Acids and Bases Questions in English

(A) The reaction is: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$.
In this reaction,$NH_3$ acts as a Bronsted-Lowry base because it accepts a proton $(H^+)$.
Water $(H_2O)$ acts as a Bronsted-Lowry acid because it donates a proton $(H^+)$ to $NH_3$ to form $OH^-$.
Therefore,water acts as an acid.

(B) According to the $Br\text{ø}nsted-Lowry$ theory,an acid is a substance that can donate a proton ($H^+$ ion).
$H_3O^+$ can donate a proton to form $H_2O$,hence it acts as a $Br\text{ø}nsted-Lowry$ acid.
$SO_4^{2-}$,$OH^-$,and $Cl^-$ are conjugate bases that can accept protons,making them $Br\text{ø}nsted-Lowry$ bases.

(B) The $CH_3COOH$ is a weak acid. According to the Bronsted-Lowry theory,the conjugate base of a weak acid is relatively strong. Since $CH_3COOH$ is a weak acid,its conjugate base $CH_3COO^-$ is a strong conjugate base.

(B) $Br\o nsted$ acid is a proton $(H^+)$ donor,and a $Br\o nsted$ base is a proton $(H^+)$ acceptor.
Compounds that can act as both are called amphoteric.
$HSO_4^-$ can donate a proton to form $SO_4^{2-}$ and accept a proton to form $H_2SO_4$.
$NH_3$ can donate a proton to form $NH_2^-$ and accept a proton to form $NH_4^+$.
$HCO_3^-$ can donate a proton to form $CO_3^{2-}$ and accept a proton to form $H_2CO_3$.
$Na_2CO_3$ dissociates into $2Na^+$ and $CO_3^{2-}$. The $CO_3^{2-}$ ion can only accept a proton to form $HCO_3^-$ or $H_2CO_3$; it cannot donate a proton because it has no hydrogen atoms.
Therefore,$Na_2CO_3$ cannot act as a $Br\o nsted$ acid.

(A) The Arrhenius theory defines acids as substances that increase the concentration of $H^+$ ions in water and bases as substances that increase the concentration of $OH^-$ ions in water.
Statement $A$ is incorrect because the Arrhenius theory is strictly limited to aqueous solutions and cannot explain acidic or basic properties in non-aqueous solvents.
Statement $B$ is correct because $AlCl_3$ is a Lewis acid,not an Arrhenius acid.
Statement $C$ is correct because $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,making its solution alkaline due to hydrolysis,even though it does not contain $OH^-$ ions in its solid state.
Statement $D$ is correct because $CO_2$ reacts with water to form $H_2CO_3$,which dissociates to provide $H^+$ ions,but $CO_2$ itself does not contain $H^+$ ions.

(C) Lewis acid is defined as a substance that can accept a lone pair of electrons.
$CO_2$ acts as a Lewis acid because the carbon atom is electron-deficient and can accept electron pairs from nucleophiles.
$NH_4Cl$ is a salt,$MgCl_2$ is an ionic compound,and $H_2O$ typically acts as a Lewis base due to the lone pairs on the oxygen atom.

(D) neutralization reaction occurs between an acid and a base to form salt and water. The net ionic equation for the neutralization of a strong acid and a strong base is: $H^{+}(aq) + OH^{-}(aq) \rightarrow H_2O(l)$. Thus,the primary product of the neutralization process is water molecules.

(B) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
Stronger acids have weaker conjugate bases,and weaker acids have stronger conjugate bases.
The corresponding acids for the given conjugate bases are:
$ClO_4^-$ comes from $HClO_4$ (a very strong acid).
$HCO_3^-$ comes from $H_2CO_3$ (a weak acid).
$F^-$ comes from $HF$ (a weak acid).
$HSO_4^-$ comes from $H_2SO_4$ (a strong acid).
Comparing the acid strengths: $HClO_4 > H_2SO_4 > HF > H_2CO_3$.
Therefore,the order of conjugate base strength is $HCO_3^- > F^- > HSO_4^- > ClO_4^-$.
Thus,$HCO_3^-$ is the strongest conjugate base among the given options.

(B) $Br\o nsted$ acid is defined as a substance that can donate a proton $(H^+)$.
$CH_3NH_3^+$ can donate a proton to form $CH_3NH_2$.
$H_2O$ can donate a proton to form $OH^-$.
$HSO_4^-$ can donate a proton to form $SO_4^{2-}$.
$CH_3COO^-$ is the conjugate base of acetic acid $(CH_3COOH)$ and cannot donate a proton; instead,it acts as a $Br\o nsted$ base by accepting a proton.

(D) An acid is a substance that donates a proton $(H^+)$.
$HCl$ acts as an acid when it reacts with a base that can accept a proton.
$1$. $HCl + NH_3 \rightarrow NH_4^+ + Cl^-$ (Here,$NH_3$ is a base,$HCl$ acts as an acid).
$2$. $HCl + C_2H_5OH \rightarrow C_2H_5OH_2^+ + Cl^-$ (Here,$C_2H_5OH$ acts as a base,$HCl$ acts as an acid).
$3$. $HCl + H_2O \rightarrow H_3O^+ + Cl^-$ (Here,$H_2O$ is a base,$HCl$ acts as an acid).
$4$. $C_6H_6$ (Benzene) is a non-polar solvent and does not have any lone pair of electrons to accept a proton from $HCl$. Therefore,$HCl$ does not act as an acid in $C_6H_6$.

(A) Lewis base is defined as a substance that can donate a lone pair of electrons.
$CH_4$ (Methane) has no lone pairs of electrons on the central carbon atom,as all its valence electrons are involved in covalent bonding.
Therefore,$CH_4$ cannot act as a Lewis base.
$C_2H_5OH$ (Ethanol) has lone pairs on the oxygen atom,Acetone has lone pairs on the oxygen atom,and Secondary amines have a lone pair on the nitrogen atom,allowing all of them to act as Lewis bases.

(A) conjugate base is formed by the removal of a proton $(H^{+})$ from the given acid.
For the species $HPO_4^{2-}$,removing one $H^{+}$ ion results in the formation of $PO_4^{3-}$.
The reaction is: $HPO_4^{2-} \rightarrow H^{+} + PO_4^{3-}$.
Therefore,the conjugate base of $HPO_4^{2-}$ is $PO_4^{3-}$.

(C) The strength of a Bronsted-Lowry base is inversely proportional to the strength of its conjugate acid.
$1$. The conjugate acids of the given bases are: $H_2$ (for $H^{-}$),$H_2O$ (for $OH^{-}$),$HCl$ (for $Cl^{-}$),and $H_2CO_3$ (for $HCO_3^{-}$).
$2$. Among these acids,$HCl$ is the strongest acid.
$3$. Since $HCl$ is the strongest acid,its conjugate base,$Cl^{-}$,must be the weakest base.

(A) In the reaction $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$,$NH_3$ accepts a proton $(H^+)$ from $H_2O$ to form $NH_4^+$.
Since $H_2O$ donates a proton $(H^+)$ to $NH_3$,it acts as a Bronsted-Lowry acid.

(C) According to the Brønsted-Lowry theory,a base is a proton $(H^+)$ acceptor.
In the given reaction: $HC_2O_4^{-} + PO_4^{3-} \rightarrow HPO_4^{2-} + C_2O_4^{2-}$
$1$. $HC_2O_4^{-}$ acts as an acid because it donates a proton to become $C_2O_4^{2-}$.
$2$. $PO_4^{3-}$ acts as a base because it accepts a proton to become $HPO_4^{2-}$.
$3$. In the reverse reaction,$HPO_4^{2-}$ acts as an acid (donates $H^+$) and $C_2O_4^{2-}$ acts as a base (accepts $H^+$).
Therefore,the two Brønsted bases are $PO_4^{3-}$ and $C_2O_4^{2-}$.

(C) $Lewis$ base is defined as a substance that can donate a lone pair of electrons.
$NH_3$ contains a nitrogen atom with one lone pair of electrons,which it can donate to an electron-deficient species.
$Cu^{2+}$,$AlCl_3$,and $BF_3$ are all electron-deficient species and act as $Lewis$ acids because they can accept electron pairs.

(C) Lewis acid is defined as an electron pair acceptor.
$SO_3$ (sulfur trioxide) has an electron-deficient sulfur atom that can accept a lone pair of electrons from a base.
$CaO$ is a basic oxide.
$CH_3NH_2$ (methylamine) acts as a Lewis base because it has a lone pair of electrons on the nitrogen atom.

(B) According to the Bronsted-Lowry theory,a conjugate base is formed when an acid loses a proton $(H^+)$.
The relationship is represented as: $\text{Acid} \rightleftharpoons \text{Conjugate Base} + H^+$.
For the chloride ion $(Cl^-)$,the corresponding acid is formed by adding a proton $(H^+)$ to it:
$Cl^- + H^+ \rightarrow HCl$.
Therefore,$Cl^-$ is the conjugate base of $HCl$.

(A) The self-ionization (auto-protolysis) of liquid ammonia is represented by the equilibrium reaction:
$2NH_3(l) \rightleftharpoons NH_4^+(am) + NH_2^-(am)$
The self-ionization constant $(K_{am})$ is defined as the product of the concentrations of the ions formed:
$K_{am} = [NH_4^+][NH_2^-]$

(C) $Br\o nsted$ acid is a proton $(H^+)$ donor,and a $Br\o nsted$ base is a proton $(H^+)$ acceptor.
Species that can both donate and accept a proton are called amphoteric.
$NH_3$ can accept a proton to form $NH_4^+$ (acting as a base) but cannot donate a proton under normal conditions.
$CN^-$ can accept a proton to form $HCN$ (acting as a base) but cannot donate a proton.
$HSO_4^-$ can donate a proton to form $SO_4^{2-}$ (acting as an acid) and can accept a proton to form $H_2SO_4$ (acting as a base).
Therefore,$HSO_4^-$ acts as both a $Br\o nsted$ acid and a $Br\o nsted$ base.

(C) In the given reaction: $HBr + H_2O \rightleftharpoons H_3O^{+} + Br^{-}$.
$HBr$ acts as an acid because it donates a proton $(H^{+})$ to $H_2O$.
$H_2O$ acts as a base because it accepts a proton.
When an acid $(HBr)$ loses a proton,it forms its conjugate base $(Br^{-})$.
Since $HBr$ is a strong acid,its conjugate base $Br^{-}$ is a very weak base.
Conversely,$H_3O^{+}$ is the conjugate acid of the base $H_2O$.

(A) An amphoteric species is one that can both donate and accept a proton $(H^+)$.
$1$. As a base: $HSO_4^- + H^+ \rightleftharpoons H_2SO_4$
$2$. As an acid: $HSO_4^- \rightleftharpoons SO_4^{2-} + H^+$
Therefore,$HSO_4^-$ acts as both an acid and a base.

(B) To determine the acidity,we look at the stability of the conjugate base or the electron-withdrawing nature of the substituents attached to the acidic proton.
$1$. $HSO_3F$ (Fluorosulfonic acid) is a superacid,making it the strongest acid among the given options.
$2$. $H_3O^+$ (Hydronium ion) is a strong acid with a $pK_a$ of approximately $-1.7$.
$3$. $HSO_4^-$ (Bisulfate ion) has a $pK_a$ of approximately $1.99$.
$4$. $HCO_3^-$ (Bicarbonate ion) is a weak acid with a $pK_a$ of approximately $10.3$.
Comparing these values,the order of acidity is: $HCO_3^- < HSO_4^- < H_3O^+ < HSO_3F$,which corresponds to $(i) < (iii) < (ii) < (iv)$.

(A) The acidic strength of a species is determined by its ability to donate a proton $(H^+)$.
Stronger acids have weaker conjugate bases.
Comparing the conjugate bases:
$NH_3$ (from $NH_4^+$),$OH^-$ (from $H_2O$),$H_2O$ (from $H_3O^+$),$F^-$ (from $HF$),and $O^{2-}$ (from $OH^-$).
The basic strength order of these conjugate bases is $O^{2-} > NH_3 > OH^- > F^- > H_2O$.
Therefore,the acidic strength order is the reverse: $OH^- < H_2O < NH_4^+ < HF < H_3O^+$.

(D) An acidic species must be able to donate a proton $(H^+)$.
$1$. $[Al(H_2O)_6]^{+3}$ acts as a Lewis acid and can release $H^+$ from coordinated water molecules.
$2$. $[Fe(H_2O)_6]^{+3}$ also acts as a Lewis acid and can release $H^+$.
$3$. $HPO_4^{-2}$ is amphoteric and can donate a proton to form $PO_4^{-3}$.
$4$. $ClO_4^-$ is the conjugate base of a strong acid $(HClO_4)$. Since $HClO_4$ is a very strong acid,its conjugate base $ClO_4^-$ has negligible basicity and no acidic properties as it cannot donate a proton.

(C) According to the Bronsted-Lowry theory,a conjugate base is formed when an acid donates a proton $(H^{+})$.
For the acid $HBr$,the reaction is:
$HBr \rightarrow H^{+} + Br^{-}$
Therefore,the conjugate base of $HBr$ is $Br^{-}$.

(C) conjugate acid-base pair differs by only one proton $(H^+)$.
For the pair $H_2SO_4$ and $HSO_4^-$,the relationship is:
$H_2SO_4 \rightleftharpoons H^+ + HSO_4^-$
Here,$H_2SO_4$ acts as the acid and $HSO_4^-$ is its conjugate base.
Therefore,$H_2SO_4, HSO_4^-$ is a conjugate acid-base pair.

(A) According to the $Brønsted-Lowry$ theory,a base is a substance that can accept a proton $(H^{+})$.
$I^{-}$ is the conjugate base of the strong acid $HI$,meaning it has a lone pair of electrons and can accept a proton to form $HI$.
$H_3O^{+}$,$HCl$,and $NH_4^{+}$ are all $Brønsted$ acids because they can donate a proton.
Therefore,$I^{-}$ acts as a base.

(B) $CO_2$ is an acidic oxide. When it dissolves in water,it forms carbonic acid $(H_2CO_3)$:
$CO_2(g) + H_2O(l) \rightleftharpoons H_2CO_3(aq)$.
The formation of carbonic acid increases the concentration of $H^+$ ions in the solution.
Therefore,the $pH$ of the resulting solution becomes less than $7$.

(B) The basic strength of conjugate bases is inversely proportional to the acidic strength of their corresponding conjugate acids.
The conjugate acids are: $(a) H_2O, (b) NH_3, (c) HF$.
The acidic strength order of these acids is: $HF > H_2O > NH_3$.
Since the acidic strength is $HF > H_2O > NH_3$,the basic strength of their conjugate bases will be the reverse: $NH_2^{-} > OH^{-} > F^{-}$.
Therefore,the correct order is $NH_2^{-} > OH^{-} > F^{-}$.

(C) The acidity of these compounds depends on the stability of their conjugate bases and the inductive effect of substituents.
$1$. $ClCH_2COOH$ is the strongest acid due to the strong electron-withdrawing effect of the $Cl$ atom.
$2$. $HCOOH$ is stronger than $CH_3COOH$ because the methyl group in $CH_3COOH$ is electron-donating,which destabilizes the carboxylate ion.
$3$. $HCN$ is a very weak acid compared to carboxylic acids.
Therefore,the correct order of acidity is: $ClCH_2COOH > HCOOH > CH_3COOH > HCN$.

(C) The conjugate base of an acid is formed by the removal of a proton $(H^+)$ from the acid.
For the acid $HSO_3^-$,the removal of one $H^+$ ion results in the formation of the sulfite ion,$SO_3^{2-}$.
The reaction is: $HSO_3^- \rightarrow H^+ + SO_3^{2-}$.

(C) According to the Lewis theory,a base is a species that can donate a lone pair of electrons.
$OH^{-}$,$H_2O$,and $NH_3$ all possess lone pairs of electrons that they can donate.
$Ag^{+}$ is a metal cation that acts as an electron pair acceptor,making it a Lewis acid,not a Lewis base.

(B) Given the relationship $pK_{b1} < pK_{b2}$.
Since $pK_b = -\log(K_b)$,a lower $pK_b$ value indicates a higher $K_b$ value.
Therefore,$K_{b1} > K_{b2}$.
$A$ higher $K_b$ value means the base is stronger.
$AOH$ is a stronger base than $BOH$.
Stronger bases produce a higher concentration of $OH^-$ ions,leading to a higher $pH$ value.
Thus,$AOH$ has a higher $pH$ than $BOH$.
The question asks which does $NOT$ represent the base with the highest $pH$,which is $BOH$.

(A) Brønsted base is a substance that can accept a proton $(H^+)$.
$NO_3^-$ is the conjugate base of the strong acid $HNO_3$ and can accept a proton to form $HNO_3$.
$H_3O^+$,$NH_4^+$,and $CH_3COOH$ are Brønsted acids because they can donate a proton.

(C) According to the $Br\o nsted-Lowry$ theory,an acid-base pair differs by a single proton $(H^+)$.
In the given reaction:
$NH_4^+$ (acid) loses a proton to form $NH_3$ (conjugate base).
$S^{2-}$ (base) gains a proton to form $HS^-$ (conjugate acid).
Therefore,$NH_4^+/NH_3$ and $S^{2-}/HS^-$ are conjugate acid-base pairs.

(D) $Br\ddot{o}nsted$ acid is a proton $(H^+)$ donor,and a $Br\ddot{o}nsted$ base is a proton $(H^+)$ acceptor.
Species that can both donate and accept a proton are called amphoteric.
In the given options,$HCO_3^-$ can act as an acid: $HCO_3^- \rightarrow H^+ + CO_3^{2-}$.
It can also act as a base: $HCO_3^- + H^+ \rightarrow H_2CO_3$.
Therefore,$HCO_3^-$ acts as both a $Br\ddot{o}nsted$ acid and a $Br\ddot{o}nsted$ base.

(C) $1$. In aqueous medium,$HCl$ is a stronger acid than $HF$ due to the higher bond dissociation energy of the $H-F$ bond compared to $H-Cl$. Thus,option $A$ is incorrect.
$2$. The acidity of oxoacids increases with the oxidation state of the central atom. In $HClO_4$,the oxidation state of $Cl$ is $+7$,while in $HClO_3$,it is $+5$. Therefore,$HClO_4$ is a stronger acid than $HClO_3$. Thus,option $B$ is incorrect.
$3$. In $HNO_3$,the oxidation state of $N$ is $+5$,and in $HNO_2$,it is $+3$. Higher oxidation state and more resonance-stabilized conjugate base make $HNO_3$ a stronger acid than $HNO_2$. Thus,option $C$ is correct.
$4$. $H_2SO_3$ is a stronger acid than $H_3PO_3$ because $S$ is more electronegative than $P$,leading to greater polarization of the $O-H$ bond. Thus,option $D$ is incorrect.

(A) According to the $Br\text{\o}nsted-Lowry$ theory,an acid is a proton $(H^+)$ donor.
In the reaction $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$:
$1$. $H_2O$ donates a proton to $NH_3$ to form $OH^-$,so $H_2O$ acts as an acid.
$2$. In the reverse reaction,$NH_4^+$ donates a proton to $OH^-$ to form $NH_3$ and $H_2O$,so $NH_4^+$ acts as an acid.
Therefore,$H_2O$ and $NH_4^+$ are the acids.

(A) $1$. $A$ $Lewis$ base is a substance that can donate a lone pair of electrons. $NH_3$ has a lone pair on the nitrogen atom,making it a strong $Lewis$ base.
$2$. $A$ $Bronsted$ acid is a proton $(H^+)$ donor. $NH_3$ can lose a proton to form $NH_2^-$,thus acting as a $Bronsted$ acid.
$3$. $A$ $Bronsted$ base is a proton $(H^+)$ acceptor. $NH_3$ can accept a proton to form $NH_4^+$,thus acting as a $Bronsted$ base.
$4$. Therefore,$NH_3$ satisfies all three criteria.

(B) Ammonium hydroxide $(NH_4OH)$ is a weak base.
In aqueous solution,it undergoes partial dissociation into $NH_4^+$ and $OH^-$ ions.
Since it does not dissociate completely in water,it is classified as a weak electrolyte.

(D) The strength of a base is inversely proportional to the strength of its conjugate acid.
$1$. The conjugate acids of the given bases are: $CH_3COOH$ (for $CH_3COO^-$),$CH_4$ (for $CH_3^-$),$CH_3OH$ (for $CH_3O^-$),and $HCl$ (for $Cl^-$).
$2$. The acid strength order is: $HCl > CH_3COOH > CH_3OH > CH_4$.
$3$. Since $HCl$ is the strongest acid,its conjugate base $Cl^-$ is the weakest base.

(A) $BF_3$ (Boron trifluoride) has an incomplete octet around the central Boron atom.
It can accept a lone pair of electrons to complete its octet.
According to the Lewis concept,an acid is an electron-pair acceptor.
Therefore,$BF_3$ is a Lewis acid.

(A) According to the Lewis acid-base theory,an acid is a substance that can accept a lone pair of electrons.
Since the $H^{+}$ ion has an empty $1s$ orbital,it can accept a pair of electrons to form a coordinate covalent bond.
Therefore,the $H^{+}$ ion acts as a Lewis acid.

(C) Water acts as a base when it accepts a proton $(H^+)$ from another substance.
In reaction $(i)$,$H_2O$ acts as a reactant to form $H_2CO_3$,but it is not acting as a Brønsted-Lowry base.
In reaction $(ii)$,$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$,water donates a proton to $NH_3$ and acts as an acid.
In reaction $(iii)$,$HCl + H_2O \rightleftharpoons Cl^- + H_3O^+$,water accepts a proton from $HCl$ to form $H_3O^+$,thus acting as a base.
Therefore,only reaction $(iii)$ demonstrates the basic property of water.

(C) $1$. $HOCl$ (hypochlorous acid) acts as a Bronsted-Lowry acid in aqueous solution by donating a proton $(H^+)$ to water: $HOCl + H_2O \rightleftharpoons H_3O^+ + OCl^-$.
$2$. $B(OH)_3$ (boric acid) acts as a Lewis acid in aqueous solution by accepting a lone pair of electrons from water: $B(OH)_3 + 2H_2O \rightleftharpoons [B(OH)_4]^- + H_3O^+$.
$3$. Since both compounds exhibit acidic behavior in aqueous solution,the correct option is $C$.

(B) The conjugate acid of a base is formed by adding a proton $(H^+)$ to the base.
For the base $NH_3$,the conjugate acid is formed by the reaction: $NH_3 + H^+ \rightarrow NH_4^+$.
Therefore,the conjugate acid of $NH_3$ is $NH_4^+$.

(A) Ammonia $(NH_3)$ is a weak base that reacts with water as follows:
$NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$
In this reaction,$NH_3$ accepts a proton from water to form the ammonium ion $(NH_4^+)$ and releases hydroxide ions $(OH^-)$.
The increase in the concentration of $OH^-$ ions makes the solution basic,which leads to a decrease in the concentration of $H_3O^+$ ions due to the ionic product of water $(K_w = [H_3O^+][OH^-])$.
Therefore,the $OH^-$ ion is responsible for the change.

(A) The conjugate base of an acid is formed by the removal of a proton $(H^+)$ from the acid.
For the bicarbonate ion $(HCO_3^-)$,the removal of one $H^+$ ion results in the carbonate ion $(CO_3^{2-})$.
The reaction is: $HCO_3^- \rightarrow H^+ + CO_3^{2-}$.
Therefore,the conjugate base of $HCO_3^-$ is $CO_3^{2-}$.

(D) conjugate acid is formed by adding a proton $(H^+)$ to the given base.
For the species $Zn(OH)_2$,adding a proton $(H^+)$ results in the formation of $[Zn(OH)_2H]^+$,which can be written as $[Zn(OH)(H_2O)]^+$.
Looking at the options,none of the provided choices represent the correct conjugate acid of $Zn(OH)_2$.