Q.The species that can form both conjugate acid and conjugate base among the following is
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Brønsted Lowry Conjugate Pairs
From Intuition to Precision: Brønsted–Lowry Conjugate Pairs
Imagine you're at a party where people are passing around a single, very special coin. The coin represents a proton (H+). The game is simple: someone can give the coin to someone else, but only if the other person is willing to take it. You can't just throw the coin; you need a willing receiver.
In chemistry, acids and bases play exactly this game. An acid is the person who gives away the proton. A base is the person who accepts it. But here's the twist: the moment the acid gives away its proton, it transforms into something new — something that can now accept a proton back. That transformed form is called the conjugate base. Similarly, the base, after accepting the proton, becomes something that can donate it back — the conjugate acid.
This is the core of the Brønsted–Lowry theory: every acid-base reaction involves a pair of substances that are linked by the gain or loss of a single proton.
The Precise Statement
Acid⇌Conjugate Base+H+
Base+H+⇌Conjugate Acid
A conjugate acid-base pair consists of two species that differ by exactly one proton (H+). The acid has the proton; the conjugate base does not.
When an acid donates a proton, it becomes its conjugate base.
When a base accepts a proton, it becomes its conjugate acid.
In any Brønsted–Lowry reaction, there are two conjugate pairs: one on the reactant side and one on the product side. They always appear together.
Seeing It in Action
Take the classic reaction between hydrochloric acid and water:
HCl+H2O→Cl−+H3O+
Let's identify the pairs:
- Pair 1: HCl (acid) and Cl− (conjugate base). They differ by one H+.
- Pair 2: H2O (base) and H3O+ (conjugate acid). They also differ by one H+.
Notice: water acted as a base here — it accepted the proton from HCl. But water can also act as an acid in other reactions. That's the beauty of the Brønsted–Lowry theory: a substance's role depends on the reaction, not on a fixed label.
To find the conjugate base of any acid, simply remove one H+ and reduce the charge by +1.
To find the conjugate acid of any base, add one H+ and increase the charge by +1.
A Quick Reference Table
| Acid | Conjugate Base | Base | Conjugate Acid |
|---|---|---|---|
| HCl | Cl− | NH3 | NH4+ |
| H2SO4 | HSO4− | H2O | H3O+ |
| NH4+ | NH3 | OH− | H2O |
| H2O | OH− | CO32− | HCO3− |
Notice how water appears in both columns — it's amphoteric, meaning it can act as either an acid or a base depending on its partner.
Why This Matters …
A species that can act as both a Bronsted proton donor and a proton acceptor is called a …
H2O can both donate a proton (acting as a Bronsted acid) and accept a proton (acting as a Bronsted base), so it is amphiprotic.
According to the Bronsted-Lowry concept, an acid is a proton donor and a base is a proton acceptor. A species that can act as both, depending on the reaction, is called amphiprotic.
- H2O can donate a proton to become OH- (its conjugate base): H2O -> H+ + OH-
- H2O can also accept a proton to become H3O+ (its conjugate acid): H2O + H+ -> H3O+
So water forms both a conjugate acid (H3O+) and a conjugate base (OH-), making it amphiprotic.
…
Showing the 12 most recent of 26 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.Which one is the conjugate base of HSO4-?(a) H2SO4(b) HSO3(c) SO3(d) SO4^2-
›Reveal solutionSolution
A conjugate base is formed by removing one H+ (proton) from an acid. HSO4- losing a proton gives SO4^2-.
By the Bronsted-Lowry definition, an acid's conjugate base is what remains after the acid donates (loses) a proton (H+).
HSO4- (hydrogen sulfate ion) donating a proton:
HSO4- -> H+ + SO4^2-
…
- CBSE 2026Set ANNUAL1 markQ.Write the conjugate base of HClO4.
›Reveal solutionSolution
Removing one H+ from HClO4 gives its conjugate base, ClO4- (perchlorate ion).
By the Bronsted-Lowry definition, an acid is a proton (H+) donor, and after it donates a proton, the resulting species (which can, in principle, accept a proton back) is called its conjugate base. HClO4 (perchloric acid) donates one H+: HClO4 → H+ + ClO4-. The species remaining, ClO4- (perchlorate ion), is the conjugate base of HCl …
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is a Lewis base ?(a) HCl(b) HNO3(c) HF(d) NH3
›Reveal solutionSolution
NH3 is the Lewis base (electron-pair donor).
A Lewis base is an electron-pair donor. NH3 has a lone pair of electrons on nitrogen which it can donate to an electron-deficient species, so it is a Lewis base. HCl, HNO3 and HF donate protons …
- CBSE 2026Set ANNUAL1 markMCQQ.The conjugate base of HBr is(a) H2Br^+(b) H^+(c) Br^-(d) Br^+
›Reveal solutionSolution
Conjugate base of HBr = Br-.
According to the Bronsted-Lowry concept, the conjugate base is what remains after an acid donates one proton. HBr → H+ + Br-, so t …
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is the strongest base ?(a) AsH3(b) NH3(c) PH3(d) SbH3
›Reveal solutionSolution
NH3 is the strongest base among NH3, PH3, AsH3 and SbH3.
Basicity of these hydrides depends on the availability of the lone pair on the central atom. As we go down group 15 (N → P → As → Sb) the atom gets larger, the lone pair spreads over a larger volume and becomes less availabl …
- CBSE 2025Set ANNUAL1 markMCQQ.Which of the following bases is weakest?(a) C2H5O-(b) NO3-(c) I-(d) CH3COO-
›Reveal solutionSolution
I- is the weakest base of the four options, being the conjugate base of the very strong acid HI.
The strength of a conjugate base is inversely related to the strength of its parent (conjugate) acid — the stronger the acid, the weaker its conjugate base.
- C2H5O- is the conjugate base of ethanol (a very weak acid), so C2H5O- is a fairly strong base.
- CH3COO- is the conjugate base of acetic acid (a weak acid), so it is a moderately weak base. …
- CBSE 2025Set ANNUAL1 markMCQQ.Which one of the following is not an amphoteric substance?(a) HNO3(b) HCO3-(c) H2O(d) NH3
›Reveal solutionSolution
HNO3 is the only substance among the options that is not amphoteric.
An amphoteric substance can donate a proton (act as an acid) as well as accept a proton (act as a base).
- HCO3- can donate a proton to become CO3^2- (acting as an acid) or accept a proton to become H2CO3 (acting as a base) — amphoteric.
- H2O can donate a proton to become OH- (acid) or accept a proton to become H3O+ (base) — amphoteric. …
- CBSE 2025Set ANNUAL1 markMCQQ.The conjugate acid of NH3 is(a) NH2^-(b) NH4^+(c) NH^2-(d) N2H4
›Reveal solutionSolution
A conjugate acid is simply the base plus one extra proton (H+); adding H+ to NH3 gives the ammonium ion, NH4+.
According to the Bronsted-Lowry theory of acids and bases: an acid is a proton (H+) donor, and a base is a proton acceptor. When a base accepts a proton, it becomes its 'conjugate acid' — a species that could, in the reverse reaction, donate that proton back.
NH3 acts as a base here (it has a lone pair on nitrogen that can accept a proton):
NH3 + H+ -> NH4+
So NH4+ (the ammonium ion) is the conjugate acid of NH3, formed by adding exactly one proton to the nitrogen's lone pair.
…
- CBSE 2025Set ANNUAL1 markQ.Fill in the blank: The conjugate base of NH2- is ___________.
›Reveal solutionSolution
Removing one proton from the amide ion NH2- gives NH^2- (imide ion), which is therefore its conjugate base. Removing a further H+ would give N^3-, which is the conjugate base of NH^2-, not of NH2-.
By the Bronsted-Lowry definition, a conjugate base is formed by removing exactly one proton (H+) from a species.
Starting from the amide ion NH2- (which has two H and a charge of -1), removing one H+ leaves NH^2- (the imide / azanide ion, charge -2):
NH2- - H+ -> NH^2- …
- CBSE 2025Set ANNUAL1 markMCQQ.Assertion (A): The conjugate base of a strong acid is a weak base. Reason (R): Strong acids completely dissociate in water, leaving their conjugate bases with low tendency to accept protons.(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).(c) (A) is true, but (R) is false.(d) (A) is false, but (R) is true.
›Reveal solutionSolution
Strong acids ionize almost completely in water. The species left behind after losing a proton (the conjugate base) has little affinity for that proton, i.e., a low tendency to accept a proton back — which is exactly the definition of a weak base. So (R) correctly explains (A).
Assertion (A): The conjugate base of a strong acid is a weak base — this is correct. For example, HCl (strong acid) dissociates completely to give Cl-, which shows essentially no tendency to accept a proton back in water (Cl- is a very weak base).
Reason (R): Strong acids completely dissociate in water, leaving their conjugate bases with a low tendency to accept protons — this is also correct and is precisely the mechanism behind (A).
…
- CBSE 2025Set hz1 markMCQQ.Select the correct one: Which of the following is a strongest conjugate base?(a) Cl-(b) SO4^2-(c) CH3COO-(d) NO3-
›Reveal solutionSolution
Acid strength and conjugate base strength are inversely related — the weaker the parent acid, the stronger its conjugate base. CH3COO- comes from the weakest acid here (CH3COOH), so it is the strongest conjugate base.
Each ion's parent acid:
- Cl- comes from HCl, a strong acid → Cl- is an extremely weak (negligible) base.
- SO4^2- comes from HSO4- (itself a fairly strong acid) → SO4^2- is a weak base.
- CH3COO- comes from CH3COOH (acetic acid), a weak acid (Ka ~ 1.8x10^-5) → CH3COO- is a comparatively strong (moderate) base. …
- CBSE 2025Set ANNUAL1 markMCQQ.Water is(a) protophobic substance(b) protophilic substance(c) amphoteric substance(d) aprotic substance
›Reveal solutionSolution
Water can both donate and accept a proton, so it is an amphoteric (amphiprotic) substance.
In the Brønsted–Lowry sense:
- Water acting as an acid (proton donor): H2O+NH3→OH−+NH4+
- Water acting as a base (proton acceptor): H2O+HCl→H3O++Cl− …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.