Acids and Bases Questions in English

(C) The strength of an acid is directly proportional to its dissociation constant $(K_a)$.
Given $K_a$ values are: $K_a(A) = 1.8 \times 10^{-5}$,$K_a(B) = 1.5 \times 10^{-4}$,and $K_a(C) = 5.2 \times 10^{-8}$.
Comparing these values,the order of acidic strength is $C < A < B$.
Since the strength of a conjugate base is inversely proportional to the strength of its parent acid,the order of basic strength is $B < A < C$.

(A) When $BCl_3$ is dissolved in water,it undergoes hydrolysis to form boric acid and hydrochloric acid according to the reaction: $BCl_3 + 3H_2O \rightarrow H_3BO_3 + 3HCl$.
$HCl$ is a strong acid that dissociates completely in water to produce $H^+$ ions,while $H_3BO_3$ is a very weak acid.
The presence of $HCl$ makes the resulting solution acidic.
Therefore,the $pH$ of the solution will be less than $7$ $(pH < 7)$.

(D) According to the Bronsted-Lowry theory,a base is a substance that accepts a proton $(H^{+})$.
In the forward reaction: $HCO_3^{-} + H_2O \rightleftharpoons CO_3^{2-} + H_3O^{+}$.
$HCO_3^{-}$ acts as a Bronsted acid because it donates a proton to form $CO_3^{2-}$.
$H_2O$ acts as a Bronsted base because it accepts a proton to form $H_3O^{+}$.
In the reverse reaction: $CO_3^{2-} + H_3O^{+} \rightleftharpoons HCO_3^{-} + H_2O$.
$CO_3^{2-}$ acts as a Bronsted base because it accepts a proton to form $HCO_3^{-}$.
$H_3O^{+}$ acts as a Bronsted acid because it donates a proton to form $H_2O$.
Thus,the two Bronsted bases in the reaction are $H_2O$ and $CO_3^{2-}$.

(A) Given that $[OH^{-}] < 10^{-8} \ M$.
Taking negative logarithm on both sides,we get $pOH > -\log(10^{-8})$,which implies $pOH > 8$.
Since $pH + pOH = 14$ at $25^{\circ} C$,we have $pH = 14 - pOH$.
Substituting the value,$pH < 14 - 8$,so $pH < 6$.
Since the $pH$ is less than $7$,the solution is acidic.

(A) Citrus fruits like lemon contain $Citric \ acid$.
This acid is responsible for the sour taste of lemon.

(D) Lewis base is defined as a substance that can donate a lone pair of electrons to form a coordinate covalent bond.
$NH_3$ (ammonia) has a nitrogen atom with one lone pair of electrons,which it can donate.
$BF_3$,$Cu^{2+}$,and $AlCl_3$ are all electron-deficient species or cations that act as Lewis acids by accepting electron pairs.

(A) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = c \alpha^2$,where $c$ is the concentration and $\alpha$ is the degree of dissociation.
Given: $c = 0.001 \ M = 10^{-3} \ M$ and $\alpha = 2 \% = 0.02 = 2 \times 10^{-2}$.
Substituting the values: $K_b = (10^{-3}) \times (2 \times 10^{-2})^2$.
$K_b = 10^{-3} \times 4 \times 10^{-4} = 4 \times 10^{-7}$.

(B) Among the given options,$CH_3COOH$ (acetic acid) is a weak acid because it partially dissociates in water to release $H^+$ ions.
$C_6H_6$ (benzene),$CH_2=CH_2$ (ethene),and $CH_3COCH_3$ (acetone) are not considered acids in the context of aqueous solutions as they do not readily donate $H^+$ ions.

(A) conjugate acid-base pair differs by only one proton $(H^{+})$.
In the reaction: $HPO_4^{2-} + H_2O \rightleftharpoons PO_4^{3-} + H_3O^{+}$
$1$. $HPO_4^{2-}$ acts as an acid and loses a proton to form its conjugate base,$PO_4^{3-}$. Thus,$(HPO_4^{2-}, PO_4^{3-})$ is a conjugate acid-base pair.
$2$. $H_2O$ acts as a base and gains a proton to form its conjugate acid,$H_3O^{+}$. Thus,$(H_3O^{+}, H_2O)$ is a conjugate acid-base pair.
Comparing this with the given options,$H_3O^{+}$ and $H_2O$ represent a conjugate acid-base pair.

(A) conjugate acid-base pair differs by only one proton $(H^+)$.
In the reaction $HSO_{3(aq)}^{-} + H_3O_{(aq)}^{+} \rightleftharpoons H_2SO_3 + H_2O$,we can identify the pairs as follows:
$1$. $H_2SO_3$ is the conjugate acid of the base $HSO_3^-$.
$2$. $H_3O^+$ is the conjugate acid of the base $H_2O$.
Therefore,$H_2SO_3$ and $HSO_3^-$ form a conjugate acid-base pair.

(D) conjugate acid is formed by adding a proton $(H^+)$ to the species. For $HCO_3{ }^{-}$,the conjugate acid is $HCO_3{ }^{-} + H^+ \rightarrow H_2CO_3$.
$A$ conjugate base is formed by removing a proton $(H^+)$ from the species. For $HCO_3{ }^{-}$,the conjugate base is $HCO_3{ }^{-} - H^+ \rightarrow CO_3{ }^{2-}$.
Therefore,the conjugate acid is $H_2CO_3$ and the conjugate base is $CO_3{ }^{2-}$.

(B) According to the Bronsted-Lowry theory,a base is defined as a proton $(H^+)$ acceptor.
In the reaction $HCl + NH_3 \rightleftharpoons NH_4^+ + Cl^-$,$NH_3$ accepts a proton from $HCl$ to form $NH_4^+$.
Therefore,$NH_3$ acts as a Bronsted-Lowry base.

(B) An amphoteric substance is one that can act as both an acid and a base.
$H_2O$ is a classic example of an amphoteric substance.
As an acid: $H_2O_{(l)} + NH_{3_{(aq)}} \rightleftharpoons OH_{(aq)}^{-} + NH_{4_{(aq)}}^{+}$
As a base: $H_2O_{(l)} + HCl_{(aq)} \rightleftharpoons H_3O_{(aq)}^{+} + Cl_{(aq)}^{-}$

(C) conjugate acid-base pair differs by a single proton $(H^{+})$.
In the reaction $HCl + NH_3 \rightleftharpoons NH_4^{+} + Cl^{-}$:
$1$. $HCl$ acts as an acid and loses a proton to form its conjugate base,$Cl^{-}$. Thus,$(HCl, Cl^{-})$ is a conjugate acid-base pair.
$2$. $NH_3$ acts as a base and gains a proton to form its conjugate acid,$NH_4^{+}$. Thus,$(NH_4^{+}, NH_3)$ is a conjugate acid-base pair.
Comparing this with the given options,option $C$ represents a correct conjugate acid-base pair.

(D) conjugate base is formed when a Bronsted-Lowry acid donates a proton $(H^{+})$.
For the acid $HClO_4$,the reaction is:
$HClO_4 \rightarrow H^{+} + ClO_4^{-}$
Therefore,the conjugate base of $HClO_4$ is $ClO_4^{-}$.

(C) conjugate acid is formed when a Bronsted-Lowry base accepts a proton $(H^{+})$.
For $NH_2^{-}$,the conjugate acid is $NH_2^{-} + H^{+} \rightarrow NH_3$.
For $NH_3$,the conjugate acid is $NH_3 + H^{+} \rightarrow NH_4^{+}$.
Therefore,the conjugate acids are $NH_3$ and $NH_4^{+}$ respectively.

(D) When $NH_3$ reacts with water,it accepts a proton $(H^+)$ from water to form $NH_4^+$ and $OH^-$ ions.
Since it accepts a proton,it acts as a Brønsted-Lowry base.
The reaction is: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$
Here,$NH_3$ acts as a base and $H_2O$ acts as an acid.

(C) According to the $Lewis$ acid-base theory,a $Lewis$ acid is defined as a chemical species that can accept a lone pair of electrons to form a coordinate covalent bond.
Therefore,the correct definition is that it accepts an electron pair.

(B) According to the $Lewis$ acid-base theory,a $Lewis$ base is defined as a substance that can donate a lone pair of electrons to form a coordinate covalent bond.
Therefore,the correct activity is to donate a pair of electrons.

(B) According to the $BRONSTED-LOWRY$ theory,an acid is a proton donor and a base is a proton acceptor.
In the reaction: $HCl_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^{+} + Cl_{(aq)}^{-}$
$HCl$ acts as $Acid-1$ and donates a proton to $H_2O$,which acts as $Base-2$.
After donating the proton,$HCl$ becomes $Cl^-$ $(Base-1)$ and $H_2O$ becomes $H_3O^+$ $(Acid-2)$.
Therefore,$H_2O_{(l)}$ is the base $2$.

(C) According to the Brønsted-Lowry theory,a species that can donate a proton $(H^+)$ acts as an acid,and a species that can accept a proton acts as a base.
$HSO_4^-$ 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,$HSO_4^-$ is an amphoteric species.

(D) conjugate acid-base pair differs by a single proton $(H^{+})$.
In the reaction $H_2O + HCl \rightarrow H_3O^{+} + Cl^{-}$:
$1$. $HCl$ acts as an acid and its conjugate base is $Cl^{-}$.
$2$. $H_2O$ acts as a base and its conjugate acid is $H_3O^{+}$.
Thus,$(H_3O^{+}, H_2O)$ and $(HCl, Cl^{-})$ are the conjugate acid-base pairs.
Therefore,the correct option is $(D)$.

(D) In the reaction $H_2O + HCl_{(aq)} \leftrightharpoons H_3O^{+} + Cl_{(aq)}^{-}$,$H_2O$ acts as a base by accepting a proton $(H^{+})$ to form its conjugate acid,$H_3O^{+}$.
$HCl$ acts as an acid by donating a proton $(H^{+})$ to form its conjugate base,$Cl^{-}$.
Therefore,the conjugate acid is $H_3O^{+}$ and the conjugate base is $Cl^{-}$.

(A) According to the Bronsted-Lowry theory,an acid is a proton $(H^+)$ donor and a base is a proton $(H^+)$ acceptor.
In the reaction $ClO_4^{-} + HCO_3^{-} \rightarrow HClO_4 + CO_3^{2-}$:
$1$. $HCO_3^{-}$ donates a proton to $ClO_4^{-}$ to form $CO_3^{2-}$,so $HCO_3^{-}$ acts as an acid.
$2$. $HClO_4$ is the conjugate acid formed from the base $ClO_4^{-}$. In the reverse reaction,$HClO_4$ acts as a proton donor,so it is an acid.
Therefore,$HCO_3^{-}$ and $HClO_4$ are the acids in the given reaction.

(D) Bronsted acid is a proton $(H^+)$ donor. $HSO_4^-$,$HNO_3$,and $NH_3$ (in specific conditions) can act as Bronsted acids because they contain hydrogen atoms that can be donated.
$BCl_3$ is an electron-deficient compound with only $6$ electrons in the valence shell of the Boron atom. Therefore,it acts as a Lewis acid (electron pair acceptor) but cannot act as a Bronsted acid because it lacks a hydrogen atom to donate.

(B) $H_2O$ is a Lewis base,not a Lewis acid,because it contains $2$ lone pairs of electrons on the oxygen atom,which it can donate.

(C) According to the Arrhenius theory,an acid is a substance that contains hydrogen and dissociates to give $H^{+}$ ions in an aqueous solution.
Lewis acid is defined as a lone pair acceptor.
Bronsted-Lowry acid is a substance that donates a proton ($H^{+}$ ion) to another compound to form a conjugate base.

(C) According to the Bronsted-Lowry theory,a base is a proton $(H^+)$ acceptor.
In the given options,$NH_{3(aq)}$ acts as a Bronsted-Lowry base because it accepts a proton from the water molecule.
The reaction is: $H_2O_{(\ell)} + NH_{3(aq)} \rightleftharpoons NH_{4(aq)}^+ + OH^-_{(aq)}$

(B) conjugate acid-base pair differs by a single proton $(H^{+})$.
In the reaction $HCl_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^{+} + Cl_{(aq)}^{-}$:
$(1)$ $HCl$ 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,$(H_2O, H_3O^{+})$ is a conjugate acid-base pair.
Comparing this with the options,$(H_2O, H_3O^{+})$ is listed as option $(B)$.

(B) conjugate acid is formed by adding a proton $(H^{+})$ to a base.
For the species $HS^{-}$,adding a proton $(H^{+})$ results in $H_2S$.
The reaction is: $HS^{-} + H^{+} \rightleftharpoons H_2S$.
Therefore,the conjugate acid of $HS^{-}$ is $H_2S$.

(B) According to the Brønsted-Lowry theory,a conjugate acid-base pair consists of two species that differ by a single proton $(H^{+})$.
An acid donates a proton to form its conjugate base,and a base accepts a proton to form its conjugate acid.

(C) species that acts as both a Bronsted acid and a base is called an amphoteric species. It must be able to donate a proton $(H^+)$ and accept a proton $(H^+)$.
$HPO_4^{2-}$ can accept a proton to form $H_2PO_4^-$ (acting as a base) and donate a proton to form $PO_4^{3-}$ (acting as an acid).
$HPO_4^{2-} + H_2O \rightleftharpoons H_2PO_4^- + OH^-$
$HPO_4^{2-} + H_2O \rightleftharpoons PO_4^{3-} + H_3O^+$
$H_2PO_2^-$ is the conjugate base of $H_3PO_2$ (a monobasic acid) and cannot donate any more protons.
$HPO_3^{2-}$ is the conjugate base of $H_2PO_3^-$ (derived from the dibasic acid $H_3PO_3$) and cannot donate any more protons.

(D) For a monoacidic base,the concentration of hydroxide ions is given by the formula: $[OH^{-}] = c \times \alpha$.
Here,$c = 0.04 \ M$ and the degree of ionisation $\alpha = 3 \% = 0.03$.
Substituting the values: $[OH^{-}] = 0.04 \times 0.03$.
$[OH^{-}] = 1.2 \times 10^{-3} \ mol \ L^{-1}$.

(D) The strength of a base is inversely proportional to the strength of its conjugate acid.
$1$. $CH_{3}COO^{-}$ has conjugate acid $CH_{3}COOH$ $(pK_{a} \approx 4.75)$.
$2$. $Cl^{-}$ has conjugate acid $HCl$ $(pK_{a} \approx -7)$.
$3$. $OH^{-}$ has conjugate acid $H_{2}O$ $(pK_{a} \approx 15.7)$.
$4$. $CH_{3}O^{-}$ has conjugate acid $CH_{3}OH$ $(pK_{a} \approx 15.5)$.
Comparing the $pK_{a}$ values,$CH_{3}OH$ is the weakest acid among the given options.
Therefore,its conjugate base,$CH_{3}O^{-}$,is the strongest base.

(D) The strength of an acid is directly proportional to its $K_{a}$ value,while the strength of its conjugate base is inversely proportional to the $K_{a}$ value of the parent acid.
Therefore,the acid with the lowest $K_{a}$ value will produce the strongest conjugate base.
Comparing the given $K_{a}$ values: $1.3 \times 10^{-2} > 4 \times 10^{-4} > 1.8 \times 10^{-5} > 4 \times 10^{-10}$.
Since $HCN$ has the lowest $K_{a}$ value $(4 \times 10^{-10})$,it produces the strongest conjugate base.

(D) The conjugate base of a species is formed by the removal of one proton $(H^{+})$ from it.
For $NH_{3}$,the reaction is:
$NH_{3} \longrightarrow H^{+} + NH_{2}^{-}$
Therefore,$NH_{2}^{-}$ is the conjugate base of $NH_{3}$.

(B) conjugate acid-base pair differs by a single proton $(H^{+})$.
To find the conjugate base of an acid,we remove one proton $(H^{+})$ from the species:
$H_{2}PO_{4}^{-} \rightarrow H^{+} + HPO_{4}^{2-}$
Therefore,the conjugate base of $H_{2}PO_{4}^{-}$ is $HPO_{4}^{2-}$.

(A) Bronsted acid is defined as a substance that has a tendency to donate a proton $(H^{+})$.
Conversely,a Lewis acid is defined as a substance that is capable of accepting a lone pair of electrons.
$H_{3}O^{+}$,$NH_{3}$ (as a conjugate acid),and $HCl$ contain $H^{+}$ ions and can act as Bronsted acids.
$BF_{3}$ is an electron-deficient molecule with an incomplete octet on the boron atom,making it an electron pair acceptor,which classifies it as a Lewis acid.

(C) $BF_3$ acts as a Lewis acid because it is an electron-deficient molecule.
The boron atom in $BF_3$ has only $6$ electrons in its valence shell,which means its octet is incomplete.
According to the Lewis concept,an acid is a substance that can accept an electron pair.
Therefore,$BF_3$ accepts an electron pair to complete its octet,classifying it as a Lewis acid.

(A) $NaHCO_{3}$ is an acidic salt (amphoteric) and $NaOH$ is a strong base.
They react with each other to form sodium carbonate and water.
The reaction is:
$NaHCO_{3} + NaOH \rightarrow Na_{2}CO_{3} + H_{2}O$.
Therefore,they cannot coexist in a solution.

(A) The basic strength of a species is inversely proportional to the stability of its conjugate acid.
$1$. The conjugate acids of the given species are:
$H_2O \rightarrow H_3O^{+}$,$NH_3 \rightarrow NH_4^{+}$,$OH^{-} \rightarrow H_2O$,$NH_2^{-} \rightarrow NH_3$.
$2$. Comparing the acidity of the conjugate acids: $H_3O^{+} > NH_4^{+} > H_2O > NH_3$.
$3$. Since the basic strength is the reverse of the acidity of the conjugate acid,the order of basic strength is: $NH_2^{-} > OH^{-} > NH_3 > H_2O$.
$4$. $NH_2^{-}$ is the strongest base because nitrogen is less electronegative than oxygen,making the lone pair more available for donation.

(C) The relationship between the dissociation constant of a weak acid $(K_{a})$ and the dissociation constant of its conjugate base $(K_{b})$ is given by the equation: $K_{a} \times K_{b} = K_{w}$.
At $25^{\circ} C$,the ionic product of water is $K_{w} = 1.0 \times 10^{-14}$.
Given $K_{a} = 1.8 \times 10^{-4}$ for formic acid $(HCOOH)$.
Therefore,$K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-4}}$.
$K_{b} = 0.555 \times 10^{-10} = 5.55 \times 10^{-11}$.

(C) The conjugate base is formed by removing one proton $(H^+)$ from the acid.
Phosphorous acid is $H_3PO_3$. Removing one $H^+$ gives $H_2PO_3^-$. Thus,$x = H_2PO_3^-$.
Oleum is $H_2S_2O_7$. Removing one $H^+$ gives $HS_2O_7^-$. Thus,$y = HS_2O_7^-$.
Therefore,the correct pair is $(x, y) = (H_2PO_3^-, HS_2O_7^-)$.

(D) The conjugate base of an acid is formed by the removal of one proton $(H^{+})$ from the acid molecule.
For chloric acid $(HClO_3)$,the reaction is:
$HClO_3 \rightleftharpoons ClO_3^{-} + H^{+}$
Thus,the conjugate base of $HClO_3$ is $ClO_3^{-}$.

(B) The conjugate acid is formed by adding a proton $(H^{+})$ to the species: $HCO_3^{-} + H^{+} \rightarrow H_2CO_3$.
The conjugate base is formed by removing a proton $(H^{+})$ from the species: $HCO_3^{-} - H^{+} \rightarrow CO_3^{2-}$.
Therefore,the conjugate acid and conjugate base of $HCO_3^{-}$ are $H_2CO_3$ and $CO_3^{2-}$ respectively.

(C) Lewis bases are species that can donate a lone pair of electrons (electron-rich species).
In the given list,the Lewis bases are: $NH_3$ (has a lone pair on $N$),$OH^{-}$ (has lone pairs on $O$),$H^{-}$ (has a lone pair),and $Cl^{-}$ (has lone pairs).
Thus,there are $4$ Lewis bases.
Lewis acids are electron-deficient species: $AlCl_3, H^{+}, Co^{3+}, Mg^{2+}, BF_3$.

(A) Lewis acids are electron-pair acceptors.
$AlCl_3$ (incomplete octet),$H^{+}$ (cation),$Co^{3+}$ (cation),$Mg^{2+}$ (cation),and $BF_3$ (incomplete octet) are all Lewis acids.
$NH_3$ (lone pair donor),$OH^{-}$ (lone pair donor),and $Cl^{-}$ (lone pair donor) are Lewis bases.
Therefore,there are $5$ Lewis acids in the given list.

(A) conjugate base is formed when an acid loses a proton $(H^{+})$.
For the acid $H_3O^{+}$,the reaction is:
$H_3O^{+} \longrightarrow H_2O + H^{+}$
Therefore,the conjugate base of $H_3O^{+}$ is $H_2O$.

(B) conjugate base is formed when an acid loses a proton $(H^{+})$.
The reaction for the dissociation of the ammonium ion $(NH_4^{+})$ is:
$NH_4^{+} \rightarrow NH_3 + H^{+}$
Therefore,the conjugate base of $NH_4^{+}$ is $NH_3$.

(C) conjugate base is formed when an acid loses a proton $(H^{+})$.
For the molecule $NH_3$,the removal of one $H^{+}$ ion results in the formation of the amide ion,$NH_2^{-}$.
Therefore,the conjugate base of $NH_3$ is $NH_2^{-}$.