Acids and Bases Questions in English

(A) conjugate acid is formed when a base accepts a proton $(H^{+})$.
$NH_{3} + H^{+} \rightarrow NH_{4}^{+}$
$HCO_{3}^{-} + H^{+} \rightarrow H_{2}CO_{3}$
$H_{2}O + H^{+} \rightarrow H_{3}O^{+}$
$HSO_{4}^{-} + H^{+} \rightarrow H_{2}SO_{4}$
Therefore,the correct matching is $(a$ $\rightarrow iii), (b$ $\rightarrow i), (c$ $\rightarrow ii), (d$ $\rightarrow iv)$.

(D) An acid loses a proton to form an anion known as its conjugate base.
According to the Bronsted-Lowry theory,a strong acid forms a weak conjugate base,and a weak acid forms a strong conjugate base.
Among the given halide ions $(F^{-}, Cl^{-}, Br^{-}, I^{-})$,the corresponding acids are $HF, HCl, HBr,$ and $HI$.
$HI$ is the strongest acid and $HF$ is the weakest acid among these hydrohalic acids.
Since $HF$ is the weakest acid,its conjugate base $F^{-}$ is the strongest conjugate base.

(D) Bronsted-Lowry acid is a proton ($H^{+}$ ion) donor,while a Bronsted-Lowry base is a proton ($H^{+}$ ion) acceptor.
$HSO_4^{-} + H_2O \rightleftharpoons H_2SO_4 + OH^{-}$ (acts as a Bronsted base)
$HSO_4^{-} + H_2O \rightleftharpoons SO_4^{2-} + H_3O^{+}$ (acts as a Bronsted acid)
Since $HSO_4^{-}$ can both donate a proton to form $SO_4^{2-}$ and accept a proton to form $H_2SO_4$,it acts as both a Bronsted acid and a Bronsted base.

(C) $BF_3$ is electron-deficient,meaning it acts as a lone-pair of electrons acceptor (Lewis acid).
In $BF_3$,the central $B$ atom is $sp^2$ hybridized and does not satisfy the octet rule because it has only $6$ bonded electrons in its valence shell.

(B) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
Among the given hydrohalic acids,the acidic strength order is $HF < HCl < HBr < HI$.
Therefore,the order of basic strength of their conjugate bases is $F^{-} > Cl^{-} > Br^{-} > I^{-}$.
Thus,$F^{-}$ is the strongest conjugate base.

(C) According to the Bronsted-Lowry theory,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 $HSO_4^{-}$,adding $H^{+}$ gives $H_2SO_4$.
For $NH_3$,adding $H^{+}$ gives $NH_4^{+}$.
Therefore,the conjugate acids are $H_2SO_4$ and $NH_4^{+}$.
Thus,the correct option is $C$.

(D) Bronsted acid is a proton $(H^{+})$ donor,and a Bronsted base is a proton $(H^{+})$ acceptor. Amphoteric species can act as both.
$1$. $HCl$: Only acts as an acid $(HCl \rightarrow H^{+} + Cl^{-})$.
$2$. $ClO_4^{-}$: Only acts as a base $(ClO_4^{-} + H^{+} \rightarrow HClO_4)$.
$3$. $OH^{-}$: Only acts as a base $(OH^{-} + H^{+} \rightarrow H_2O)$.
$4$. $H^{+}$: Only acts as an acid.
$5$. $H_2O$: Acts as an acid $(H_2O \rightarrow H^{+} + OH^{-})$ and as a base $(H_2O + H^{+} \rightarrow H_3O^{+})$.
$6$. $HSO_4^{-}$: Acts as an acid $(HSO_4^{-} \rightarrow H^{+} + SO_4^{2-})$ and as a base $(HSO_4^{-} + H^{+} \rightarrow H_2SO_4)$.
$7$. $SO_4^{2-}$: Only acts as a base $(SO_4^{2-} + H^{+} \rightarrow HSO_4^{-})$.
$8$. $H_2SO_4$: Only acts as an acid $(H_2SO_4 \rightarrow H^{+} + HSO_4^{-})$.
$9$. $Cl^{-}$: Only acts as a base $(Cl^{-} + H^{+} \rightarrow HCl)$.
Thus,only $H_2O$ and $HSO_4^{-}$ can act as both Bronsted acids and bases. The total number is $2$.

(D) The acidic strength of an acid is directly proportional to its dissociation constant ($K_{a}$ value).
Given $K_{a}$ values are:
$A = 1.8 \times 10^{-4}$
$B = 5 \times 10^{-10}$
$C = 3 \times 10^{-8}$
Comparing the values: $1.8 \times 10^{-4} > 3 \times 10^{-8} > 5 \times 10^{-10}$.
Therefore,the order of acidic strength is $A > C > B$.

(B) The $pK_a$ value is inversely proportional to the acid strength. Stronger acids have lower $pK_a$ values.
The order of acid strength for hydrohalic acids is $HF < HCl < HBr < HI$.
The corresponding $pK_a$ values are:
$HF$ $(pK_a \approx 3.1)$
$HCl$ $(pK_a \approx -6.0)$
$HBr$ $(pK_a \approx -9.0)$
$HI$ $(pK_a \approx -9.5)$
Thus,$HF$ is the weakest acid among the given options and possesses the highest $pK_a$ value.
Therefore,the correct option is $(B)$.

(B) Both lemon juice and orange juice are rich sources of citric acid.
Citric acid is the primary organic acid found in citrus fruits,including lemons,limes,oranges,and grapefruits.
Therefore,both contain citric acid.

(A) The relative strength of an acid is directly proportional to its ionisation constant $(K_a)$.
$A$ larger $K_a$ value indicates a stronger acid because it ionises to a greater extent in aqueous solution.
Comparing the given $K_a$ values:
$HCN: 4.9 \times 10^{-10}$ $(4)$
$HClO: 3.0 \times 10^{-8}$ $(2)$
$HCOOH: 1.8 \times 10^{-4}$ $(1)$
$HNO_2: 4.5 \times 10^{-4}$ $(3)$
Arranging them in increasing order of $K_a$: $4.9 \times 10^{-10} < 3.0 \times 10^{-8} < 1.8 \times 10^{-4} < 4.5 \times 10^{-4}$.
Therefore,the increasing order of acid strength is $4 < 2 < 1 < 3$.

(A) The acid dissociation constant $(K_a)$ values for the given acids are as follows:
$A$. Phenol $(C_6H_5OH)$: $K_a \approx 1.0 \times 10^{-10}$ (Match $III$)
$B$. Benzoic acid $(C_6H_5COOH)$: $K_a \approx 6.5 \times 10^{-5}$ (Match $IV$)
$C$. Hypochlorous acid $(HClO)$: $K_a \approx 3.0 \times 10^{-8}$ (Match $II$)
$D$. Acetic acid $(CH_3COOH)$: $K_a \approx 1.75 \times 10^{-5}$ (Match $V$)
Thus,the correct matching is $A-III, B-IV, C-II, D-V$.

(A) According to the Lewis concept,a base is a substance that can donate an electron pair to form a coordinate covalent bond.
In the $NH_3$ molecule,the nitrogen atom has one lone pair of electrons.
Because $NH_3$ can donate this lone pair of electrons to an electron-deficient species,it acts as a Lewis base.

(D) conjugate acid-base pair differs by only one proton $(H^+)$.
$(a)$ $HPO_3^{2-}$ and $PO_3^{3-}$ differ by one $H^+$,so they form a conjugate pair.
$(b)$ $H_2PO_4^{-}$ and $HPO_4^{2-}$ differ by one $H^+$,so they form a conjugate pair.
$(c)$ $H_3PO_4$ and $H_2PO_4^{-}$ differ by one $H^+$,so they form a conjugate pair.
$(d)$ $H_2PO_4^{-}$ and $PO_4^{3-}$ differ by two protons $(2H^+)$,therefore they do not form a conjugate acid-base pair.

(A) $1$. Option $A$: The Bronsted-Lowry theory defines acids as proton donors. $BCl_3$ is a Lewis acid because it accepts an electron pair,but it does not contain a proton to donate. Thus,the theory cannot explain its acidity. This statement is correct.
$2$. Option $B$: For $0.01 \ M \ NaOH$,$[OH^-] = 10^{-2} \ M$. Thus,$pOH = -\log(10^{-2}) = 2$. Therefore,$pH = 14 - 2 = 12$. The statement is incorrect.
$3$. Option $C$: The ionic product of water $(K_w)$ at $25^{\circ}C$ is $10^{-14} \ mol^2 \ L^{-2}$,not $10^{-10}$. The statement is incorrect.
$4$. Option $D$: The correct equation is $pH = -\log [H^+]$. The statement is incorrect.

(D) The acid dissociation constants $(K_a)$ for the given acids are as follows:
$HCN$: $K_a = 4.9 \times 10^{-10}$ $(III)$
$H_2C_2O_4$ (Oxalic acid): $K_a = 5.6 \times 10^{-2}$ $(IV)$
$H_2S$: $K_a = 8.9 \times 10^{-8}$ $(II)$
Niacin (Vitamin $B_3$): $K_a = 1.5 \times 10^{-5}$ $(V)$
Therefore,the correct matching is $A-III, B-IV, C-II, D-V$.

(D) For a tribasic acid like $H_3PO_4$,there are three ionisation constants. The overall ionisation constant $(K)$ is the product of the ionisation constants of the three individual steps.
Step $1$: $H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-$; $K_1 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}$ $(i)$
Step $2$: $H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$; $K_2 = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$ $(ii)$
Step $3$: $HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}$; $K_3 = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ $(iii)$
Adding the three equations gives the overall reaction: $H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-}$.
The overall equilibrium constant $K$ is given by $K = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]}$.
Multiplying the three stepwise equilibrium constants:
$K_1 \times K_2 \times K_3 = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} \times \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} \times \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$
$K_1 \times K_2 \times K_3 = \frac{[H^+]^3[PO_4^{3-}]}{[H_3PO_4]} = K$
Therefore,$K = K_1 \times K_2 \times K_3$.

(C) $Cl^{-}$ is a Lewis base,not a Lewis acid,because its octet is complete and it has a lone pair of electrons to donate.
For $1.0 \times 10^{-8} \ M \ HCl$,the contribution of $H^{+}$ from water must be considered,resulting in a $pH$ slightly less than $7$ (approx $6.98$).
The ionic product of water $(K_w)$ at $25^{\circ} C$ is $1.0 \times 10^{-14} \ mol^2 \ L^{-2}$.
$AlCl_{3}$ acts as a Lewis acid due to an incomplete octet,which is not explained by the Bronsted-Lowry theory.

(A) The acid strength is directly proportional to the acid dissociation constant $(K_a)$. The order of acid strength for hydrohalic acids is $HI > HBr > HCl > HF$.
Comparing the given $K_a$ values:
$HI: 3.2 \times 10^9$ $(i)$
$HBr: 1.0 \times 10^9$ $(iv)$
$HCl: 1.3 \times 10^6$ $(iii)$
$HF: 3.5 \times 10^{-4}$ $(ii)$
Therefore,the correct matching is $A-(i), B-(ii), C-(iii), D-(iv)$.

(D) Among hydrogen halides,the strongest acid is $HI$.
As the size of the halogen increases,the bond length increases,and consequently,the bond dissociation energy decreases.
Therefore,$HI$ releases $H^{\oplus}$ ions most easily,exhibiting the highest acidic nature.

(C) diprotic acid is an acid that can donate two protons ($H^+$ ions) per molecule.
$1$. Citric acid $(C_6H_8O_7)$ is a triprotic acid.
$2$. Chromic acid $(H_2CrO_4)$ is a diprotic acid.
$3$. Oxalic acid $(H_2C_2O_4)$ is a diprotic acid.
$4$. Pyrosulfuric acid $(H_2S_2O_7)$ is a diprotic acid.
$5$. Sulfurous acid $(H_2SO_3)$ is a diprotic acid.
Thus,there are $4$ diprotic acids in the given list.

(A) Orthophosphoric acid is $H_3PO_4$. It dissociates as follows:
$(I)$. $H_3PO_4 \rightleftharpoons H_2PO_4^{-} + H^{+}$
$(II)$. $H_2PO_4^{-} \rightleftharpoons HPO_4^{2-} + H^{+}$
$(III)$. $HPO_4^{2-} \rightleftharpoons PO_4^{3-} + H^{+}$
Since it releases $3$ $H^{+}$ ions upon complete dissociation,the number of dissociable protons is $3$.

(A) The strength of oxo-acids increases with the increase in the oxidation number of the central atom.
For $HClO_{3}$,the oxidation number of $Cl$ is $+5$.
For $HClO_{2}$,the oxidation number of $Cl$ is $+3$.
For $HOCl$,the oxidation number of $Cl$ is $+1$.
For $HOBr$,the oxidation number of $Br$ is $+1$.
Since $HClO_{3}$ has the highest oxidation number of the central atom,it is the strongest Bronsted acid among the given compounds.