Ionisation constant of a weak base MOH, is given by the expression Kb = [M^+][OH^-] / [MOH] Values of ionisati

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Q. Ionisation constant of a weak base MOH, is given by the expression Kb = [M^+][OH^-] / [MOH] Values of ionisation constant of some weak bases at a particular temperature are given below: Base: Dimethylamine, Urea, Pyridine, Ammonia Kb: 5.4 \times 10^-4, 1.3 \times 10^-14, 1.77 \times 10^-9, 1.77 \times 10^-5 Arrange the bases in decreasing order of the extent of their ionisation at equilibrium. Which of the above base is the strongest?

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Concept understanding — Weak Acid Ionization

Weak Acid Ionization: From Intuition to Precision

Imagine you drop a spoonful of sugar into a glass of water. Some sugar dissolves, but a lot just sits at the bottom. Now imagine you drop a spoonful of salt — it all dissolves completely. Acids behave the same way. Some acids, like hydrochloric acid (HCl), dissolve completely in water — every single molecule breaks apart. Others, like acetic acid (vinegar), only partially break apart. Most of the acid molecules stay intact, and only a few actually ionize.

That's the core idea: weak acids are shy about giving away their hydrogen ion. They don't fully commit.

The Precise Statement

A weak acid ( $HA$ ) in water establishes an equilibrium between the intact acid molecule and its ions:

HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}

The double arrow ( $\rightleftharpoons$ ) is the key. It tells you the reaction happens in both directions simultaneously. Some $HA$ molecules break apart to form $H_3O^+$ and $A^-$ , while some $H_3O^+$ and $A^-$ recombine back into $HA$ . At equilibrium, both processes happen at the same rate — so the concentrations stop changing.

Important

For a weak acid, most of the acid remains as $HA$ at equilibrium. Only a tiny fraction exists as ions. This is the opposite of a strong acid, where the forward reaction goes to completion (single arrow: $\rightarrow$ ).

The Quantitative Measure: $K_a$

Every weak acid has a number that tells you exactly how "shy" it is — the acid dissociation constant, $K_a$ :

K_a = \frac{[H_3O^+][A^-]}{[HA]}

K_a = \frac{[H_3O^+][A^-]}{[HA]}

The smaller the $K_a$ , the weaker the acid. For acetic acid (vinegar), $K_a \approx 1.8 \times 10^{-5}$ . That tiny number means the numerator (ions) is very small compared to the denominator (intact acid). For a strong acid like HCl, $K_a$ is effectively infinite — the denominator is essentially zero because all the acid has ionized.

A Concrete Example

Suppose you dissolve 0.10 mol of acetic acid ( $CH_3COOH$ ) in 1 L of water. At equilibrium, you'll find:

$[CH_3COOH] \approx 0.0998$ M (almost all of it is still intact)
$[H_3O^+] \approx 0.0013$ M (only about 1.3% has ionized)
$[CH_3COO^-] \approx 0.0013$ M

Note

|---------|-------------|------------|-----------------|

| $CH_3COOH$ | 0.10 | $-x$ | $0.10 - x$ |

| $H_3O^+$ | 0 | $+x$ | $x$ |

| $CH_3COO^-$ | 0 | $+x$ | $x$ |

Plugging into $K_a = \frac{x^2}{0.10 - x} = 1.8 \times 10^{-5}$ and solving gives $x \approx 0.0013$ M.

Why This Matters …

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The extent of ionisation of a weak base is directly proportional to its $K_b$ value. The decreasing order of ionisation is: Dimethylamine > Ammonia > Pyridine > Urea, making Dimethylamine the strongest base.

The key idea here is that the ionisation constant $K_b$ is a direct measure of how far the equilibrium $\text{MOH} \rightleftharpoons \text{M}^+ + \text{OH}^-$ lies to the right. A larger $K_b$ means a greater fraction of the base molecules have dissociated into ions at equilibrium — that is, a greater extent of ionisation. The strength of a base is also judged by the same constant: the larger the $K_b$ , the stronger the base.

Let’s work through the reasoning step by step.

Understand what $K_b$ tells us. For a weak base MOH, the equilibrium constant is

K_b = \frac{[\text{M}^+][\text{OH}^-]}{[\text{MOH}]}

If $K_b$ is large, the numerator (product of ion concentrations) is large relative to the denominator (concentration of unionised base). This directly implies that a larger proportion of the base has ionised. So, higher $K_b$ → greater extent of ionisation → stronger base.

List the given $K_b$ values in order.

From the data:
- Dimethylamine: $5.4 \times 10^{-4}$
- Ammonia: $1.77 \times 10^{-5}$
- Pyridine: $1.77 \times 10^{-9}$
- Urea: $1.3 \times 10^{-14}$
Notice that $5.4 \times 10^{-4}$ is the largest, followed by $1.77 \times 10^{-5}$ , then $1.77 \times 10^{-9}$ , and finally $1.3 \times 10^{-14}$ is the smallest.
Arrange in decreasing order of ionisation.

Since ionisation extent follows $K_b$ , we simply write the bases from the largest $K_b$ to the smallest:

$\text{Dimethylamine} > \text{Ammonia} > \text{Pyridine} > \text{Urea}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
$^-$
›Reveal solutionSolution
The conjugate base of an acid is what remains after the acid donates a proton; hypochlorous acid (HOCl) loses H⁺ to give OCl⁻, so Statement I is correct. The ion-product constant of water, $K_w$ , changes with temperature because the autoionization of water is endothermic, so Statement II is false. Therefore, only Statement I is correct.

Concept & Intuition

This question tests two fundamental ideas in acid–base chemistry:

Conjugate acid–base pairs: When an acid donates a proton (H⁺), the species left behind is its conjugate base. For hypochlorous acid (HOCl), removing H⁺ yields OCl⁻.

Temperature dependence of $K_w$ : The autoionization of water ( $2H_2O \rightleftharpoons H_3O^+ + OH^-$ ) is an equilibrium process. Like all equilibrium constants, $K_w = [H_3O^+][OH^-]$ changes with temperature because the reaction is endothermic (absorbs heat). At 25 °C, $K_w = 1.0 \times 10^{-14}$ , but it increases at higher temperatures and decreases at lower temperatures.

Step-by-step reasoning

Analyze Statement I Hypochlorous acid has the formula HOCl (or HClO). When it acts as an acid, it donates a proton (H⁺):

$\text{HOCl} \rightleftharpoons \text{H}^+ + \text{OCl}^-$
The species that remains after losing H⁺ is OCl⁻, which is the conjugate base. Hence Statement I is correct.

Analyze Statement II The value of $K_w$ is the equilibrium constant for the autoionization of water: $2H_2O \rightleftharpoons H_3O^+ + OH^-$ …
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TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
$_2$
›Reveal solutionSolution

Equal moles $n$ of each give $190n=5$ ; the mixture needs $3n$ mol of HCl, so $789$ mL of $0.1$ M HCl. Correct option: (B).

Moles of each salt. Let $n$ = moles of $\text{Na}_2\text{CO}_3$ = moles of $\text{NaHCO}_3$ .

Molar masses: $\text{Na}_2\text{CO}_3=106,\ \text{NaHCO}_3=84\ \text{g mol}^{-1}$ .
$106n+84n=190n=5\ \Rightarrow\ n=\frac{5}{190}=\frac{1}{38}\ \text{mol}.$
HCl needed for complete reaction (to CO $_2$ ).

$\text{Na}_2\text{CO}_3+2\text{HCl}\rightarrow2\text{NaCl}+\text{H}_2\text{O}+\text{CO}_2\quad(2n\ \text{mol HCl})$ …
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TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
$_2$
›Reveal solutionSolution
The key is that both Na₂CO₃ and NaHCO₃ react with HCl, but in different stoichiometric ratios. Since they are present in equal moles, we find the total moles of each, then the total moles of HCl needed, and finally the volume of 0.1 M HCl. The answer is 789.0 mL, option (C).

Concept & Intuition

When a mixture of Na₂CO₃ (sodium carbonate) and NaHCO₃ (sodium bicarbonate) reacts with HCl, each undergoes a distinct neutralization:

Na₂CO₃ + 2 HCl → 2 NaCl + H₂O + CO₂

(Each mole of carbonate consumes 2 moles of HCl)

NaHCO₃ + HCl → NaCl + H₂O + CO₂

(Each mole of bicarbonate consumes 1 mole of HCl)

The problem gives a solid mixture of known total mass (5 g) that contains equal number of moles of the two compounds. This is the crucial constraint: we can set up an equation relating the total mass to the unknown common mole number, solve for that number, then compute the total HCl needed.

Step-by-step solution

Define the unknown Let $n$ be the number of moles of Na₂CO₃ and also the number of moles of NaHCO₃ in the 5 g mixture. So:

$\text{moles of Na}_2\text{CO}_3 = n, \quad \text{moles of NaHCO}_3 = n$

Write the mass equation Molar mass of Na₂CO₃ = $2 \times 23 + 12 + 3 \times 16 = 106$ g/mol Molar mass of NaHCO₃ = $23 + 1 + 12 + 3 \times 16 = 84$ g/mol Total mass:

$106n + 84n = 190n = 5 \text{ g}$
Therefore:
$n = \frac{5}{190} = \frac{1}{38} \text{ mol}$

Calculate total moles of HCl required

From Na₂CO₃: $n$ moles require $2n$ moles of HCl

From NaHCO₃: $n$ moles require $n$ moles of HCl Total HCl needed:

$2n + n = 3n = 3 \times \frac{1}{38} = \frac{3}{38} \text{ mol}$

Find volume of 0.1 M HCl Molarity $M = 0.1$ mol/L means:
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TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
$27^\circ \mathrm{C}$
›Reveal solutionSolution
The key idea is to find the excess strong acid after neutralization, then account for dilution to get the final H⁺ concentration and pH. The final pH is 2.0, so the correct option is (B).

Concept and intuition:

When a strong acid (HCl) and a strong base (NaOH) are mixed, they neutralize each other completely. The pH of the final solution depends on which reactant is in excess. Here, we have more moles of HCl than NaOH, so after reaction, leftover HCl remains. Adding water dilutes this leftover acid. Since HCl is a strong acid, the H⁺ concentration equals the concentration of excess HCl. The pH is then simply $-\log[\text{H}^+]$ .

Step-by-step solution:

Calculate moles of HCl and NaOH initially.

Moles of HCl = volume (L) × molarity = $0.100 \, \text{L} \times 0.5 \, \text{M} = 0.050 \, \text{mol}$ .

Moles of NaOH = $0.100 \, \text{L} \times 0.4 \, \text{M} = 0.040 \, \text{mol}$ .

Determine the excess after neutralization.

The reaction is: $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ .

HCl and NaOH react in a 1:1 ratio.

Moles of HCl used = moles of NaOH = 0.040 mol.

Moles of HCl remaining = $0.050 - 0.040 = 0.010 \, \text{mol}$ .

Account for dilution.

Initial total volume after mixing = $100 + 100 = 200 \, \text{mL}$ .

Then 800 mL of water is added, so final volume = $200 + 800 = 1000 \, \text{mL} = 1.00 \, \text{L}$ . …
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TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
$\mathrm{H_2Te, H_2Se}$
›Reveal solutionSolution
The boiling points of group 16 hydrides are dominated by two competing effects: increasing London dispersion forces down the group (which raises boiling points) and the exceptional hydrogen bonding in water (which gives it the highest boiling point). The lowest boiling point belongs to H₂S, and the highest to H₂O, so the correct pair is H₂S and H₂O — option (D).

Concept & Intuition

For the hydrides of group 16 (O, S, Se, Te), boiling points generally increase as you go down the group because the molecules get larger and more polarizable, leading to stronger London dispersion forces. However, water (H₂O) is a dramatic exception: it forms strong intermolecular hydrogen bonds, which require much more energy to break. So while H₂Te is heavy and has strong dispersion forces, H₂O’s hydrogen bonding gives it the highest boiling point of all. The lowest boiling point is not at the top (water) but at H₂S, because H₂S is too small for strong dispersion forces and too weakly polar to hydrogen-bond effectively. This is a classic “anomaly of water” question.

Step-by-step reasoning

Recall the trend in boiling points for group 16 hydrides

The actual boiling points (in °C) are:

H₂O: 100°C

H₂S: –60.3°C

H₂Se: –41.3°C

H₂Te: –2.2°C

So the order is: H₂S < H₂Se < H₂Te << H₂O.

Identify the hydride with the lowest boiling point

From the data, H₂S has the lowest boiling point. Why? Because it is not heavy enough to have strong London forces (unlike H₂Se and H₂Te), and it does not form significant hydrogen bonds (unlike H₂O). So X = H₂S.

Identify the hydride with the highest boiling point …
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TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
$K_1$
›Reveal solutionSolution
The overall equilibrium constant for a reaction that is the sum of several stepwise reactions is the product of the equilibrium constants for those individual steps. For a tribasic acid, the overall dissociation constant is the product of its three stepwise dissociation constants. The final result is $\boxed{K_1 \times K_2 \times K_3}$ .

Concept and Intuition

Acids can dissociate in water to release hydrogen ions ( $H^+$ ). A tribasic acid is an acid that can donate three protons (hydrogen ions) per molecule. This dissociation typically occurs in successive steps, not all at once. Each step has its own equilibrium constant, reflecting the extent of dissociation at that particular stage.

Consider a generic tribasic acid, $H_3A$ . Its dissociation proceeds as follows:

First dissociation: $H_3A$ loses one proton to form $H_2A^-$ .

Second dissociation: $H_2A^-$ loses another proton to form $HA^{2-}$ .

Third dissociation: $HA^{2-}$ loses the final proton to form $A^{3-}$ .

Each of these steps is an equilibrium reaction, and each has an associated equilibrium constant ( $K_1, K_2, K_3$ ). The question asks for the equilibrium constant for the overall dissociation, which represents the complete loss of all three protons from $H_3A$ to form $A^{3-}$ .

The fundamental principle here is that when chemical equations are added together to yield an overall reaction, their individual equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction. This is because the concentrations of intermediate species (like $H_2A^-$ and $HA^{2-}$ in this case) cancel out when the equilibrium constant expressions are multiplied, leaving only the initial reactants and final products.

Step-by-step Solution

Write the first dissociation step and its equilibrium constant:

The first proton is lost from the tribasic acid $H_3A$ :

$H_3A \rightleftharpoons H^+ + H_2A^-$

The equilibrium constant for this step, $K_1$ , is given by:

$K_1 = \frac{[H^+][H_2A^-]}{[H_3A]}$

Write the second dissociation step and its equilibrium constant:

The dihydrogen anion $H_2A^-$ then loses its second proton:

$H_2A^- \rightleftharpoons H^+ + HA^{2-}$

The equilibrium constant for this step, $K_2$ , is given by:

$K_2 = \frac{[H^+][HA^{2-}]}{[H_2A^-]}$

Write the third dissociation step and its equilibrium constant:

Finally, the hydrogen anion $HA^{2-}$ loses its third proton:

$HA^{2-} \rightleftharpoons H^+ + A^{3-}$

The equilibrium constant for this step, $K_3$ , is given by:

$K_3 = \frac{[H^+][A^{3-}]}{[HA^{2-}]}$

Write the overall dissociation reaction:

The overall dissociation represents the complete loss of all three protons from $H_3A$ :

$H_3A \rightleftharpoons 3H^+ + A^{3-}$

Let the equilibrium constant for this overall reaction be $K_{overall}$ . Its expression would be:

$K_{overall} = \frac{[H^+]^3[A^{3-}]}{[H_3A]}$

Combine the stepwise reactions to obtain the overall reaction: We can obtain the overall reaction by adding the three stepwise reactions: $(H_3A \rightleftharpoons H^+ + H_2A^-)$ $(H_2A^- \rightleftharpoons H^+ + HA^{2-})$ $(HA^{2-} \rightleftharpoons H^+ + A^{3-})$

Adding these equations, we get:

$H_3A + H_2A^- + HA^{2-} \rightleftharpoons (H^+ + H^+ + H^+) + H_2A^- + HA^{2-} + A^{3-}$

By cancelling the intermediate species ( $H_2A^-$ and $HA^{2-}$ ) that appear on both sides of the equation, we arrive at the overall reaction:

$H_3A \rightleftharpoons 3H^+ + A^{3-}$

Apply the rule for combining equilibrium constants:

Since the overall reaction is the sum of the three stepwise reactions, the overall equilibrium constant is the product of the individual equilibrium constants:

$K_{overall} = K_1 \times K_2 \times K_3$

›Proof

To confirm this, let's multiply the expressions for $K_1$ , $K_2$ , and $K_3$ : …
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