Acids and Bases Questions in English

(C) Water $(H_2O)$ can act as both a proton donor ($H^+$ donor) and a proton acceptor ($H^+$ acceptor).
According to the Brønsted-Lowry theory,substances that can both donate and accept protons are called amphiprotic or amphoteric.
Therefore,water is an amphiprotic solvent.

(D) $Lewis$ acid-base reaction involves the donation of a lone pair of electrons from a $Lewis$ base to a $Lewis$ acid to form a coordinate covalent bond.
In the reaction $Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$,the $Cu^{2+}$ ion acts as a $Lewis$ acid (electron pair acceptor) and $NH_3$ acts as a $Lewis$ base (electron pair donor).
This reaction does not involve the transfer of a proton $(H^+)$,which is the defining characteristic of a $Br\o nsted$ acid-base reaction. Therefore,it is a $Lewis$ acid-base reaction but not a $Br\o nsted$ acid-base reaction.

(B) In the given reaction $HCl + HF \rightleftharpoons H_2Cl^+ + F^-$,$HCl$ accepts a proton $(H^+)$ from $HF$ to form $H_2Cl^+$.
According to the Bronsted-Lowry theory,a substance that accepts a proton is a base.
Since $HF$ is a stronger acid than $HCl$ in this specific solvent system,$HF$ donates a proton to $HCl$.
Therefore,$HCl$ acts as a base.

(D) Lewis base is defined as a substance that can donate a lone pair of electrons.
$CN^{-}$,$ROH$,and $NH_3$ all possess lone pairs of electrons that can be donated.
$AlCl_3$ is an electron-deficient molecule (it has an incomplete octet) and acts as a Lewis acid because it can accept a pair of electrons.
Therefore,$AlCl_3$ is not a Lewis base.

(D) In the reaction $HCl + CH_3COOH \rightleftharpoons Cl^- + CH_3COOH_2^+$,$HCl$ acts as an acid by donating a proton to form its conjugate base $Cl^-$.
Thus,$(HCl, Cl^-)$ is one conjugate acid-base pair.
$CH_3COOH$ acts as a base by accepting a proton to form its conjugate acid $CH_3COOH_2^+$.
Thus,$(CH_3COOH_2^+, CH_3COOH)$ is the other conjugate acid-base pair.

(A) According to the Bronsted-Lowry theory,an acid is a proton $(H^+)$ donor and a base is a proton acceptor.
$NH_4^+$ can donate a proton to form $NH_3$ $(NH_4^+ \rightarrow NH_3 + H^+)$,therefore it acts as a conjugate acid of the base $NH_3$.
It cannot accept a proton to form $NH_5^{2+}$,so it does not act as a base in this context.
Thus,$NH_4^+$ acts as a conjugate acid.

(C) Given that the $pK_b$ of the fluoride ion $(F^-)$ is $10$.
We know that $pK_a + pK_b = pK_w$.
At $25^{\circ}C$,$pK_w = 14$.
Therefore,$pK_a = 14 - pK_b = 14 - 10 = 4$.
Since $pK_a = -\log(K_a)$,we have $K_a = 10^{-pK_a} = 10^{-4}$.
Thus,the ionization constant of hydrofluoric acid is $1 \times 10^{-4}$.

(A) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
The order of acidic strength of the given oxyacids is: $HClO < HClO_2 < HClO_3 < HClO_4$.
Since $HClO$ is the weakest acid among the given options,its conjugate base $ClO^-$ is the strongest Bronsted base.

(B) The conjugate base of an acid is formed by the removal of a proton $(H^+)$.
For the acid $H_2PO_4^-$,the reaction is $H_2PO_4^- \rightarrow H^+ + HPO_4^{2-}$.
Thus,the conjugate base of $H_2PO_4^-$ is $HPO_4^{2-}$.
Option $A$ is incorrect because the $pH$ of $10^{-8} \ M \ HCl$ is slightly less than $7$ due to the contribution of $H^+$ from water.
Option $C$ is incorrect because the auto-protolysis constant $(K_w)$ of water increases with temperature.
Option $D$ is incorrect because the $pH$ at the equivalence point for a weak acid-strong base titration is greater than $7$.

(B) According to the $Lewis$ acid-base theory,a $Lewis$ acid is defined as a substance that can accept a lone pair of electrons to form a coordinate covalent bond.
Therefore,it is an electron pair acceptor.

(C) The reaction is $HNO_3 + H_2O \rightleftharpoons H_3O^{+} + NO_3^{-}$.
According to the Bronsted-Lowry theory,an acid donates a proton $(H^{+})$ to form its conjugate base.
$HNO_3$ acts as an acid and loses a proton to form $NO_3^{-}$.
Therefore,$NO_3^{-}$ is the conjugate base of $HNO_3$.

(C) The conjugate base of an acid is formed by the removal of one proton $(H^+)$ from the acid molecule.
For the given acid,${(CH_3)_2}\mathop N\limits^ + {H_2}$,removing one $H^+$ ion results in the formation of $(CH_3)_2NH$.
Therefore,the conjugate base is $(CH_3)_2NH$.

(A) Lewis base is a substance that can donate a lone pair of electrons. The strength of a Lewis base is inversely proportional to the electronegativity of the atom carrying the lone pair.
Comparing the atoms $C$,$N$,$O$,and $F$ in the same period of the periodic table,electronegativity increases in the order $C < N < O < F$.
Therefore,the basicity (ability to donate electrons) decreases in the order $CH_3^- > NH_2^- > OH^- > F^-$.
$CH_3^-$ is the strongest Lewis base because carbon is the least electronegative among the given atoms,making its lone pair most available for donation.

(B) $Br\text{ø}nsted$ acid is a proton $(H^{+})$ donor,and a $Br\text{ø}nsted$ base is a proton $(H^{+})$ acceptor.
Species that can both donate and accept a proton are called amphoteric.
$HCO_3^{-}$ can act as an acid: $HCO_3^{-} \rightarrow H^{+} + CO_3^{2-}$.
$HCO_3^{-}$ can act as a base: $HCO_3^{-} + H^{+} \rightarrow H_2CO_3$.
Therefore,$HCO_3^{-}$ acts as both a $Br\text{ø}nsted$ acid and a $Br\text{ø}nsted$ base.

(C) The reaction is ${K_2}O + H_2O \to 2KOH$.
Since $KOH$ is a strong base,it dissociates completely in water to produce $OH^-$ ions.
The presence of $OH^-$ ions makes the solution basic.

(A) During digestion,the stomach produces $HCl$,which can lead to acidity.
Excess acidity in the stomach causes discomfort such as ulcers and gastric reflux.
Compounds that neutralize or reduce this acidity are known as antacids.
Both $NaHCO_3$ (sodium bicarbonate) and $Mg(OH)_2$ (magnesium hydroxide) are commonly used as antacids because they react with excess $HCl$ to neutralize it.
The reactions are:
$NaHCO_3 + HCl \to NaCl + H_2O + CO_2$
$Mg(OH)_2 + 2HCl \to MgCl_2 + 2H_2O$

(D) $HN_3$ is known as hydrazoic acid. It acts as an acid in aqueous solution,not as a base.

(C) In the nitrating mixture,concentrated $H_2SO_4$ acts as a strong acid and $HNO_3$ acts as a base. The reaction is as follows:
$HNO_3 + H_2SO_4 \rightleftharpoons H_2NO_3^+ + HSO_4^-$
$H_2NO_3^+ \rightleftharpoons NO_2^+ + H_2O$
Here,$HNO_3$ accepts a proton from $H_2SO_4$,therefore it behaves as a base.

(A) The basic strength of a conjugate base is inversely proportional to the acidic strength of its conjugate acid.
The conjugate acids are $HCl, CH_3COOH, H_2O,$ and $HF$.
The acidic strength order is $HCl > HF > CH_3COOH > H_2O$.
Therefore,the basic strength order of their conjugate bases is $Cl^{-} < F^{-} < CH_3COO^{-} < OH^{-}$.

(D) The dissociation constant $(K_a)$ is a measure of the acidity of a compound. Higher acidity corresponds to a higher $K_a$ value.
$1$. $C_6H_5OH$ (Phenol) is a weak acid $(pK_a \approx 10)$.
$2$. $C_6H_5CH_2OH$ (Benzyl alcohol) is less acidic than phenol $(pK_a \approx 15)$.
$3$. $CH_3C \equiv CH$ (Propyne) is a very weak acid $(pK_a \approx 25)$.
$4$. $CH_3NH_3^+ Cl^-$ (Methylammonium chloride) acts as a conjugate acid of a weak base,making it significantly more acidic than the others ($pK_a \approx 10.6$ for the ammonium ion,but in aqueous solution,the salt provides $CH_3NH_3^+$,which is a stronger acid compared to the neutral organic molecules listed).
Comparing the acidity,the methylammonium ion is the strongest acid among the given options,thus it has the highest dissociation constant.

(A) The basicity of a conjugate base is inversely proportional to the acidity of its corresponding acid.
$1$. The corresponding acids are: $RCOOH$ (carboxylic acid),$HC \equiv CH$ (alkyne),$NH_3$ (ammonia),and $RH$ (alkane).
$2$. The acidity order of these acids is: $RCOOH > HC \equiv CH > NH_3 > RH$.
$3$. Therefore,the basicity order of their conjugate bases is the reverse: $RCOO^- < HC \equiv C^- < NH_2^- < R^-$.
$4$. Thus,the correct order is $RCOO^- < HC \equiv C^- < NH_2^- < R^-$.

(D) Ammonium sulphate,$(NH_4)_2SO_4$,is a physiological acidic fertilizer.
When it is added to the soil,the ammonium ion $(NH_4^+)$ undergoes nitrification by soil bacteria to form nitrate $(NO_3^-)$ and releases hydrogen ions $(H^+)$ into the soil.
These $H^+$ ions increase the acidity of the soil,thereby lowering its $pH$.

(B) Tomatoes contain $Oxalic \ acid$.

(A) The strength of an acid is inversely proportional to its $pK_a$ value.
Therefore,the lower the $pK_a$ value,the stronger the acid.
Among the given options,$1$ is the lowest value,representing the strongest acid.

(A) The strength of an acid is directly proportional to its dissociation constant $(K_a)$.
Comparing the given values:
$A = 2 \times 10^{-2} = 0.02$
$B = 0.02 \times 10^{-1} = 0.002$
$C = 3 \times 10^{-4} = 0.0003$
$D = 2 \times 10^{-4} = 0.0002$
Since $0.02$ is the largest value,the acid with $K_a = 2 \times 10^{-2}$ is the strongest.

(C) Given $pK_b = 10.83$ for $F^-$.
The base dissociation constant $K_b = 10^{-pK_b} = 10^{-10.83} = 1.479 \times 10^{-11} \approx 1.48 \times 10^{-11}$.
At $25\,^{\circ}C$,the ionic product of water $K_w = 10^{-14}$.
The relationship between $K_a$ and $K_b$ for a conjugate acid-base pair is $K_a \times K_b = K_w$.
Therefore,$K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{1.48 \times 10^{-11}} = 6.756 \times 10^{-4} \approx 6.75 \times 10^{-4}$.

(C) The relative strength of bases is inversely proportional to the strength of their conjugate acids.
$1$. The conjugate acid of $OH^{-}$ is $H_2O$ $(K_a \approx 10^{-14})$.
$2$. The conjugate acid of $CH_3COO^{-}$ is $CH_3COOH$ $(K_a \approx 1.8 \times 10^{-5})$.
$3$. The conjugate acid of $Cl^{-}$ is $HCl$ (a strong acid,$K_a$ is very high).
Since the strength of conjugate acids follows the order $HCl > CH_3COOH > H_2O$,the strength of the corresponding conjugate bases follows the reverse order: $OH^{-} > CH_3COO^{-} > Cl^{-}$.

(A) In the reaction $HSO_4^- + OH^- \to SO_4^{2-} + H_2O$,the species $HSO_4^-$ acts as an acid because it donates a proton $(H^+)$ to $OH^-$.
After donating the proton,$HSO_4^-$ becomes $SO_4^{2-}$,which is its conjugate base.
Therefore,$HSO_4^-$ is the conjugate acid of the base $SO_4^{2-}$.

(B) The strength of a base is directly proportional to its dissociation constant $(K_b)$.
Comparing the given values:
$1.6 \times 10^{-6} > 6.3 \times 10^{-10} > 3.8 \times 10^{-10} > 5.6 \times 10^{-4}$ is incorrect; the correct order of magnitude is $5.6 \times 10^{-4} > 1.6 \times 10^{-6} > 6.3 \times 10^{-10} > 3.8 \times 10^{-10}$.
The smallest value of $K_b$ is $3.8 \times 10^{-10}$,which corresponds to $C_6H_5NH_2$.
Therefore,$C_6H_5NH_2$ is the weakest base.

(A) Non-metallic oxides are acidic in nature. When they dissolve in water,they form acidic solutions.
For example,$P_2O_5 + 3H_2O \to 2H_3PO_4$ (phosphoric acid).

(A) Proton affinity is directly related to the basicity of the species.
In a period,as electronegativity increases,the ability of the atom to hold its lone pair increases,thus decreasing basicity. Comparing $NH_{2}^{-}$ and $F^{-}$,nitrogen is less electronegative than fluorine,so $NH_{2}^{-}$ is a stronger base.
In a group,as atomic size increases,the charge density decreases,which reduces the basicity. Comparing $F^{-}$ and $I^{-}$,$F^{-}$ is a stronger base.
Comparing $F^{-}$ and $HS^{-}$,$F^{-}$ is a stronger base because oxygen (in $OH^{-}$) is more electronegative than sulfur (in $HS^{-}$),and $F^{-}$ is smaller than $HS^{-}$.
Therefore,$NH_{2}^{-}$ has the highest proton affinity among the given species.

(D) For a polyprotic acid $H_2A$,the dissociation steps are:
$H_2A \rightleftharpoons H^{+} + HA^{-}$; $K_1 = \frac{[H^{+}][HA^{-}]}{[H_2A]} = 1.0 \times 10^{-5}$
$HA^{-} \rightleftharpoons H^{+} + A^{2-}$; $K_2 = \frac{[H^{+}][A^{2-}]}{[HA^{-}]} = 5.0 \times 10^{-10}$
The overall dissociation reaction is the sum of these two steps:
$H_2A \rightleftharpoons 2H^{+} + A^{2-}$
The overall dissociation constant $K$ is the product of the individual dissociation constants:
$K = K_1 \times K_2$
$K = (1.0 \times 10^{-5}) \times (5.0 \times 10^{-10})$
$K = 5.0 \times 10^{-15}$

(C) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^{+})$.
$1. \, HSO_3F$ is a superacid,which is the strongest among the given species.
$2. \, H_3O^{+}$ is a strong acid $(pK_a \approx -1.7)$.
$3. \, HSO_4^-$ is a moderately strong acid $(pK_a \approx 1.99)$.
$4. \, HCO_3^-$ is a weak acid $(pK_a \approx 10.3)$.
Therefore,the correct order of acidic strength is:
$HCO_3^- < HSO_4^- < H_3O^{+} < HSO_3F$
Which corresponds to the sequence: $i < iii < ii < iv$.

(A) An acid is a substance that donates a proton $(H^{+})$.
In reaction $(i)$,$H_2PO_4^-$ is formed as a product,so it is not acting as an acid.
In reaction $(ii)$,$H_2PO_4^- + H_2O \rightarrow HPO_4^{2-} + H_3O^{+}$,$H_2PO_4^-$ donates a proton to $H_2O$,thus it acts as an acid.
In reaction $(iii)$,$H_2PO_4^- + OH^{-} \rightarrow H_2O + HPO_4^{2-}$,$H_2PO_4^-$ donates a proton to $OH^{-}$,thus it also acts as an acid.
Therefore,$H_2PO_4^-$ acts as an acid in both $(ii)$ and $(iii)$.

(D) The strength of a conjugate base is inversely proportional to the strength of its corresponding conjugate acid.
First,we determine the acidic strength order of the corresponding conjugate acids: $RCOOH > HC \equiv CH > NH_3 > RH$.
Since $RCOOH$ is the strongest acid,its conjugate base $RCOO^{-}$ is the weakest base.
Since $RH$ (alkane) is the weakest acid,its conjugate base $R^{-}$ is the strongest base.
Therefore,the order of increasing basicity is: $RCOO^{-} < HC \equiv C^{-} < NH_2^{-} < R^{-}$.

(B) The acidity of oxoacids depends on the stability of their conjugate bases.
$HNO_3$ (nitric acid) forms the nitrate ion $(NO_3^-)$ as its conjugate base,which is stabilized by resonance.
$HNO_2$ (nitrous acid) forms the nitrite ion $(NO_2^-)$ as its conjugate base.
Since the conjugate base of $HNO_3$ is more resonance-stabilized than that of $HNO_2$,$HNO_3$ is a stronger acid than $HNO_2$.
Other options are incorrect because:
$HClO_4$ is a stronger acid than $HClO_3$ due to higher oxidation state of $Cl$.
$H_2SO_3$ is a stronger acid than $H_3PO_3$.
$HCl$ is a stronger acid than $HF$ in aqueous medium due to the high bond dissociation energy of $H-F$.

(D) Zinc oxide $(ZnO)$ is an amphoteric oxide,meaning it can react with both acids and bases.
In reaction $(A)$,$ZnO + Na_2O \rightarrow Na_2ZnO_2$,$ZnO$ acts as an acid because it reacts with the base $Na_2O$.
In reaction $(B)$,$ZnO + CO_2 \rightarrow ZnCO_3$,$ZnO$ acts as a base because it reacts with the acidic oxide $CO_2$.
Therefore,$ZnO$ acts as an acid in $(A)$ and a base in $(B)$.

(A) In any acid-base equilibrium,the equilibrium lies towards the side of the weaker acid and weaker base.
$1$. For $N_2H_5^+ + NH_3 \rightleftharpoons NH_4^+ + N_2H_4$,since the equilibrium lies to the right,$N_2H_5^+$ is a stronger acid than $NH_4^+$.
$2$. For $N_2H_4 + HBr \rightleftharpoons N_2H_5^+ + Br^{-}$,since the equilibrium lies to the right,$HBr$ is a stronger acid than $N_2H_5^+$.
Combining these two observations,we get the order of acidic strength as $HBr > N_2H_5^+ > NH_4^+$.

(B) The conjugate acid of a species is formed by adding a proton $(H^{+})$ to it.
For the species $HPO_4^{2-}$,the conjugate acid is obtained by adding $H^{+}$:
$HPO_4^{2-} + H^{+} \rightarrow H_2PO_4^-$
Therefore,the correct option is $B$.

(B) The conjugate acid of a base is formed by adding a proton $(H^{+})$ to the base.
For the species $HPO_4^{2-}$,the addition of $H^{+}$ results in:
$HPO_4^{2-} + H^{+} \rightarrow H_2PO_4^-$
Therefore,the conjugate acid of $HPO_4^{2-}$ is $H_2PO_4^-$.

(C) When $K_2O$ is added to water,it reacts to form potassium hydroxide: $K_2O + H_2O \rightarrow 2KOH$.
$KOH$ is a strong base that dissociates completely in water to produce $K^{+}$ and $OH^{-}$ ions: $KOH \rightarrow K^{+} + OH^{-}$.
The presence of a significant concentration of $OH^{-}$ ions makes the solution basic.
Hence,option $C$ is correct.

(B) reaction is favoured in the forward direction if a stronger acid reacts with a stronger base to form a weaker acid and a weaker base.
In option $B$,the reaction is:
$C_6H_5NH_3^+ + o-CH_3C_6H_4NH_2 \rightleftharpoons C_6H_5NH_2 + o-CH_3C_6H_4NH_3^+$.
Here,$o$-toluidine $(o-CH_3C_6H_4NH_2)$ is a stronger base than aniline $(C_6H_5NH_2)$ due to the electron-donating $+I$ effect of the $-CH_3$ group.
Therefore,$o$-toluidine will accept the proton more readily than aniline,shifting the equilibrium to the right (forward direction).

(A) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
$K_a$ values are given as: $HSO_4^- (1.2 \times 10^{-2}) > H_2PO_4^- (6.3 \times 10^{-8}) > HCO_3^- (4.7 \times 10^{-11})$.
Since $HCO_3^-$ is the weakest acid among the given species,its conjugate base,$CO_3^{2-}$,will be the strongest base.
Therefore,the correct option is $A$.

(B) The relative strength of two acids is given by the ratio of the square roots of their dissociation constants: $\text{Relative strength} = \sqrt{\frac{K_{a1}}{K_{a2}}}$
Given $K_{a1} = 2.88 \times 10^{-4}$ and $K_{a2} = 1.8 \times 10^{-5}$.
$\text{Relative strength} = \sqrt{\frac{2.88 \times 10^{-4}}{1.8 \times 10^{-5}}} = \sqrt{\frac{28.8 \times 10^{-5}}{1.8 \times 10^{-5}}} = \sqrt{16} = 4$.
Thus,the ratio is $4 : 1$.

(D) The strength of an acid is directly proportional to its acid dissociation constant,$K_a$.
From the given data,$HI$ has the highest $K_a$ value $(3 \times 10^{11})$,which makes it the strongest Bronsted acid among the series.
According to the Bronsted-Lowry theory,the conjugate base of a strong acid is a weak base,and the conjugate base of a very strong acid is the weakest.
Therefore,since $HI$ is the strongest acid,its conjugate base $(I^-)$ is the weakest conjugate base.

(C) conjugate acid is formed when a base accepts a proton $(H^+)$.
Conversely,a conjugate base is formed when an acid loses a proton $(H^+)$.
To find the base for which $HCO_3^-$ is the conjugate acid,we remove a proton $(H^+)$ from $HCO_3^-$.
$HCO_3^- - H^+ \rightarrow CO_3^{2-}$.
Therefore,$CO_3^{2-}$ is the base,and $HCO_3^-$ is its conjugate acid.

(C) For a conjugate acid-base pair,the relationship between $pK_a$ and $pK_b$ at $25^{\circ}C$ is given by:
$pK_a + pK_b = 14$
Given $pK_b = 5$,we can calculate $pK_a$:
$pK_a = 14 - 5 = 9$
Since $K_a = 10^{-pK_a}$,we substitute the value of $pK_a$:
$K_a = 10^{-9}$

(B) In the given equilibrium reactions,the species on the right side are the conjugate acids and conjugate bases.
$HF + HClO_4 \rightleftharpoons H_2F^{+} + ClO_4^-$
Here,$H_2F^{+}$ is the conjugate acid of $HF$.
$H_2O + H_2PO_4^- \rightleftharpoons H_3O^{+} + HPO_4^{2-}$
Here,$H_3O^{+}$ is the conjugate acid of $H_2O$.
Therefore,the acidic species on the right side are $H_2F^{+}$ and $H_3O^{+}$.