Acids and Bases Questions in English

(A) Lewis acid is an electron-pair acceptor.
In $BF_3$,the central Boron atom has only $6$ valence electrons in its outer shell,making it electron-deficient.
Therefore,it can accept a lone pair of electrons to complete its octet,acting as a Lewis acid.
Thus,the correct option is $A$.

(B) Ammonium hydroxide $(NH_4OH)$ is a weak base because it is only slightly ionized in aqueous solution,making it a weak electrolyte.

(D) The theory of electrolytic dissociation,commonly known as the theory of ionization,was proposed by $Svante \ Arrhenius$ in $1887$.
Therefore,the correct option is $(D)$.

(C) Acetic acid $(CH_3COOH)$ is a weak electrolyte because it undergoes partial dissociation in aqueous solution,meaning its degree of ionization is very low.

(A) The acidic strength of an acid is directly proportional to its dissociation constant $(K_a)$.
Greater the value of $K_a$,the stronger the acid.
Comparing the given values:
$1.4 \times 10^{-2} > 1.6 \times 10^{-4} > 4.3 \times 10^{-7} > 4.4 \times 10^{-10}$.
Therefore,the acid with $K_a = 1.4 \times 10^{-2}$ is the strongest acid.
Thus,the correct option is $(A)$.

(D) In the given reaction: $HClO_4 + H_2O \rightleftharpoons H_3O^{+} + ClO_4^-$
$HClO_4$ acts as an acid and donates a proton $(H^+)$ to form its conjugate base,$ClO_4^-$.
$H_2O$ acts as a base and accepts a proton $(H^+)$ to form its conjugate acid,$H_3O^+$.
Therefore,the conjugate acid-base pairs are $(HClO_4, ClO_4^-)$ and $(H_3O^+, H_2O)$.
Comparing this with the options,$ClO_4^-$ is the conjugate base of $HClO_4$ is the correct statement.

(B) Bronsted base is a substance that can accept a proton $(H^{+})$,while a Bronsted acid is a substance that can donate a proton $(H^{+})$.
$HCO_3^{-}$ can accept a proton to form carbonic acid: $HCO_3^{-} + H^{+} \rightleftharpoons H_2CO_3$ (acting as a Bronsted base).
$HCO_3^{-}$ can donate a proton to form carbonate ion: $HCO_3^{-} \rightleftharpoons H^{+} + CO_3^{2-}$ (acting as a Bronsted acid).
Therefore,$HCO_3^{-}$ acts as both a Bronsted acid and a Bronsted base.

(D) According to the Lewis concept,a Lewis acid is defined as a substance that can accept a lone pair of electrons to form a coordinate covalent bond.
Lewis acids are electrophilic in nature,meaning they are electron-deficient and seek to attract electrons.
When bonding with a Lewis base,the acid utilizes its lowest unoccupied molecular orbital $(LUMO)$ to accommodate the electron pair.

(B) The strength of an acid is inversely proportional to its $pK_a$ value.
Since the $pK_a$ of acid $A$ $(1.2)$ is less than the $pK_a$ of acid $B$ $(2.8)$,acid $A$ is stronger than acid $B$.

(B) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
$1$. The conjugate acids of the given bases are: $HCl$ $(Cl^{-})$,$CH_3COOH$ $(CH_3COO^{-})$,$HSO_4^{-}$ $(SO_4^{2-})$,and $HNO_2$ $(NO_2^{-})$.
$2$. Among these acids,$CH_3COOH$ is the weakest acid.
$3$. Since the conjugate base of a weak acid is a strong base,$CH_3COO^{-}$ is the strongest conjugate base among the given options.
$CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$

(D) strong base is a substance that completely dissociates into its constituent ions in an aqueous solution.
$NaOH$ (sodium hydroxide) is a strong base because it undergoes complete ionization in water to provide $OH^{-}$ ions.
Therefore,both the ability to ionize easily and the production of $OH^{-}$ ions are characteristic of its strength as a base.

(A) According to the Bronsted-Lowry theory,a Bronsted base is a substance that can accept a proton $(H^{+})$.
$NO_3^-$ can accept a proton to form $HNO_3$ $(NO_3^- + H^{+} \rightarrow HNO_3)$.
Therefore,$NO_3^-$ acts as a Bronsted base.
$H_3O^{+}$,$NH_4^{+}$,and $CH_3COOH$ are Bronsted acids because they can donate a proton.

(C) Proton affinity is directly related to the basicity of a substance.
Among the given compounds,$NH_3$ is the strongest Lewis base because the lone pair on the nitrogen atom is more available for donation compared to the lone pairs on $O$,$S$,or $P$.
Therefore,$NH_3$ has the highest affinity for hydrogen ions $(H^+)$.

(D) Ammonia $(NH_3)$ is a weak base that reacts with water to form ammonium ions and hydroxide ions. The equilibrium reaction is:
$NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$
Therefore,the aqueous solution contains both $NH_4^+$ and $OH^-$ ions.

(B) The conjugate base of a species is formed by removing one proton $(H^+)$ from it.
For $NH_2^-$,removing one $H^+$ gives $NH^{2-}$.
The reaction is: $NH_2^- \rightarrow NH^{2-} + H^+$.
Therefore,the conjugate base of $NH_2^-$ is $NH^{2-}$.

(A) The reaction is $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$.
In this reaction,$NH_3$ accepts a proton $(H^+)$ to become $NH_4^+$,acting as a Bronsted-Lowry base.
Conversely,$H_2O$ donates a proton $(H^+)$ to $NH_3$,thereby acting as a Bronsted-Lowry acid.

(B) In the reaction $CH_3COOH + HF \rightleftharpoons CH_3COOH_2^+ + F^{-}$,$HF$ acts as an acid because it donates a proton $(H^+)$ to $CH_3COOH$.
After losing a proton,$HF$ becomes $F^{-}$,which is its conjugate base.
Similarly,$CH_3COOH$ acts as a base by accepting a proton to form $CH_3COOH_2^+$,which is its conjugate acid.
Therefore,$F^{-}$ is the conjugate base of $HF$.

(D) Lewis acid is defined as an electron-pair acceptor.
$AlCl_3$ and $BeCl_2$ are electron-deficient compounds (incomplete octet),so they act as Lewis acids.
$HSO_4^-$ can accept a proton or an electron pair in certain conditions.
$NH_3$ has a lone pair of electrons on the nitrogen atom,which it can donate.
Therefore,$NH_3$ acts as a Lewis base,not a Lewis acid.

(B) The strength of an acid is determined by its ability to donate protons ($H^+$ ions) in an aqueous solution.
$H_2SO_4$ (sulfuric acid) is a strong mineral acid that undergoes complete dissociation in water.
$H_3PO_4$ and $HNO_2$ are weak acids,and $CH_3COOH$ is a weak organic acid.
Therefore,$H_2SO_4$ is the strongest acid among the given options.

(C) Lewis acid is defined as an electron pair acceptor.
$AlCl_3$ acts as a Lewis acid because the central $Al$ atom has an incomplete octet (only $6$ electrons in its valence shell),allowing it to accept a lone pair of electrons.
While $SnCl_4$ can also act as a Lewis acid due to vacant $d$-orbitals,$AlCl_3$ is the most classic textbook example of an electron-deficient Lewis acid.

(D) According to the Brønsted-Lowry theory,an acid-base pair differs by a single proton $(H^+)$.
In the reaction $HClO_4 + H_2O \rightleftharpoons H_3O^{+} + ClO_4^-$,$HClO_4$ acts as an acid and loses a proton to form its conjugate base,$ClO_4^-$.
Similarly,$H_2O$ acts as a base and gains a proton to form its conjugate acid,$H_3O^{+}$.
Therefore,$ClO_4^-$ is the conjugate base of $HClO_4$.

(D) According to the Bronsted-Lowry theory,an acid is a proton $(H^+)$ donor and a base is a proton acceptor.
In the reaction $NH_3 + H_2O \to NH_4^+ + OH^-$,$H_2O$ donates a proton to $NH_3$ to form $NH_4^+$ and $OH^-$.
Since $H_2O$ acts as a proton donor,it is the acid in this reaction.

(B) base is classified as weak if it does not dissociate completely into its constituent ions in an aqueous solution.
$NH_4OH$ is a weak base because it undergoes partial dissociation in water,represented by the equilibrium:
$NH_4OH(aq) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$.
Since the degree of ionization is very low,it is considered a weak base.

(B) $HF$ undergoes self-ionization as follows:
$2 HF \rightleftharpoons [H_2F]^+ + F^-$
When $HNO_3$ is dissolved in liquid $HF$,it acts as a base because it accepts a proton $(H^+)$ from the solvent $HF$.
The reaction is:
$HNO_3 + HF \rightleftharpoons H_2NO_3^+ + F^-$
Since $HNO_3$ accepts a proton from $HF$,it behaves as a base in this solvent.

(D) According to the Brønsted-Lowry theory,a base is a proton $(H^+)$ acceptor.
In the forward reaction: $PO_4^{3-}$ accepts a proton to become $HPO_4^{2-}$,so $PO_4^{3-}$ is a Brønsted base.
In the reverse reaction: $C_2O_4^{2-}$ accepts a proton to become $HC_2O_4^-$,so $C_2O_4^{2-}$ is a Brønsted base.
Therefore,the two Brønsted bases are $PO_4^{3-}$ and $C_2O_4^{2-}$.

(A) $H^{+}$ is a $Lewis$ acid because it is an electron pair acceptor according to the $Lewis$ theory of acids and bases.

(C) $H_3PO_4$ is a triprotic acid. Its dissociation occurs in three steps:
$I$. step: $H_3PO_4 \rightleftharpoons H^{+} + H_2PO_4^-$
$II$. step: $H_2PO_4^- \rightleftharpoons H^{+} + HPO_4^{2-}$
$III$. step: $HPO_4^{2-} \rightleftharpoons H^{+} + PO_4^{3-}$
Therefore,the total number of steps is $3$.

(C) The conjugate acid of a species is formed by the addition of a proton $(H^+)$ to that species.
For the species $HPO_3^{2-}$,the conjugate acid is obtained by adding one $H^+$ ion:
$HPO_3^{2-} + H^+ \rightleftharpoons H_2PO_3^-$.
Therefore,the correct option is $C$.

(D) The reaction is $H^{+} + OH^{-} \rightleftharpoons H_2O$. This is a neutralization reaction where an acid $(H^{+})$ and a base $(OH^{-})$ combine to form water.

(C) Lewis base is a species that can donate a lone pair of electrons.
According to the Bronsted-Lowry theory,the conjugate base of a strong acid is a weak base.
Among the given options,$HCl$ is a strong acid,so its conjugate base $Cl^-$ is the weakest base.
$H^-$ is the conjugate base of $H_2$,$OH^-$ is the conjugate base of $H_2O$,and $HCO_3^-$ is the conjugate base of $H_2CO_3$.
Since $HCl$ is the strongest acid among the corresponding conjugate acids,$Cl^-$ is the weakest base.

(A) The relationship between $pK_a$ and acid strength is given by the equation $pK_a = -\log(K_a)$.
As the $K_a$ value increases,the acid strength increases,and the $pK_a$ value decreases.
Since $pK_a(A) > pK_a(B)$,it implies that $K_a(A) < K_a(B)$.
Therefore,acid $B$ is stronger than acid $A$.

(A) The conjugate acid of a base is formed by the addition of a proton $(H^+)$ to the base.
For the base $NH_2^-$,the conjugate acid is formed as follows:
$NH_2^- + H^+ \rightarrow NH_3$
Therefore,the correct option is $(A)$.

(C) . $CH_3COOH$ is a weak acid,which shows dissociation equilibrium as follows:
$CH_3COOH ⇌ CH_3COO^{-} + H^{+}$
Explanation of other options:
$A$. $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it forms an acidic solution.
$B$. $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,so it forms an alkaline solution.
$D$. $NH_4OH$ is a weak base.

(C) By definition,the $pK_a$ value is the negative logarithm of the acid dissociation constant $(K_a)$.
Mathematically,$pK_a = -\log_{10} K_a$.
Since $-\log_{10} K_a = \log_{10} (K_a)^{-1} = \log_{10} \frac{1}{K_a}$,the correct expression is $\log_{10} \frac{1}{K_a}$.

(A) An acid salt is a salt that contains at least one replaceable hydrogen atom attached to an electronegative atom like oxygen.
In $NaH_2PO_2$ (sodium hypophosphite),the structure is $Na[H_2PO_2]$,where both hydrogen atoms are directly bonded to the phosphorus atom ($P-H$ bonds).
Since these hydrogen atoms are not bonded to oxygen,they are not acidic and cannot be replaced by a base.
Therefore,$NaH_2PO_2$ is not an acid salt.

(B) diprotic acid is an acid that can donate two protons ($H^+$ ions) per molecule in an aqueous solution.
$H_3PO_4$ is triprotic as it can donate $3H^+$ ions.
$H_2S$ is diprotic as it can donate $2H^+$ ions.
$HClO_3$ is monoprotic as it can donate $1H^+$ ion.
Therefore,the question as stated is slightly ambiguous regarding $H_3PO_4$. However,if we consider the definition of diprotic,$H_2S$ is strictly diprotic. Given the options,if the question implies compounds that can act as diprotic acids,$H_2S$ is the correct choice. If the question intended to ask for acids that are at least diprotic,both $(a)$ and $(b)$ would be considered. Based on standard chemistry nomenclature,$H_2S$ is the primary diprotic acid here.

(D) Lewis acid is an electron pair acceptor,while a Lewis base is an electron pair donor.
$BF_3$,$AlCl_3$,and $FeCl_3$ are electron-deficient compounds that can accept an electron pair,thus acting as Lewis acids.
$PH_3$ contains a phosphorus atom with a lone pair of electrons,which it can donate,making it a Lewis base.
Therefore,the correct answer is $D$.

(B) The conjugate base of an acid is formed by the removal of a proton $(H^{+})$ from the acid.
For an acid $HA$,the conjugate base is $A^{-}$.
In this case,for $Cl^{-}$,the corresponding acid is formed by adding a proton $(H^{+})$ to $Cl^{-}$,which results in $HCl$.
Therefore,$HCl \rightleftharpoons H^{+} + Cl^{-}$.
Thus,$Cl^{-}$ is the conjugate base of $HCl$.

(B) Bronsted base is a proton $(H^{+})$ acceptor,and a Lewis base is an electron pair donor.
$Cl^{-}$ can accept a proton to form $HCl$ $(Cl^{-} + H^{+} \to HCl)$,thus acting as a Bronsted base.
$Cl^{-}$ also possesses lone pairs of electrons that it can donate to an electron-deficient species,thus acting as a Lewis base.
Therefore,$Cl^{-}$ behaves as both a Lewis base and a Bronsted base.

(B) According to the Brønsted-Lowry theory,there is an inverse relationship between the strength of an acid and its conjugate base,and vice versa.
If a base is strong,it has a very low tendency to accept a proton.
Therefore,its conjugate acid will have a very low tendency to donate a proton,making it a very weak acid.
Thus,the conjugate acid of a strong base is a weak acid.

(A) The acidic strength of these acids follows the order: $HClO_4 > H_2SO_4 > HBr > HNO_3$.
Among the given options,$HNO_3$ is the weakest acid.

(A) conjugate base is formed when an acid loses a proton $(H^+)$.
For the species $HPO_4^{2-}$,removing one $H^+$ ion results in the formation of $PO_4^{3-}$.
Therefore,the conjugate base of $HPO_4^{2-}$ is $PO_4^{3-}$.

(D) Lewis acid is an electron pair acceptor,while a Lewis base is an electron pair donor.
$FeCl_3$,$AlCl_3$,and $BCl_3$ have incomplete octets or vacant d-orbitals,allowing them to accept electron pairs,thus acting as Lewis acids.
$NH_3$ contains a nitrogen atom with a lone pair of electrons,which it can donate,making it a Lewis base.
Therefore,the correct option is $D$.

(A) Statement $(i)$ is correct: According to the Bronsted-Lowry theory,a strong acid has a weak conjugate base because the strong acid readily donates a proton,and the resulting conjugate base has little tendency to accept it back.
Statement $(ii)$ is correct: According to the Lewis acid-base theory,an acid is defined as a substance that can accept an electron pair to form a coordinate covalent bond.
Therefore,both statements are correct.

(B) According to the $Br\o nsted-Lowry$ theory,an acid is defined as a substance that has the tendency to donate a proton ($H^+$ ion) to a base.
Therefore,the strength of an acid is directly proportional to its tendency to donate protons.

(D) Bronsted acid must be a proton $(H^+)$ donor. $CCl_4$ does not contain any hydrogen atoms,so it cannot act as a Bronsted acid.
$A$ Lewis acid is an electron-pair acceptor. In $CCl_4$,the carbon atom has a complete octet and cannot accept an electron pair. Therefore,$CCl_4$ cannot act as a Lewis acid.
$BF_3$,$AlCl_3$,and $SnCl_4$ have incomplete octets or vacant d-orbitals,allowing them to act as Lewis acids.
Thus,the correct option is $(D)$.

(A) The species which can accept as well as donate $H^{+}$ ions are known as amphoteric species and can act both as an acid and a base.
$HSO_4^- + H^{+} \rightleftharpoons H_2SO_4$ (Acting as a base)
$HSO_4^- \rightleftharpoons SO_4^{2-} + H^{+}$ (Acting as an acid)

(A) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
Comparing the conjugate acids: $NH_3$ (weakest acid),$H_2O$,$OH^-$,and $HS^-$.
Since $NH_3$ is the weakest acid among the corresponding conjugate acids,its conjugate base $NH_2^-$ is the strongest base.

(B) Lewis acid is defined as an electron-pair acceptor.
$Ag^{+}$ is a metal cation with an incomplete valence shell,making it electron-deficient.
Therefore,$Ag^{+}$ can accept an electron pair from a Lewis base to form a coordinate bond,acting as a Lewis acid.
$Cl^{-}$,$C_2H_5OH$,and $S^{2-}$ possess lone pairs of electrons and act as Lewis bases.