Q.What will be the conjugate bases for the following Brönsted acids: HF, H 2SO4 and HCO3 – ?

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.In the reaction given below, water behaves as $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$ (A) A Lewis base (B) A Bronsted–Lowry acid (C) A Bronsted–Lowry base (D) A Lewis acid

   ›Reveal solutionSolution

   Water acts as a proton donor to ammonia in this reaction, which by definition makes it a Bronsted-Lowry acid.

   Concept and Intuition

   A Bronsted-Lowry acid is any species that donates a proton ($H^+$); a Bronsted-Lowry base accepts one. Water is amphoteric — it can act as either, depending on what it's reacting with. With $NH_3$ (a base, since $N$ has a lone pair that readily accepts $H^+$), water plays the acid role.

   Step-by-Step Solution

   1. Look at the reaction: $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$.
   2. $NH_3$ gains a proton to become $NH_4^+$ — so $NH_3$ is acting as the base (proton acceptor).
   3. $H_2O$ loses a proton (an $H$ atom leaves as $H^+$) to become $OH^-$ — so $H_2O$ is donating a proton.
   4. By the Bronsted-Lowry definition, donating a proton = acting as an acid.

   Common Mistakes

   • Confusing this with the Lewis acid/base framework — here water is NOT accepting an electron pair (it's giving up a proton), so "Lewis acid" would be the wrong classification; the correct Bronsted-Lowry role is what's being tested.
   • Assuming water is "always a base" because it's often shown accepting protons from strong acids — water's role is reaction-dependent; here the partner ($NH_3$) is itself a base, forcing water into the acid role.

   ✓Final answer

   The correct option is (B) — A Bronsted–Lowry acid.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Identify the correct statements from the following I) Conjugate base of nitrous acid is $NO_2^-$ II) The concentration of $OH^-$ ions in 0.1 M aqueous pyridine solution is $\sqrt{2} \times 10^{-5}$M. ($K_b$ of pyridine $= 2 \times 10^{-9}$) III) The acid strength of aqueous hydrogen halides follow the order $HF > HCl > HBr > HI$ The correct answer is (A) I, II, III (B) I, II only (C) I, III only (D) II, III only

   ›Reveal solutionSolution

   Statements I and II are correct; statement III reverses the true acid-strength order of hydrogen halides in water.

   Concept and Intuition

   A conjugate base is simply the acid minus one $H^+$. For a weak base at low ionization, $[OH^-]=\sqrt{K_bC}$ follows directly from the equilibrium expression. Acid strength of the hydrogen halides in water is governed mainly by H–X bond dissociation energy (weakest bond = strongest acid once in water); H–F is by far the strongest bond, making HF the weakest acid, not the strongest.

   Step-by-Step Solution

   1. Statement I: $HNO_2 \rightleftharpoons H^+ + NO_2^-$; the conjugate base is $NO_2^-$. TRUE.
   2. Statement II: For a weak base, $[OH^-] = \sqrt{K_bC} = \sqrt{(2\times10^{-9})(0.1)} = \sqrt{2\times10^{-10}} = \sqrt{2}\times10^{-5}$ M. Matches exactly. TRUE.
   3. Statement III: The real trend of acid strength for aqueous hydrogen halides is $HF \ll HCl < HBr < HI$ (HI strongest, HF weakest), because H–F's very high bond energy resists ionization despite F's high electronegativity. The statement's order $HF>HCl>HBr>HI$ is the reverse of reality. FALSE.
   4. Correct statements: I and II only.

   Common Mistakes

   • Assuming higher electronegativity of F implies HF is the strongest acid (confusing electronegativity trend with actual ionization ease, which is bond-energy dominated here).
   • Arithmetic slip turning $\sqrt{2\times10^{-10}}$ into $2\times10^{-5}$ instead of $\sqrt2\times10^{-5}$.

   ✓Final answer

   The correct option is (B) — I, II only.

   ANSWER: B

3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.What is the conjugate acid of $\mathrm{H_3P_2O_6^-}$? (A) Orthophosphorus acid (B) Hypophosphorus acid (C) Pyrophosphoric acid (D) Hypophosphoric acid

   ›Reveal solutionSolution

   The conjugate acid of an anion is obtained by adding one $H^+$ back to it. $H_3P_2O_6^-$ + $H^+$ gives $H_4P_2O_6$, the formula of hypophosphoric acid.

   Concept and Intuition

   A conjugate acid–base pair differs by exactly one proton: the conjugate acid has one more $H^+$ than the species given. To identify it, add a proton to the anion and match the resulting neutral formula to the known oxoacid of phosphorus.

   Step-by-Step Solution

   1. Given ion: $H_3P_2O_6^-$.
   2. Its conjugate acid = $H_3P_2O_6^- + H^+ = H_4P_2O_6$.
   3. Recall the phosphorus oxoacid formulas: orthophosphorous acid $H_3PO_3$; hypophosphorous acid $H_3PO_2$; pyrophosphoric acid $H_4P_2O_7$; hypophosphoric acid $H_4P_2O_6$.
   4. $H_4P_2O_6$ matches hypophosphoric acid exactly.

   Common Mistakes

   • Confusing pyrophosphoric acid ($H_4P_2O_7$, one more oxygen) with hypophosphoric acid ($H_4P_2O_6$) — easy to mix up since both are diprotic-per-P dimeric acids.
   • Forgetting that "conjugate acid" means adding a proton, not removing one.

   ✓Final answer

   The correct option is (D) — Hypophosphoric acid.

   ANSWER: D

4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.At $25°C$, $K_a$ of formic acid is $1.8 \times 10^{-4}$. What is the $K_b$ of $\mathrm{HCOO^-}$? (A) $1.8 \times 10^{-10}$ (B) $5.55 \times 10^{-4}$ (C) $5.55 \times 10^{-11}$ (D) $5.55 \times 10^{-12}$

   ›Reveal solutionSolution

   This tests the conjugate acid-base pair relation $K_a\cdot K_b = K_w$; the conjugate base $HCOO^-$ of formic acid has $K_b \approx 5.55\times10^{-11}$.

   Concept and Intuition

   For any weak acid HA and its conjugate base $A^-$, their equilibrium constants are linked through water's autoionisation constant, because adding the acid-dissociation and base-hydrolysis equilibria together reproduces the autoionisation of water.

   Step-by-Step Solution

   1. Conjugate pair relation: $K_a(\text{HCOOH}) \times K_b(\text{HCOO}^-) = K_w = 1\times10^{-14}$ (at 25°C).
   2. Given $K_a = 1.8\times10^{-4}$.
   3. $K_b = \dfrac{1\times10^{-14}}{1.8\times10^{-4}} = 0.5555\times10^{-10} = 5.55\times10^{-11}$.

   Common Mistakes

   • Using $K_w = 1\times10^{-14}$ but forgetting to adjust the power of ten correctly when dividing.
   • Confusing which species is the acid and which is the conjugate base in the relation.

   ✓Final answer

   The correct option is (C) — $5.55\times10^{-11}$.

   ANSWER: C

5. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The conjugate base of phosphorus acid is $x$. The conjugate base of oleum is $y$. What are $x$ and $y$, respectively? (A) $\mathrm{H_2PO_4^{-}, HS_2O_7^{-}}$ (B) $\mathrm{H_2PO_4^{-}, HSO_5^{-}}$ (C) $\mathrm{H_2PO_3^{-}, HS_2O_7^{-}}$ (D) $\mathrm{H_2PO_3^{-}, HSO_4^{-}}$

   ›Reveal solutionSolution

   Tests knowledge of the structures of phosphorus acid and oleum, and correctly identifying which protons are acidic. Answer: (C).

   Concept and Intuition

   A conjugate base is what remains after a Brønsted acid donates one proton. To find it correctly you must know which H atoms in the parent acid are actually ionisable (O–H, acidic) versus which are non-ionisable (P–H or similar, not acidic).

   Phosphorus acid, $\mathrm{H_3PO_3}$, is written with three H's in its formula but structurally it is $\mathrm{HPO(OH)_2}$ — one H is bonded directly to the central P atom (non-acidic) and only the two O–H hydrogens are ionisable. That is why phosphorus acid is diprotic, not triprotic.

   Oleum is fuming sulfuric acid, essentially $\mathrm{H_2S_2O_7}$ — two $\mathrm{SO_3}$/$\mathrm{SO_4}$ units bridged by an oxygen $(\mathrm{HO})(\mathrm{O_2}S)$–O–$(\mathrm{SO_2})(\mathrm{OH})$, with two acidic O–H protons.

   Step-by-Step Solution

   1. Write phosphorus acid structurally: $\mathrm{HPO(OH)_2}$, i.e. one P–H (non-acidic) plus two acidic O–H groups.
   2. Removing one acidic proton from $\mathrm{H_3PO_3}$ gives $x=\mathrm{H_2PO_3^-}$ (the P–H proton is retained, since it never ionises).
   3. Write oleum as $\mathrm{H_2S_2O_7}$ (disulfuric acid), with two acidic O–H protons.
   4. Removing one proton from $\mathrm{H_2S_2O_7}$ gives $y=\mathrm{HS_2O_7^-}$.
   5. Match to the options: $x=\mathrm{H_2PO_3^-}$, $y=\mathrm{HS_2O_7^-}$ is option (C).

   Common Mistakes

   • Treating $\mathrm{H_3PO_3}$ as triprotic and writing $x=\mathrm{H_2PO_4^-}$ (this is actually the conjugate base of phosphoric acid, $\mathrm{H_3PO_4}$, a different acid).
   • Confusing oleum's conjugate base $\mathrm{HS_2O_7^-}$ with $\mathrm{HSO_4^-}$ (the conjugate base of ordinary sulfuric acid) or $\mathrm{HSO_5^-}$ (peroxymonosulfuric/Caro's acid's conjugate base).

   ✓Final answer

   The correct option is (C) — $x=\mathrm{H_2PO_3^-}$, $y=\mathrm{HS_2O_7^-}$.

   ANSWER: C

6. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.What is the conjugate base of $\mathrm{OH^-}$? (A) $\mathrm{H_2O}$ (B) $\mathrm{O^{2-}}$ (C) $\mathrm{H_3O^+}$ (D) $\mathrm{H^+}$

   ›Reveal solutionSolution

   Removing a proton from $\mathrm{OH^-}$ (acting as a Bronsted acid) leaves the oxide ion $\mathrm{O^{2-}}$, its conjugate base.

   Concept and Intuition

   In Bronsted-Lowry theory, an acid's conjugate base is what remains after donating one $H^+$. $\mathrm{OH^-}$ can itself act as a very weak Bronsted acid, donating a proton to become $\mathrm{O^{2-}}$.

   Step-by-Step Solution

   1. Write $\mathrm{OH^-} \rightleftharpoons H^+ + \text{(conjugate base)}$.
   2. Removing $H^+$ from $\mathrm{OH^-}$ (charge $-1$) leaves a species of charge $-1 - (+1) = -2$, i.e. one oxygen atom with charge $-2$: $\mathrm{O^{2-}}$.
   3. So the conjugate base of $\mathrm{OH^-}$ is $\mathrm{O^{2-}}$.

   Common Mistakes

   • Confusing conjugate acid ($\mathrm{H_2O}$, formed by ADDING $H^+$ to $\mathrm{OH^-}$) with conjugate base (formed by REMOVING $H^+$) — $\mathrm{H_2O}$ and $\mathrm{H_3O^+}$ are distractors from this confusion.

   ✓Final answer

   The correct option is (B) — $\mathrm{O^{2-}}$.

   ANSWER: B

7. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.Observe the following stoichiometric equation $\mathrm{P_4 + 3\,OH^- + 3H_2O \rightarrow PH_3 + 3x^-}$ What is the conjugate acid of $x^-$? (A) Phosphorous acid (B) Hypophosphorous acid (C) Phosphoric acid (D) Pyrophosphoric acid

   ›Reveal solutionSolution

   This tests recognizing the disproportionation reaction of white phosphorus with hot alkali and identifying the conjugate acid of the resulting anion. The answer is (B).

   Concept and Intuition

   White phosphorus ($P_4$) undergoes a base-induced disproportionation with hot concentrated NaOH solution: one phosphorus atom is reduced to $PH_3$ (phosphine, oxidation state -3) while the other three are oxidized to the hypophosphite ion, $H_2PO_2^-$ (oxidation state +1). This is a classic NCERT reaction used to illustrate disproportionation in p-block chemistry.

   Step-by-Step Solution

   1. Write the known balanced equation: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
   2. In ionic form, this matches the given equation with $x^- = H_2PO_2^-$ (hypophosphite ion, from sodium hypophosphite $NaH_2PO_2$).
   3. The conjugate acid of an anion $A^-$ is $HA$ -- so the conjugate acid of $H_2PO_2^-$ is $H_3PO_2$, hypophosphorous acid.
   4. Hypophosphorous acid ($H_3PO_2$) is a well-known monobasic acid (despite three H atoms, only one H is acidic -- bonded to O; the other two are directly bonded to P and are non-ionizable).

   Common Mistakes

   • Confusing hypophosphorous acid ($H_3PO_2$, +1 oxidation state of P) with phosphorous acid ($H_3PO_3$, +3 oxidation state) -- the oxidation state of P in the product here is +1, matching hypophosphite/hypophosphorous acid specifically.

   ✓Final answer

   The correct option is (B) — Hypophosphorous acid.

   ANSWER: B

8. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.What is the conjugate base of chloric acid? (A) $\mathrm{ClO_4^-}$ (B) $\mathrm{ClO^-}$ (C) $\mathrm{ClO_2^-}$ (D) $\mathrm{ClO_3^-}$

   ›Reveal solutionSolution

   This tests the simple Bronsted-Lowry definition of a conjugate base applied to one of the common oxoacids of chlorine. The answer is (D).

   Concept and Intuition

   A conjugate base is what remains after an acid donates one proton ($H^+$). Chlorine forms a family of oxoacids -- hypochlorous ($HClO$), chlorous ($HClO_2$), chloric ($HClO_3$), and perchloric ($HClO_4$) -- each with a corresponding conjugate base (hypochlorite, chlorite, chlorate, perchlorate).

   Step-by-Step Solution

   1. Identify chloric acid: $HClO_3$.
   2. Removing $H^+$: $HClO_3 \rightarrow H^+ + ClO_3^-$.
   3. The conjugate base is therefore $ClO_3^-$ (chlorate ion).
   4. Cross-check the other options: $ClO_4^-$ is the conjugate base of perchloric acid ($HClO_4$); $ClO^-$ is that of hypochlorous acid ($HClO$); $ClO_2^-$ is that of chlorous acid ($HClO_2$) -- none of these match chloric acid.

   Common Mistakes

   • Confusing the naming pattern for -ic acids (which give -ate ions) with that for -ous acids (which give -ite ions), which could lead to mismatching chloric acid with chlorite instead of chlorate.

   ✓Final answer

   The correct option is (D) — $\mathrm{ClO_3^-}$.

   ANSWER: D

9. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.Conjugate acid and conjugate base of $\mathrm{HCO_3^-}$ are respectively (A) $\mathrm{H_2CO_3,\ H_3CO_3^+}$ (B) $\mathrm{H_2CO_3,\ CO_3^{2-}}$ (C) $\mathrm{CO_3^{2-},\ H_2CO_3}$ (D) $\mathrm{CO_3^{2-},\ CO_2}$

   ›Reveal solutionSolution

   $\mathrm{HCO_3^-}$ is amphiprotic — its conjugate acid is formed by adding H⁺, its conjugate base by removing H⁺.

   Concept and Intuition

   A conjugate acid–base pair differs by exactly one proton ($H^+$). The conjugate acid of a species is what you get by adding a proton; the conjugate base is what you get by removing a proton. Amphiprotic species like $\mathrm{HCO_3^-}$ can do both.

   Step-by-Step Solution

   1. Conjugate acid of $\mathrm{HCO_3^-}$: add $H^+$ → $\mathrm{H_2CO_3}$.
   2. Conjugate base of $\mathrm{HCO_3^-}$: remove $H^+$ → $\mathrm{CO_3^{2-}}$.
   3. The question asks for conjugate acid, then conjugate base, in that order: $\mathrm{H_2CO_3}$, $\mathrm{CO_3^{2-}}$.

   Common Mistakes

   • Swapping the order (giving base then acid) — the question explicitly asks acid first, then base.
   • Writing the conjugate acid as $\mathrm{H_3CO_3^+}$ (adding a proton to the wrong atom/over-protonating) instead of the correct neutral $\mathrm{H_2CO_3}$.

   ✓Final answer

   The correct option is (B) — $\mathrm{H_2CO_3}$, $\mathrm{CO_3^{2-}}$.

   ANSWER: B

10. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.The conjugate base of $H_2PO_4^{-}$ is (A) $HPO_4^{-}$ (B) $PO_4^{2-}$ (C) $H_2PO_4^{2-}$ (D) $HPO_4^{2-}$

    ›Reveal solutionSolution

    Removing one $H^+$ from $H_2PO_4^-$ gives its conjugate base, $HPO_4^{2-}$.

    Concept and Intuition

    By the Brønsted–Lowry definition, the conjugate base of any acid is what remains after the acid donates exactly one proton. Phosphate species form a stepwise proton-relay chain: $H_3PO_4 \rightleftharpoons H_2PO_4^- \rightleftharpoons HPO_4^{2-} \rightleftharpoons PO_4^{3-}$, each step losing one $H^+$ and gaining one unit of negative charge.

    Step-by-Step Solution

    1. Start with $H_2PO_4^-$ (charge $-1$).
    2. Remove one proton: charge becomes $-1-(+1)\to$ the species loses an $H^+$, so charge decreases by one more unit of negative charge added, giving $-2$.
    3. The resulting species is $HPO_4^{2-}$.

    Common Mistakes

    • Removing two protons instead of one, which would give $PO_4^{3-}$ (this is two steps down the chain, not the immediate conjugate base).
    • Miscounting the charge after proton removal.

    ✓Final answer

    The correct option is (D) — $HPO_4^{2-}$.

    ANSWER: D

11. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.The conjugate base of $H_3O^+$ is (A) $H_2O$ (B) $OH^-$ (C) $H^+$ (D) $H^-$

    ›Reveal solutionSolution

    Removing one $H^+$ from the acid $H_3O^+$ leaves its conjugate base, $H_2O$.

    Concept and Intuition

    In Bronsted-Lowry theory, an acid and its conjugate base differ by exactly one proton: Acid $\rightleftharpoons$ Conjugate base $+ H^+$.

    Step-by-Step Solution

    1. $H_3O^+$ acting as an acid donates a proton: $H_3O^+ \rightleftharpoons H_2O + H^+$.
    2. The species left behind, $H_2O$, is the conjugate base of $H_3O^+$.

    Common Mistakes

    • Confusing conjugate base with conjugate acid, or picking $OH^-$ (which is the conjugate base of $H_2O$, one step further removed).

    ✓Final answer

    The correct option is (A) — $H_2O$.

    ANSWER: A

12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.The conjugate base of $\mathrm{NH_4^+}$ is (A) $\mathrm{NH^{2-}}$ (B) $\mathrm{NH_3}$ (C) $\mathrm{NH_2^-}$ (D) $\mathrm{NH_4OH}$

    ›Reveal solutionSolution

    Removing one proton from $\mathrm{NH_4^+}$ (a Brønsted-Lowry acid) directly gives its conjugate base, $\mathrm{NH_3}$.

    Concept and Intuition

    In Brønsted-Lowry acid-base theory, a conjugate acid-base pair differs by exactly one proton ($\mathrm{H^+}$). The conjugate base of any acid HA is simply $\mathrm{A^-}$ (or, if the acid is a cation, the neutral species left after losing $\mathrm{H^+}$).

    Step-by-Step Solution

    1. Write the proton-donation equilibrium: $\mathrm{NH_4^+ \rightleftharpoons NH_3 + H^+}$.
    2. The species remaining on the right after $\mathrm{NH_4^+}$ has donated its proton is $\mathrm{NH_3}$.
    3. By definition, this remaining species is the conjugate base of $\mathrm{NH_4^+}$.
    4. Checking the distractors: $\mathrm{NH^{2-}}$ and $\mathrm{NH_2^-}$ would require losing 2 or more protons/electrons inconsistently, and $\mathrm{NH_4OH}$ is not related to $\mathrm{NH_4^+}$ by simple proton loss.

    Common Mistakes

    • Confusing conjugate base with conjugate acid (which would instead involve gaining a proton).
    • Mistakenly picking $\mathrm{NH_4OH}$, which is a different (undissociated) species, not a proton-loss product of $\mathrm{NH_4^+}$.

    ✓Final answer

    The correct option is (B) — $\mathrm{NH_3}$.

    ANSWER: B