since the acid with the lowest $pK_a$ is the strongest,$HA_3$ is the strongest acid.

(A) The strength of a conjugate base is inversely proportional to the strength of the corresponding acid.
$(A)$ $H_2S$ $(pK_a \approx 7)$ is a stronger acid than $H_2O$ $(pK_a \approx 15.7)$,so $H_2O$ $(2)$ has a stronger conjugate base.
$(B)$ Glutarimide $(3)$ has two carbonyl groups withdrawing electron density,making it a stronger acid than Piperidin$-2-$one $(4)$. Thus,Piperidin$-2-$one $(4)$ has a stronger conjugate base.
$(C)$ $CH_3C \equiv CH$ $(5)$ ($sp$ hybridized carbon) is a stronger acid than $CH_3CH = CH_2$ $(6)$ ($sp^2$ hybridized carbon). Thus,$CH_3CH = CH_2$ $(6)$ has a stronger conjugate base.
Therefore,the acids with stronger conjugate bases are $2, 4, 6$.

(A) The equilibrium reaction is: $(CH_3)_3CO^- + H_2O \rightleftharpoons (CH_3)_3COH + OH^-$.
Acid-base reactions always proceed in the direction of forming the weaker acid and the weaker base.
Comparing the $pK_a$ values: $H_2O$ $(pK_a = 15.7)$ is a stronger acid than $(CH_3)_3COH$ $(pK_a = 18)$.
Since the reaction proceeds from the stronger acid to the weaker acid,the equilibrium lies to the right (towards the products).
Therefore,$(CH_3)_3CO^-$ is a stronger base than $OH^-$,and the equilibrium favours the formation of the weaker base $(OH^-)$ and the weaker acid $((CH_3)_3COH)$.
Thus,$t$-butoxide is the dominant anionic species is incorrect,and the equilibrium favours the products.

(D) The given reaction is: $KF + HCOOH \rightleftharpoons HCOOK + HF$.
Comparing the $pK_a$ values: $pK_a(HCOOH) = 3.8$ and $pK_a(HF) = 3.2$.
Since $pK_a(HF) < pK_a(HCOOH)$,$HF$ is a stronger acid than $HCOOH$.
Therefore,statement $(B)$ is true.
Stronger acids have weaker conjugate bases. Since $HF$ is a stronger acid,its conjugate base $F^-$ is weaker than the conjugate base of formic acid,$HCOO^-$.
Thus,$HCOO^-$ (present in $KO_2CH$) is a stronger base than $F^-$ (present in $KF$). Therefore,statement $(D)$ is true.
Statement $(H)$ is true because $F^-$ is the conjugate base of the stronger acid $HF$.
Acid-base reactions always proceed in the direction of the formation of the weaker acid and weaker base.
Since $HF$ (stronger acid) is on the product side and $HCOOH$ (weaker acid) is on the reactant side,the equilibrium will shift towards the reactants. Therefore,statement $(E)$ is true.
The correct statements are $(B), (D), (E),$ and $(H)$.

(A) The basic strength of anions is inversely proportional to the acidic strength of their conjugate acids.
The conjugate acids are: $(p) \text{CH}_4, (q) \text{NH}_3, (r) \text{H}_2\text{O}, (s) \text{HF}$.
The acidic strength order of these acids is: $\text{HF} > \text{H}_2\text{O} > \text{NH}_3 > \text{CH}_4$.
Therefore,the basic strength order of their conjugate bases is: $\text{CH}_3^{-} > \text{NH}_2^{-} > \text{OH}^{-} > \text{F}^{-}$,which corresponds to $p > q > r > s$.

(A) The equilibrium reaction is: $CH_3OH + (CH_3)_2NH \rightleftarrows CH_3O^- + (CH_3)_2NH_2^+$.
Comparing the $pK_a$ values: $pK_a$ of $CH_3OH = 15.2$ and $pK_a$ of $(CH_3)_2NH_2^+ = 10.7$.
Since $15.2 > 10.7$,$CH_3OH$ is a weaker acid $(WA)$ than $(CH_3)_2NH_2^+$.
Consequently,$(CH_3)_2NH_2^+$ is the stronger acid $(SA)$.
In an acid-base equilibrium,the side with the weaker acid and weaker base is favored.
Here,$CH_3OH$ $(1)$ is the $WA$ and $(CH_3)_2NH$ $(2)$ is the $WB$.

(D) An acid-base reaction always proceeds in the direction of the formation of a weaker acid and a weaker base.
In option $A$,$EtO^-$ is a stronger base than $C_6H_5O^-$ and $C_6H_5OH$ is a stronger acid than $EtOH$,so the forward reaction is favoured.
In option $B$,$KH$ is a very strong base,so the forward reaction is favoured.
In option $C$,$Me_3CO^-$ is a stronger base than $OH^-$,so the forward reaction is favoured.
In option $D$,$CH_3O^-$ is a stronger base than $C_{10}H_7O^-$ (naphthoxide ion),and $C_{10}H_7OH$ (naphthol) is a stronger acid than $CH_3OH$. Therefore,the equilibrium lies to the left,meaning the backward reaction is favoured.

(A) The basic strength of an anion is inversely proportional to the acidic strength of its conjugate acid.
The conjugate acids of the given halides are $HF$,$HCl$,$HBr$,and $HI$.
The acidic strength order of these hydrohalic acids is $HI > HBr > HCl > HF$.
Since the acidic strength decreases from $HI$ to $HF$,the basic strength of the corresponding conjugate bases increases in the order $I^{-} < Br^{-} < Cl^{-} < F^{-}$.
Therefore,the order of decreasing basic strength is $F^{-} > Cl^{-} > Br^{-} > I^{-}$.
Thus,the correct option is $A$.

(C) Basic strength is inversely proportional to the stability of the conjugate base (anion).
$1$. $CH_3 - CH_2 - CO_2^-$: The negative charge is delocalized over two oxygen atoms via resonance,making it the most stable and least basic.
$2$. $CH_3 - CH_2 - C \equiv C^-$: The negative charge is on an $sp$-hybridized carbon atom. Due to high electronegativity of $sp$ carbon,it is relatively stable.
$3$. $CH_3 - CH_2 - S^-$: The negative charge is on a large sulfur atom. Due to the large size of $S$,the charge is dispersed,making it more stable than the oxygen anion.
$4$. $CH_3 - CH_2 - O^-$: The negative charge is on a small,highly electronegative oxygen atom with no resonance stabilization,making it the least stable and most basic.
Comparing the stability of the conjugate acids: $CH_3-CH_2-OH$ $(pKa \approx 16)$ < $CH_3-CH_2-SH$ $(pKa \approx 10.5)$ < $CH_3-CH_2-C \equiv CH$ $(pKa \approx 25)$ < $CH_3-CH_2-COOH$ $(pKa \approx 4.8)$.
Correct order of basicity: $D > B > A > C$.

(A) The acidic strength of dicarboxylic acids depends on the distance between the two carboxylic acid groups.
As the number of carbon atoms between the two $-COOH$ groups increases,the electron-withdrawing inductive effect ($-I$ effect) of one $-COOH$ group on the other decreases.
Oxalic acid $(HOOC-COOH)$ has $0$ carbon atoms between the groups.
Malonic acid $(HOOC-CH_2-COOH)$ has $1$ carbon atom between the groups.
Heptanedioic acid $(HOOC-(CH_2)_5-COOH)$ has $5$ carbon atoms between the groups.
Acidic strength decreases as the distance increases: $\text{Oxalic} > \text{Malonic} > \text{Heptanedioic}$.
Since $pK_a = -\log(K_a)$,a higher $K_a$ corresponds to a lower $pK_a$.
Therefore,the order of $pK_a$ values is $pK_1 < pK_2 < pK_3$.
Option $(a)$ states $pK_1 > pK_2 > pK_3$,which is the incorrect order.

(A) An acid-base reaction always proceeds in the direction of the formation of the weaker acid and weaker base.
For reaction $(I)$:
Acid on left is $CH_3NH_2$ $(pK_a = 35)$ and acid on right is $CH_3CH_3$ $(pK_a = 50)$.
Since $50 > 35$,$CH_3CH_3$ is a weaker acid than $CH_3NH_2$. Thus,the equilibrium is favoured to the right.
For reaction $(II)$:
Acid on left is $H_2O$ $(pK_a = 15.7)$ and acid on right is $HF$ $(pK_a = 3.2)$.
Since $15.7 > 3.2$,$H_2O$ is a weaker acid than $HF$. Thus,the equilibrium is favoured to the left.

(D) Acid-base equilibrium always favors the formation of the weaker acid and weaker base.
In option $(d)$,the reactant acid is the anilinium ion ($C_6H_5NH_3^+$,$pK_a \approx 4.6$) and the product acid is the phenyloxonium ion ($C_6H_5OH_2^+$,$pK_a < 0$).
Since the anilinium ion is a much weaker acid than the phenyloxonium ion,the equilibrium lies to the left (reactant side).
Therefore,the backward reaction is favoured.

(D) According to the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[A^-]}{[HA]}$.
If $pH > pK_a$,the concentration of the conjugate base $[A^-]$ is greater than the concentration of the acid $[HA]$.
If $pH < pK_a$,the concentration of the acid $[HA]$ is greater than the concentration of the conjugate base $[A^-]$.
For acetic acid $(CH_3COOH)$ with $pK_a = 4.8$ at $pH = 7.0$:
Since $7.0 > 4.8$,the conjugate base $CH_3COO^{-}$ is the major species.
For ethanol $(CH_3CH_2OH)$ with $pK_a = 16.0$ at $pH = 7.0$:
Since $7.0 < 16.0$,the undissociated acid $CH_3CH_2OH$ is the major species.
Therefore,the major species present are $CH_3COO^{-}$ and $CH_3CH_2OH$.

(A) The acidity of a compound is inversely proportional to its $pK_a$ value.
Picric acid ($2,4,6$-trinitrophenol) is a very strong acid due to the strong electron-withdrawing effect of three $-NO_2$ groups,which stabilizes the phenoxide ion,resulting in a low $pK_a$ of approximately $0.4$.
Acetic acid is a moderately strong organic acid with a $pK_a$ of approximately $4.75$.
Phenol is a weak acid with a $pK_a$ of approximately $10.0$.
Therefore,the correct order of $pK_a$ values is: Picric acid $(0.4)$ < Acetic acid $(4.75)$ < Phenol $(10.0)$.

(B) The dissociation of hydrazoic acid $(HN_3)$ is given by the equation:
$HN_3 \rightleftharpoons N_3^{-} + H^{+}$
By definition,a conjugate base is formed when an acid loses a proton $(H^{+})$.
Therefore,the conjugate base of hydrazoic acid $(HN_3)$ is the azide ion,$N_3^{-}$.

(C) The strength of an acid is directly proportional to its dissociation constant $K_a$.
Given $K_a$ values: $HCN$ $(6.2 \times 10^{-10})$,$HNO_2$ $(4.0 \times 10^{-4})$,$HF$ $(7.2 \times 10^{-4})$.
Thus,the order of acidic strength is $HCN < HNO_2 < HF$.
Since the conjugate base of a stronger acid is a weaker base,the order of base strength is the inverse of the acidic strength.
Therefore,the order of increasing base strength is $F^{-} < NO_2^- < CN^{-}$.

(A) The proton affinity of a species is directly related to its basicity. $A$ stronger base has a higher proton affinity.
To determine the order,we look at the strength of the corresponding conjugate acids: $HI$,$HF$,$H_2S$,and $NH_3$.
The acidity order is $HI > HF > H_2S > NH_3$.
The basicity (and thus proton affinity) of the conjugate bases is the inverse of the acidity of their conjugate acids.
Therefore,the order of proton affinity is $I^{-} < F^{-} < HS^{-} < NH_2^{-}$.

(B) The strength of a base is determined by its ability to accept a proton $(H^{+})$.
In reaction $(ii)$,$CN^{-}$ reacts with $H_2O$ to form $OH^{-}$. The equilibrium constant $K_b = 1.6 \times 10^{-5}$ indicates the extent of this reaction.
Since $OH^{-}$ is the product of the base dissociation of $CN^{-}$,and $K_b$ is relatively large,$OH^{-}$ is a stronger base than $CN^{-}$.
Comparing the conjugate bases: $OH^{-}$ is the conjugate base of $H_2O$,and $CN^{-}$ is the conjugate base of $HCN$.
Since $HCN$ is a weak acid $(K_a = 6.2 \times 10^{-10})$ and $H_2O$ is an extremely weak acid,the conjugate base of the weaker acid is stronger.
Thus,the order of relative base strength is $OH^{-} > CN^{-} > H_2O$.

(C) The basic strength of a species is inversely proportional to the stability of its conjugate acid.
$(I)$ Conjugate acid is $CH_3COOH$ $(pK_a \approx 4.75)$
$(II)$ Conjugate acid is $CH_3CH_3$ $(pK_a \approx 50)$
$(III)$ Conjugate acid is $NH_3$ $(pK_a \approx 38)$
$(IV)$ Conjugate acid is $C_6H_5OH$ $(pK_a \approx 10)$
Since the order of acidic strength is $CH_3COOH > C_6H_5OH > NH_3 > CH_3CH_3$,the order of basic strength is the reverse: $CH_3CH_2^- > NH_2^- > C_6H_5O^- > CH_3COO^-$.
Thus,the correct order is $(II) > (III) > (IV) > (I)$.

(A) The degree of ionisation of an acid in water depends on its acid dissociation constant $(K_a)$. $A$ higher $K_a$ value indicates a stronger acid,which ionises to a greater extent.
Picric acid ($2,4,6$-trinitrophenol) is a very strong organic acid due to the strong electron-withdrawing effect of three $-NO_2$ groups at the ortho and para positions,which stabilizes the phenoxide ion.
Its acidity is comparable to that of strong mineral acids,making its ionisation in water significantly higher than that of carboxylic acids like acetic acid,formic acid,or benzoic acid.

(B) The chemical formula for chloroplatinic acid is $H_2PtCl_6 \cdot xH_2O$.
In aqueous solution,it dissociates to release two $H^+$ ions,as represented by the formula $H_2[PtCl_6]$.
Since it provides two replaceable hydrogen ions per molecule,it is classified as a dibasic acid.

(A) According to the Bronsted-Lowry concept,an acid is a proton $(H^+)$ donor and a base is a proton acceptor.
$A$. $H_2S$ can donate a proton to form $HS^-$,but it cannot accept a proton to form $H_3S^+$. Thus,it acts only as a Bronsted acid.
$B$. $H_2O$ can donate a proton to form $OH^-$ and accept a proton to form $H_3O^+$. Thus,it is amphoteric.
$C$. $HCO_3^-$ can donate a proton to form $CO_3^{2-}$ and accept a proton to form $H_2CO_3$. Thus,it is amphoteric.
$D$. $NH_3$ can accept a proton to form $NH_4^+$,but it does not act as a Bronsted acid in aqueous solution. Thus,it acts only as a Bronsted base.
Therefore,$H_2S$ is a Bronsted acid but not a Bronsted base.

(D) $Al(OH)_3$ is amphoteric in nature,which means it can react with both acids and bases.
For example,it reacts with $HCl$ (acid) to form $AlCl_3$ and with $NaOH$ (base) to form $Na[Al(OH)_4]$.

(D) $HCl$ is highly soluble in water because it ionizes in water and forms $ion-dipole$ interactions with water molecules. Therefore,it cannot be collected over water.

(B) $1$. In cis-butene dioic acid (maleic acid),the first dissociation leads to an intramolecular hydrogen bond in $X_1^-$,which stabilizes it. This makes the first dissociation easier,so $K_{a_1} > K'_{a_1}$.
$2$. The second dissociation of $X_1^-$ to $X_2^{2-}$ is more difficult because it involves breaking the intramolecular hydrogen bond and overcoming electrostatic repulsion,so $K_{a_2} < K'_{a_2}$.
$3$. Since $K_{a_2} < K'_{a_2}$,the conjugate base $X_2^{2-}$ is stronger than $Y_2^{2-}$.
$4$. $X_1^-$ is stabilized by intramolecular $H$-bonding,making it a weaker base than $Y_1^-$. Therefore,the statement '$X_1^-$ species is more basic than $Y_1^-$ species' is incorrect.

(D) The reaction of $Na_2CO_3$ with an acid produces $CO_2$.
Since the oxide in water produces $CO_2$ upon addition of $Na_2CO_3$,the oxide must be acidic in nature.
Acidic oxides are typically oxides of non-metals.
For example,$SO_2$ is a non-metal oxide that forms an acidic solution in water,which reacts with $Na_2CO_3$ as follows:
$Na_2CO_3 + SO_2 \longrightarrow Na_2SO_3 + CO_2$.

(B) The function of $Ca(H_2PO_4)_2$ (calcium dihydrogen phosphate) in baking powder is to act as an acid source.
When the mixture is moistened,$Ca(H_2PO_4)_2$ reacts with $NaHCO_3$ (sodium bicarbonate) to release $CO_2$ gas.
The reaction is: $Ca(H_2PO_4)_2 + 2NaHCO_3 \rightarrow CaHPO_4 + Na_2HPO_4 + 2H_2O + 2CO_2$.
This $CO_2$ gas helps the dough to rise.

(D) Compounds that react with each other cannot coexist in an aqueous solution.
$I.$ $NaH_2PO_4$ (acidic salt) and $NaHCO_3$ (amphoteric salt) can coexist as they do not react significantly.
$II.$ $Na_2CO_3$ and $NaHCO_3$ are both salts of a weak acid and strong base and can coexist.
$III.$ $NaOH$ (strong base) and $NaH_2PO_2$ (salt of hypophosphorous acid) can coexist as $NaH_2PO_2$ is not acidic enough to react with $NaOH$.
$IV.$ $NaHCO_3$ (acidic salt) and $NaOH$ (strong base) react to form $Na_2CO_3$ and $H_2O$:
$NaHCO_3 + NaOH \rightarrow Na_2CO_3 + H_2O$
Therefore,only the pair in $IV$ cannot exist together.

(A) When $KHSO_4$ is added to a concentrated solution of $H_2SO_4$,it dissociates as follows: $KHSO_4 \rightarrow K^+ + HSO_4^-$.
The $HSO_4^-$ ion acts as a weak acid and further dissociates in the solution: $HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$.
The release of additional $H^+$ ions into the solution increases the total concentration of $H^+$ ions.
Since acidity is directly proportional to the concentration of $H^+$ ions,the acidity of the solution increases.

(B) Amphoteric substances react with both acids and bases.
$Zn(OH)_2$ acts as an acid when reacting with a base:
$Zn(OH)_{2(s)} + 2OH^{-}_{(aq)} \longrightarrow [Zn(OH)_4]^{2-}_{(aq)}$ (or $ZnO_2^{2-} + 2H_2O$)
$Zn(OH)_2$ acts as a base when reacting with an acid:
$Zn(OH)_{2(s)} + 2H^{+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + 2H_2O_{(l)}$
Therefore,Set $1$ and Set $3$ represent the amphoteric nature of $Zn(OH)_2$.

(C) The proton affinity depends on the availability of the lone pair of electrons on the central atom.
$NH_3$ has a lone pair on the nitrogen atom which is more available for donation compared to the lone pairs on $H_2O$,$H_2S$,or $PH_3$.
Due to the smaller size of nitrogen and higher electron density,$NH_3$ acts as a stronger Lewis base.
Therefore,the reaction is: $NH_3 + H^+ \to NH_4^+$.

(A) The acidic strength of an acid is directly proportional to its ability to release $H^+$ ions,which is inversely proportional to its $pKa$ value $(Acidic \ Strength \ \propto \frac{1}{pKa})$.
The order of acidic strength is: $HClO_4 > HNO_3 > H_2CO_3 > B(OH)_3$.
$HClO_4$ is a very strong acid due to the high oxidation state of $Cl$ and the stability of the perchlorate ion.
$HNO_3$ is a strong acid,$H_2CO_3$ is a weak acid,and $B(OH)_3$ (boric acid) is a very weak Lewis acid.
Since the $pKa$ value is inversely proportional to acidic strength,the order of increasing $pKa$ values is the reverse of the order of acidic strength:
$HClO_4 < HNO_3 < H_2CO_3 < B(OH)_3$.

(A) strong base corresponds to a weak conjugate acid.
The conjugate acids of $F^{-}$,$Cl^{-}$,$Br^{-}$,and $I^{-}$ are $HF$,$HCl$,$HBr$,and $HI$ respectively.
The order of acidic strength is $HF < HCl < HBr < HI$.
Therefore,the order of basic strength of their conjugate bases is $F^{-} > Cl^{-} > Br^{-} > I^{-}$.
Thus,$F^{-}$ is the most basic ion.

(A) The acidity of oxyacids of chlorine increases with the oxidation state of the central chlorine atom. Therefore,$HClO$ (oxidation state $+1$) is weaker than $HClO_3$ (oxidation state $+5$).
Comparing $HCl$ and $HBr$,the bond dissociation energy of $H-Cl$ is higher than $H-Br$ due to the smaller size of $Cl$,making $HCl$ a weaker acid than $HBr$ in terms of proton release in certain solvents,but generally,$HClO$ is considered the weakest among the given options due to the low oxidation state of chlorine and the nature of the $O-H$ bond.

(C) In the context of the reaction between $HF$ and $HNO_3$,$HF$ acts as a base because it accepts a proton from $HNO_3$ to form $H_2F^+$ and $NO_3^-$.
$HF + HNO_3 \rightarrow H_2F^+ + NO_3^-$.
Among the given options,$HF$ is the only substance that can act as a base in the presence of a stronger acid like $HNO_3$.

(B) According to the Bronsted-Lowry theory,a base is a proton acceptor.
In the given reaction,$HNO_3$ accepts a proton $(H^+)$ from $HF$ to form $H_2NO_3^+$.
Therefore,$HNO_3$ acts as a base.

(C) Ammonium sulphate $(NH_4)_2SO_4$ is a physiological acidic fertilizer.
When it is added to the soil,it undergoes hydrolysis to produce $H^+$ ions,which increase the acidity of the soil.
Therefore,regular use of ammonium sulphate makes the soil acidic.

(C) An Arrhenius acid is a substance that increases the concentration of $H^+$ ions in aqueous solution.
$HNO_3$,$HClO_4$,and $H_3PO_4$ are strong or moderate acids that dissociate to release $H^+$ ions directly.
$H_3BO_3$ (Boric acid) is a Lewis acid. It does not dissociate to give $H^+$ ions in water; instead,it accepts an $OH^-$ ion from water to form $[B(OH)_4]^-$ and releases $H^+$ ions from the water molecule itself.
Therefore,$H_3BO_3$ is not an Arrhenius acid in the classical sense as it does not contain ionizable $H^+$ protons.

(C) Amphiprotic species are those that can act as both a Bronsted-Lowry acid (by donating a proton) and a Bronsted-Lowry base (by accepting a proton).
$1$. $H_2PO_4^-$: Can accept $H^+$ to form $H_3PO_4$ and donate $H^+$ to form $HPO_4^{2-}$.
$2$. $H_2PO_3^-$: Can accept $H^+$ to form $H_3PO_3$ and donate $H^+$ to form $HPO_3^{2-}$.
$3$. $H_2O$: Can accept $H^+$ to form $H_3O^+$ and donate $H^+$ to form $OH^-$.
Thus,the set $H_2PO_4^-, H_2PO_3^-, H_2O$ consists of amphiprotic species.

(C) According to the Bronsted-Lowry theory,an acid is defined as a proton $(H^{+})$ donor,while a base is defined as a proton $(H^{+})$ acceptor.
Therefore,an acid is a substance that gives a proton.

(B) The relative strength of two acids is given by the ratio of the square roots of their dissociation constants $(K_a)$:
$\text{Relative Strength} = \sqrt{\frac{K_{a1}}{K_{a2}}} = \sqrt{\frac{3.0 \times 10^{-4}}{1.8 \times 10^{-5}}} = \sqrt{\frac{30 \times 10^{-5}}{1.8 \times 10^{-5}}} = \sqrt{\frac{30}{1.8}} = \sqrt{16.66} \approx 4.08 \approx 4$.
Thus,the ratio of the strengths of $HA_1$ to $HA_2$ is $4 : 1$.

(C) Lewis acid is defined as an electron pair acceptor.
$NH_3$ has a lone pair on nitrogen and acts as a Lewis base.
$H_2O$ has lone pairs on oxygen and acts as a Lewis base.
$HCl$ is a Bronsted-Lowry acid but does not have an incomplete octet to accept an electron pair in the Lewis sense.
$SO_2$ has a central sulfur atom with vacant $d$-orbitals and can accept an electron pair,making it a Lewis acid.