Acids and Bases Questions in English

(A) $HCl$ is a strong acid that dissociates completely in water according to the reaction: $HCl \rightarrow H^{+} + Cl^{-}$.
Since it is a strong electrolyte,the concentration of $H^{+}$ ions is equal to the concentration of the acid,which is $0.1 \ M$.
$NH_4Cl$ is a salt of a weak base and strong acid,which undergoes hydrolysis to produce a small amount of $H^{+}$ ions.
$NaHCO_3$ and $Na_2CO_3$ are basic salts that produce $OH^{-}$ ions in solution,thereby decreasing the concentration of $H^{+}$ ions.
Therefore,the $0.1 \ M \ HCl$ solution contains the maximum number of $H^{+}$ ions.

(B) According to the Bronsted-Lowry theory,a Bronsted acid is a substance that can donate a proton $(H^+)$.
$CH_3NH_3^+$,$H_2O$,and $HSO_4^-$ all contain at least one hydrogen atom bonded to an electronegative atom,allowing them to act as proton donors.
$CH_3COO^-$ is the conjugate base of acetic acid and does not have any acidic protons to donate.
Therefore,$CH_3COO^-$ is not a Bronsted acid.

(A) The concentration of hydroxide ions is $[OH^-] = 0.05 \ mol \ L^{-1} = 5 \times 10^{-2} \ mol \ L^{-1}$.
$pOH = - \log[OH^-] = - \log(5 \times 10^{-2}) = 2 - \log 5 = 2 - 0.699 = 1.301$.
Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pH = 14 - 1.301 = 12.699$.
Since the $pH > 7$,the solution is basic.

(D) An acid salt is a salt that contains at least one replaceable hydrogen atom in its structure.
$Na_2HPO_4$ is an acid salt because it contains an acidic hydrogen atom that can be released as an $H^{+}$ ion in an aqueous solution.
Therefore,the correct option is $(d)$.

(B) The $pH$ scale measures the acidity or basicity of an aqueous solution.
Distilled water is neutral with a $pH$ of $7$.
$NH_3$ (ammonia) is a weak base. When dissolved in water,it undergoes the reaction: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$.
This reaction increases the concentration of $OH^-$ ions,making the solution basic $(pH > 7)$.
$NH_3$ gas itself does not have a $pH$ until dissolved in water.
Water saturated with $Cl_2$ forms $HCl$ and $HOCl$,making the solution acidic $(pH < 7)$.
Therefore,the $NH_3$ solution in water has the highest $pH$ among the given options.

(B) 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.
Therefore,$NH_3$ acts as a Lewis base.

(A) The strength of a conjugate base is inversely proportional to the strength of its corresponding acid.
The order of acidic strength of the oxyacids of chlorine 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) protonic acid is a substance that can donate a proton ($H^+$ ion) in an aqueous solution.
$SO_2(OH)_2$ (Sulfuric acid),$PO(OH)_3$ (Phosphoric acid),and $SO(OH)_2$ (Sulfurous acid) all contain ionizable $H$ atoms attached to oxygen,which can be released as $H^+$ ions.
$B(OH)_3$ (Boric acid) acts as a Lewis acid by accepting an $OH^-$ ion from water to form $[B(OH)_4]^-$,releasing $H^+$ from the water molecule,not from itself. Therefore,it is not a protonic acid.

(A) The $pK_a$ value is defined as $pK_a = -\log(K_a)$.
For acid $A$,$pK_a = 4$,so $K_a(A) = 10^{-4}$.
For acid $B$,$pK_a = 5$,so $K_a(B) = 10^{-5}$.
The ratio of the strengths of the acids is given by the ratio of their dissociation constants: $\frac{K_a(A)}{K_a(B)} = \frac{10^{-4}}{10^{-5}} = 10$.
Therefore,acid $A$ is $10$ times stronger than acid $B$.

(B) The relative strength of two acids is given by the ratio of their degrees of dissociation,which is equal to the square root of the ratio of their dissociation constants:
$\frac{\text{Strength}_1}{\text{Strength}_2} = \sqrt{\frac{K_{a_1}}{K_{a_2}}}$
Given $K_{a_1} = 3.14 \times 10^{-4}$ and $K_{a_2} = 1.96 \times 10^{-5}$.
$\frac{\text{Strength}_1}{\text{Strength}_2} = \sqrt{\frac{3.14 \times 10^{-4}}{1.96 \times 10^{-5}}} = \sqrt{\frac{31.4 \times 10^{-5}}{1.96 \times 10^{-5}}} \approx \sqrt{16} = 4$.
Thus,the relative strength is $4:1$.

(C) The dissociation of a weak acid is given by: $HA \rightleftharpoons H^{+} + A^{-}$,where $K_a = \frac{[H^{+}][A^{-}]}{[HA]} = 1.0 \times 10^{-5}$ $(I)$.
The reaction of a weak acid with a strong base is: $HA + OH^{-} \rightleftharpoons A^{-} + H_2O$,with equilibrium constant $K = \frac{[A^{-}]}{[HA][OH^{-}]}$ $(II)$.
We know that the auto-ionization of water is: $H_2O \rightleftharpoons H^{+} + OH^{-}$,where $K_w = [H^{+}][OH^{-}] = 1.0 \times 10^{-14}$ $(III)$.
Dividing $(I)$ by $(III)$: $\frac{K_a}{K_w} = \frac{[H^{+}][A^{-}]}{[HA]} \times \frac{1}{[H^{+}][OH^{-}]} = \frac{[A^{-}]}{[HA][OH^{-}]} = K$.
Therefore,$K = \frac{K_a}{K_w} = \frac{1.0 \times 10^{-5}}{1.0 \times 10^{-14}} = 1.0 \times 10^{9}$.

(C) The strength of oxoacids is determined by the number of non-hydroxylated oxygen atoms attached to the central atom.
Greater the number of such oxygen atoms,higher is the oxidation state of the central atom and stronger is the acid.
$A$: $SO(OH)_2$ is $H_2SO_3$ (Sulfurous acid).
$B$: $SO_2(OH)_2$ is $H_2SO_4$ (Sulfuric acid).
$C$: $ClO_3(OH)$ is $HClO_4$ (Perchloric acid).
$D$: $PO(OH)_3$ is $H_3PO_4$ (Phosphoric acid).
Among these,$HClO_4$ has the highest oxidation state of chlorine $(+7)$ and the most non-hydroxylated oxygen atoms,making it the strongest acid.

(D) According to the solvent system definition of acids and bases,an acid is a substance that increases the concentration of the characteristic cation of the solvent.
In liquid ammonia,the self-ionisation is $2NH_3 \rightleftharpoons NH_4^+ + NH_2^-$.
The characteristic cation is $NH_4^+$.
Therefore,any substance that increases the concentration of $NH_4^+$ ions in liquid ammonia acts as an acid.
$NH_4^+$ itself is the conjugate acid of the solvent $NH_3$ and acts as an acid in this system.
Thus,both $NH_4^+$ and any species that generates $NH_4^+$ in this solvent are considered acids.

(B) $Li_2O$ and $K_2O$ are basic oxides that react with water to form $OH^{-}$ ions ($Li_2O + H_2O \rightarrow 2LiOH$; $K_2O + H_2O \rightarrow 2KOH$).
$Fe_2O_3$ is an amphoteric oxide that is practically insoluble in water.
$SO_3$ is an acidic oxide that reacts with water to form $H_2SO_4$,which provides $H^{+}$ ions,not $OH^{-}$ ions.
Since the question asks for an oxide that does not produce $OH^{-}$ ions,$SO_3$ is the correct answer.

(C) According to the Bronsted-Lowry theory,an acid is a proton $(H^+)$ donor.
In the forward reaction: $HCO_3^- (aq) + OH^- (aq) \rightarrow CO_3^{2-} (aq) + H_2O (l)$,$HCO_3^-$ donates a proton to $OH^-$,so $HCO_3^-$ acts as a Bronsted acid.
In the reverse reaction: $CO_3^{2-} (aq) + H_2O (l) \rightarrow HCO_3^- (aq) + OH^- (aq)$,$H_2O$ donates a proton to $CO_3^{2-}$,so $H_2O$ acts as a Bronsted acid.
Therefore,the Bronsted acids are $HCO_3^-$ and $H_2O$.

(C) The acidic strength depends on the dissociation constant $(K_a)$ of the acids.
$H_2SO_4$ is a strong mineral acid with a very high $K_a$ value.
$CH_3COOH$ is a weak organic acid with $K_a \approx 1.8 \times 10^{-5}$.
$H_2CO_3$ is a weaker acid than acetic acid with $K_a \approx 4.3 \times 10^{-7}$.
Therefore,the decreasing order of acidic strength is $H_2SO_4 > CH_3COOH > H_2CO_3$.
Thus,the increasing order is $H_2CO_3 < CH_3COOH < H_2SO_4$.

(D) The enthalpy of neutralization for the reaction between a strong acid and a strong base is constant and is approximately $-13.7 \ kcal$ (or $-57.1 \ kJ$).
$HCl$ is a strong acid and $KOH$ and $NaOH$ are strong bases.
Therefore,both $(b)$ and $(c)$ represent the neutralization of a strong acid with a strong base.

(D) The heat of neutralisation is defined as the amount of heat released when $1 \ gram$ equivalent of an acid is neutralised by $1 \ gram$ equivalent of a base.
For the reaction between a strong acid and a strong base,the enthalpy change is constant at approximately $-57.1 \ kJ \ mol^{-1}$ or $-13.7 \ kcal \ mol^{-1}$,because the reaction is essentially the formation of water from $H^+$ and $OH^-$ ions.
When a weak acid or a weak base is involved,some of the heat released is consumed in the dissociation of the weak electrolyte.
Since $NaOH$ is a strong base and $HCl$ is a strong acid,their neutralisation reaction releases the maximum amount of heat compared to reactions involving weak acids $(CH_3COOH)$ or weak bases $(NH_4OH)$.
Therefore,the correct option is $(D)$.

(B) The heat of neutralisation for a strong acid and a strong base is constant (approximately $-57.1 \ kJ \ mol^{-1}$) because both are completely dissociated in aqueous solution.
Therefore,the net chemical reaction in every such case is simply the combination of hydrogen ions and hydroxide ions to form water:
$H^{+}(aq) + OH^{-}(aq) \rightarrow H_2O(l)$.

(D) The neutralization of a strong acid by a strong base involves the reaction between $H^{+}$ ions and $OH^{-}$ ions to form water: $H^{+}(aq) + OH^{-}(aq) \rightarrow H_2O(l)$.
Since strong acids and strong bases are completely dissociated in aqueous solution,the net reaction is always the same regardless of the specific acid or base used.
Therefore,the enthalpy of neutralization for any strong acid and strong base is constant,approximately $-57.1 \ kJ \ mol^{-1}$.

(C) The enthalpy of neutralization for any strong acid and strong base is constant because the net reaction is the formation of water from $H^{+}$ and $OH^{-}$ ions:
$H^{+} (aq) + OH^{-} (aq) \to H_{2}O (l)$,$\Delta H = -13.7 \, kcal$.
Since both $HCl$ and $H_{2}SO_{4}$ are strong acids and $NaOH$ is a strong base,the heat liberated per mole of water formed remains the same,i.e.,$13.7 \, kcal$.

(D) An amphoteric substance (or amphiprotic species) is one that can act as both an acid (by donating a proton) and a base (by accepting a proton).

Species	As an Acid (Donates $H^+$)	As a Base (Accepts $H^+$)
$HClO_3^-$	$ClO_3^{2-}$	$H_2ClO_3$
$H_2PO_4^-$	$HPO_4^{2-}$	$H_3PO_4$
$HS^-$	$S^{2-}$	$H_2S$

Since all the given species can donate and accept a proton,the correct answer is $D$.

(A) protonic acid is a substance that acts as a proton $(H^+)$ donor in a polar solvent.
For example,$HCl$ is a protonic acid because it dissociates in water to release a proton,which then forms a hydronium ion:
$HCl + H_2O \rightleftharpoons [H_3O]^+ + Cl^-$
Therefore,the correct definition is that it is a compound that forms a solvated hydrogen ion in a polar solvent.

(A) $NaHCO_3$ is an amphoteric substance (acidic salt) and $NaOH$ is a strong base. \\ When they are present in the same solution,they undergo an acid-base neutralization reaction: \\ $NaHCO_3 + NaOH \rightarrow Na_2CO_3 + H_2O$. \\ Therefore,they cannot coexist in a solution as they react to form sodium carbonate and water.

(C) $H_3PO_4$ is a triprotic acid,meaning it has three ionizable hydrogen atoms. The dissociation occurs in three successive stages:
$1$. $H_3PO_4 ⇌ H^{+} + H_2PO_4^-$
$2$. $H_2PO_4^- ⇌ H^{+} + HPO_4^{2-}$
$3$. $HPO_4^{2-} ⇌ H^{+} + PO_4^{3-}$

(D) When ammonia $(NH_3)$ is dissolved in water,it undergoes a reversible reaction to form ammonium hydroxide $(NH_4OH)$,which further dissociates into ammonium ions $(NH_4^+)$ and hydroxide ions $(OH^-)$.
The equilibrium is represented as: $NH_3 + H_2O \rightleftharpoons NH_4OH \rightleftharpoons NH_4^+ + OH^-$.
Therefore,the solution contains $NH_4^+$,$OH^-$,and undissociated $NH_4OH$ molecules.

(D) An acid salt is a salt that contains at least one replaceable hydrogen atom attached to an electronegative atom (usually oxygen).
$NaH_2PO_3$ contains two replaceable $H$ atoms.
$NaH_2PO_2$ contains two replaceable $H$ atoms.
$Na_3HP_2O_6$ contains one replaceable $H$ atom.
$Na_4P_2O_7$ (Sodium pyrophosphate) is a normal salt formed by the complete neutralization of pyrophosphoric acid $(H_4P_2O_7)$ with sodium hydroxide $(NaOH)$. It contains no replaceable $H$ atoms.
Therefore,$Na_4P_2O_7$ is not an acid salt.

(B) $HNO_3$ is a strong acid that dissociates completely in water.
The reaction is: $HNO_3(aq) + H_2O(l) \rightarrow H_3O^+(aq) + NO_3^-(aq)$.
Thus,it yields nitrate ion $(NO_3^-)$ and hydronium ion $(H_3O^+)$.

(D) Urea is preferred as a nitrogenous fertilizer because it does not cause acidity in the soil.
Ammonium sulphate,$(NH_4)_2SO_4$,leaves behind acidic residues in the soil,which can be harmful to crops over time.
Urea,$NH_2CONH_2$,is neutral and contains approximately $46.6 \%$ nitrogen by weight,making it a highly efficient fertilizer.

(D) An acid salt is a salt that contains at least one replaceable hydrogen atom in its structure,which can dissociate to release $H^{+}$ ions in an aqueous solution.
$Na_2HPO_4$ dissociates as follows: $Na_2HPO_4 \rightarrow 2Na^{+} + HPO_4^{2-}$.
The $HPO_4^{2-}$ ion can further dissociate: $HPO_4^{2-} \rightleftharpoons H^{+} + PO_4^{3-}$.
Since it can release $H^{+}$ ions,$Na_2HPO_4$ is classified as an acid salt.

(C) $NH_3$ and $PH_3$ both exhibit basic nature due to the presence of a lone pair of electrons on the central atom ($N$ and $P$ respectively),which can be donated to a Lewis acid.

(C) Sulphurous acid is $H_{2}SO_{3}$.
The salts derived from sulphurous acid are known as sulphites.
For example,the reaction of sulphurous acid with sodium hydroxide is:
$H_{2}SO_{3} + 2NaOH \to Na_{2}SO_{3} + 2H_{2}O$
Here,$Na_{2}SO_{3}$ is sodium sulphite.

(A) The strength of an acid depends on the stability of its conjugate base and the polarity of the $H-X$ bond.
$HBr$ and $HCl$ are strong mineral acids that dissociate completely in water.
Among the oxyacids of chlorine,the acidic strength increases with the number of oxygen atoms attached to the central chlorine atom due to the increase in the oxidation state of chlorine and the stability of the conjugate base.
$HClO_3$ is a strong acid,whereas $HClO$ (hypochlorous acid) is a weak acid because the $ClO^-$ ion is less stable compared to the conjugate bases of the other listed acids.
Therefore,$HClO$ is the weakest acid among the given options.

(D) The basicity of an anion is inversely proportional to the strength of its conjugate acid.
$HF$,$HCl$,$HBr$,and $HI$ are the conjugate acids of $F^{-}$,$Cl^{-}$,$Br^{-}$,and $I^{-}$ respectively.
The acidic strength order is $HF < HCl < HBr < HI$.
Therefore,the basicity order is $F^{-} > Cl^{-} > Br^{-} > I^{-}$.
Thus,$F^{-}$ is the most basic among the given options.

(A) The acidic strength depends on the stability of the conjugate base formed after the loss of a proton $(H^+)$.
In $HClO_4$,the negative charge on the conjugate base $ClO_4^-$ is delocalized over four oxygen atoms,making it the most stable conjugate base.
Therefore,the order of acidic strength is: $HClO_4 > H_2SO_4 > HNO_3 > HCl$.
Thus,$HClO_4$ is the strongest acid among the given options.

(A) An amphoteric substance is one that can act as both an acid and a base.
$HCO_3^-$ can accept a proton to form $H_2CO_3$ or donate a proton to form $CO_3^{2-}$.
$H_2O$ can accept a proton to form $H_3O^+$ or donate a proton to form $OH^-$.
$NH_3$ can accept a proton to form $NH_4^+$ or donate a proton to form $NH_2^-$.
$HNO_3$ is a strong acid and acts only as a proton donor; it does not exhibit amphoteric behavior.
Therefore,the correct option is $A$.

(D) The dissociation constant $(K_a)$ is a measure of the acidity of a substance.
$1$. $CH_3NH_3^+Cl^-$ is a salt of a weak base $(CH_3NH_2)$ and a strong acid $(HCl)$,making it acidic.
$2$. $C_6H_5OH$ (Phenol) has a $pK_a$ of approximately $10$.
$3$. $C_6H_5CH_2OH$ (Benzyl alcohol) is very weakly acidic $(pK_a \approx 15-16)$.
$4$. $CH_3C \equiv CH$ (Propyne) is very weakly acidic $(pK_a \approx 25)$.
Comparing these,the conjugate acid of the weak base $CH_3NH_2$,which is $CH_3NH_3^+$,is significantly more acidic than the others. Therefore,$CH_3NH_3^+Cl^-$ has the highest dissociation constant.

(D) The normality $(N)$ of a solution is given by the formula: $N = M \times n$-factor,where $M$ is molarity and $n$-factor is the basicity of the acid.
Since $1 \ mole$ of each acid is dissolved in $1 \ L$ of water,the molarity $(M)$ for all is $1 \ M$.
For $HCl$,$n$-factor = $1$,so $N = 1 \times 1 = 1 \ N$.
For $HClO_4$,$n$-factor = $1$,so $N = 1 \times 1 = 1 \ N$.
For $HNO_3$,$n$-factor = $1$,so $N = 1 \times 1 = 1 \ N$.
For $H_3PO_4$,it is a triprotic acid ($n$-factor = $3$),so $N = 1 \times 3 = 3 \ N$.
Therefore,$H_3PO_4$ does not give a solution of strength $1 \ N$.

(D) Citric acid $(HOOC-CH_2-C(OH)(COOH)-CH_2-COOH)$ contains three carboxylic acid $(-COOH)$ groups,which can donate three protons $(H^+)$ in an aqueous solution,making it a tribasic acid.
- Oxalic acid $(HOOC-COOH)$ is dibasic.
- Tartaric acid $(HOOC-CH(OH)-CH(OH)-COOH)$ is dibasic.
- Lactic acid $(CH_3-CH(OH)-COOH)$ is monobasic.

(B) The strength of an acid is determined by its ability to donate protons ($H^+$ ions) in an aqueous solution.
$CH_3COOH$ (acetic acid) is a weak acid because it undergoes partial dissociation in water.
In contrast,$H_2SO_4$ (sulphuric acid) is a strong acid that undergoes almost complete dissociation.
Therefore,acetic acid has a lower degree of ionisation compared to sulphuric acid.

(C) The juice of the lemon contains approximately $5\%$ to $6\%$ citric acid.
This acid provides the characteristic sour taste and results in a $pH$ value of around $2.2$.

(D) Explanation: $HN_3$ (hydrazoic acid) dissociates into $N_3^-$ and $H^+$. Due to the release of $H^+$ ions,it acts as an acid.
$N_2H_4$ (hydrazine) has a lone pair on nitrogen atoms,which it can donate,making it a base.
$NH_2OH$ (hydroxylamine) is basic because it can accept a proton.
$(CH_3)_3N$ (trimethylamine) is a tertiary amine and acts as a weak base due to the lone pair on the nitrogen atom.

(C) An oxyacid is an acid that contains an oxygen atom bonded to a hydrogen atom and at least one other element.
$H_3PO_3$ (phosphorous acid) contains oxygen,hydrogen,and phosphorus,making it an oxyacid.
$Ba(OH)_2$ and $Mg(OH)_2$ are bases (hydroxides).
$HCl$ is a hydracid (binary acid) as it does not contain oxygen.

(B) The enthalpy of neutralization is the heat released when $1 \text{ gram equivalent}$ of an acid is neutralized by $1 \text{ gram equivalent}$ of a base.
For strong acid and strong base,the enthalpy of neutralization is constant at $-57.1 \text{ kJ/mol}$.
When either the acid or the base is weak,some energy is consumed in the dissociation of the weak electrolyte.
Therefore,the enthalpy of neutralization is lowest (least negative) for the reaction between a weak acid and a weak base,such as $NH_4OH$ and $CH_3COOH$.

(B) $NaOH$ is hygroscopic and reacts with atmospheric $CO_2$ to form $Na_2CO_3$.
The reaction is: $2NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O$.
Due to this reaction,the amount of $NaOH$ in the solution decreases,which leads to a decrease in the strength (molarity) of the $NaOH$ solution.

(B) The conjugate base of an acid is formed by removing a proton $(H^+)$ from the acid molecule.
For hydrazoic acid $(N_3H)$,the reaction is:
$N_3H \rightarrow H^+ + N_3^-$
Thus,the conjugate base of $N_3H$ is $N_3^-$.

(A) Lewis base is defined as a substance that can donate a lone pair of electrons to form a coordinate covalent bond.
$NH_3$ has a lone pair on the nitrogen atom.
$H_2O$ has two lone pairs on the oxygen atom.
$OH^{-}$ has lone pairs on the oxygen atom.
$CO_3^{2-}$ has lone pairs on the oxygen atoms.
However,in the provided options,all species listed are technically Lewis bases. If we consider the option $CO_3^{2-}$ as the intended answer for a potential typo in the question (e.g.,if it were a Lewis acid),we must note that all listed species possess lone pairs. Given the standard classification,all are Lewis bases. If the question implies a species that cannot act as a base,none of these fit. Assuming a typo in option $A$ where it might have been intended as a Lewis acid like $BF_3$,but based on the provided list,all are Lewis bases.

(C) An acid is defined as a substance that can donate a proton $(H^{+})$.
In the reaction $NH_3 + Na \rightarrow NaNH_2 + \frac{1}{2}H_2$,the ammonia molecule loses a proton to form the amide ion $(NH_2^-)$.
Since $NH_3$ acts as a proton donor here,it behaves as a Brønsted-Lowry acid.
In the other reactions,$NH_3$ acts as a base by accepting a proton.

(D) An amphoteric substance is one that can act as both an acid and a base.
$H_2O$ can accept a proton to form $H_3O^+$ (acting as a base) and can donate a proton to form $OH^-$ (acting as an acid).
Therefore,$H_2O$ is an amphoteric substance.