The ionisation constant of an acid, Ka, is the measure of strength of an acid. The Ka values of acetic acid, h

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Q. The ionisation constant of an acid, Ka, is the measure of strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are 1.74 \times 10^-5, 3.0 \times 10^-8 and 1.8 \times 10^-4 respectively. Which of the following orders of pH of 0.1 mol dm^-3 solutions of these acids is correct?

(i) acetic acid > hypochlorous acid > formic acid

(ii) hypochlorous acid > acetic acid > formic acid

(iii) formic acid > hypochlorous acid > acetic acid

(iv) formic acid > acetic acid > hypochlorous acid

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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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For weak acids, pH is inversely related to $K_a$ : the smaller the $K_a$ , the higher the pH. Since $K_a$ values are $1.8 \times 10^{-4}$ (formic), $1.74 \times 10^{-5}$ (acetic), $3.0 \times 10^{-8}$ (hypochlorous), the pH order is hypochlorous acid > acetic acid > formic acid, which is option (ii).

The key idea is simple: a weak acid’s strength is measured by its ionization constant $K_a$ . A larger $K_a$ means more dissociation, more $H^+$ in solution, and therefore a lower pH. For a given concentration, the pH of a weak acid solution depends directly on $K_a$ .

Let’s see why this works.

The relationship between $K_a$ and $[H^+]$ For a weak acid $HA$ of initial concentration $c$ , the equilibrium is:

HA \rightleftharpoons H^+ + A^-

If the degree of ionization is $\alpha$ , then $[H^+] = c\alpha$ and $[A^-] = c\alpha$ , while $[HA] = c(1-\alpha)$ . The ionization constant is:

K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(c\alpha)^2}{c(1-\alpha)}

For a weak acid, $\alpha$ is small, so $1-\alpha \approx 1$ . This gives the approximation:

K_a \approx c\alpha^2 \quad \Rightarrow \quad \alpha \approx \sqrt{\frac{K_a}{c}}

Hence:

[H^+] = c\alpha \approx \sqrt{K_a c}

And pH is:

\text{pH} = -\log[H^+] \approx -\log\sqrt{K_a c} = \frac{1}{2}(-\log K_a - \log c)

c

Since $c$ is the same for all three acids (0.1 mol dm $^{-3}$ ), the pH depends only on $K_a$ : larger $K_a$ → smaller pH.

Compare the $K_a$ values
- Formic acid: $K_a = 1.8 \times 10^{-4}$ (largest)
- Acetic acid: $K_a = 1.74 \times 10^{-5}$ (middle)
- Hypochlorous acid: $K_a = 3.0 \times 10^{-8}$ (smallest)
So formic acid is the strongest among the three, and hypochlorous acid is the weakest.
Translate to pH order

Stronger acid → lower pH. Therefore:
- Formic acid has the lowest pH. …

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

Showing the 12 most recent of 16 on this concept.

AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$H_2A$
›Reveal solutionSolution
A classic diprotic-acid trick: combining $K_{a1}$ and $K_{a2}$ eliminates the intermediate species $HA^-$ when $[A^{2-}]=[H_2A]$ , letting $[H^+]$ be found directly from $\sqrt{K_{a1}K_{a2}}$ .

Concept and Intuition

For a diprotic acid $H_2A$ , the two successive dissociations are:
$H_2A \rightleftharpoons H^+ + HA^- \qquad K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}$ $HA^- \rightleftharpoons H^+ + A^{2-} \qquad K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]}$
Multiplying these two expressions together makes the intermediate concentration $[HA^-]$ cancel out entirely:
$K_{a1}K_{a2} = \frac{[H^+]^2[A^{2-}]}{[H_2A]}$
This is a very useful identity whenever you're told something about the relationship between $[A^{2-}]$ and $[H_2A]$ directly, bypassing the need to know $[HA^-]$ at all.

Step-by-Step Solution

Write $K_{a1}K_{a2} = \dfrac{[H^+]^2[A^{2-}]}{[H_2A]}$ .

Given $[A^{2-}] = [H_2A]$ , the fraction $\dfrac{[A^{2-}]}{[H_2A]} = 1$ .

So $K_{a1}K_{a2} = [H^+]^2$ .

Substitute: $[H^+]^2 = (6\times10^{-2})(6\times10^{-5}) = 36\times10^{-7} = 3.6\times10^{-6}$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$pH$
›Reveal solutionSolution
Weak-acid pH formula $[H^+]=\sqrt{K_aC}$ gives $[H^+]=5\times10^{-6}$ M and $pH=5.3$ .

Concept and Intuition

For a weak monoprotic acid HA with small degree of ionisation, the equilibrium $[H^+]\approx[A^-]$ and $[HA]\approx C$ (initial concentration), so $K_a = \dfrac{[H^+]^2}{C}$ , giving the shortcut $[H^+]=\sqrt{K_aC}$ .

Step-by-Step Solution

$[H^+] = \sqrt{K_a \times C} = \sqrt{5\times10^{-10}\times 0.05}$ .

$5\times10^{-10}\times0.05 = 2.5\times10^{-11}$ .

$\sqrt{2.5\times10^{-11}} = \sqrt{25\times10^{-12}} = 5\times10^{-6}$ M. …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$1.0\times10^{-8}$
›Reveal solutionSolution
Standard weak-acid dissociation calculation from pH: $K_a = [H^+]^2/C$ . Answer: $1.0\times10^{-7}$ .

Concept and Intuition

For a weak monobasic acid HA dissociating as $HA \rightleftharpoons H^+ + A^-$ , if $\alpha$ is the small degree of dissociation, $[H^+]=C\alpha$ and $K_a = \dfrac{C\alpha^2}{1-\alpha}\approx C\alpha^2$ (since $\alpha\ll1$ for a weak acid). This simplifies to $K_a\approx[H^+]^2/C$ .

Step-by-Step Solution

$pH=4.0 \Rightarrow [H^+]=10^{-4}$ M.

Concentration $C=0.1$ M.

$K_a = \dfrac{[H^+]^2}{C} = \dfrac{(10^{-4})^2}{0.1} = \dfrac{10^{-8}}{10^{-1}} = 10^{-7}$ .

Check: $\alpha = [H^+]/C = 10^{-4}/0.1 = 10^{-3}$ , which is indeed $\ll 1$ , validating the weak-acid approximation.

Common Mistakes …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$25°C$
›Reveal solutionSolution
This tests Ostwald's dilution law for a weak acid: using $K_a = C\alpha^2$ to back out the concentration from the given percentage ionisation, then computing $[H^+]=C\alpha$ and finally the pH, gives 3.37.

Concept and Intuition

For a weak monoprotic acid HA that ionises to a small fraction $\alpha$ (degree of dissociation), the equilibrium constant is $K_a = \dfrac{C\alpha^2}{1-\alpha} \approx C\alpha^2$ for small $\alpha$ . Given $K_a$ and $\alpha$ , we can solve backward for the concentration $C$ , and from there for $[H^+] = C\alpha$ and hence pH. The provided log values are the exact hint that this problem wants a logarithmic (not linear) route to the answer.

Step-by-Step Solution

Convert percentage ionisation to a fraction: $\alpha = 4.242\% = 0.04242$ .

Apply Ostwald's dilution law: $K_a = C\alpha^2 \Rightarrow C = \dfrac{K_a}{\alpha^2}$ .

Take logs: $\log C = \log K_a - 2\log\alpha$ .

$\log K_a = \log(1.8\times10^{-5}) = \log1.8 - 5 \approx 0.2553 - 5 = -4.7447$ .

$\log\alpha = \log(0.04242) = -1.372$ (given).

$\log C = -4.7447 - 2(-1.372) = -4.7447 + 2.744 = -2.0007 \approx -2.00$ , so $C \approx 1.0\times10^{-2}$ M. …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$K_a = 1.8\times10^{-5}$
›Reveal solutionSolution
Tests the weak-acid degree-of-ionization formula $\alpha = \sqrt{K_a/C}$ ; solving for concentration gives $x = 0.01$ M.

Concept and Intuition

For a weak monoprotic acid $HA \rightleftharpoons H^+ + A^-$ with initial concentration $C$ and degree of dissociation $\alpha$ (small, so $1-\alpha \approx 1$ ), the equilibrium constant is approximately $K_a \approx C\alpha^2$ , giving $\alpha \approx \sqrt{K_a/C}$ . Rearranging lets us find the concentration from a known percentage ionization.

Step-by-Step Solution

Convert percentage ionization to a fraction: $\alpha = \dfrac{4.242}{100} = 0.04242$ .

Use the approximate relation: $K_a = C\alpha^2 \Rightarrow x = \dfrac{K_a}{\alpha^2}$ .

Compute $\alpha^2 = (0.04242)^2 \approx 1.8\times10^{-3}$ . …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$25\,^\circ\mathrm{C}$
›Reveal solutionSolution
Ostwald's dilution law gives the degree of ionisation of a weak acid as $\sqrt{K_a/C}$ ; here it works out to $3\%$ .

Concept and Intuition

For a weak monoprotic acid $\mathrm{HA}\rightleftharpoons H^++A^-$ with initial concentration $C$ and degree of dissociation $\alpha$ (small, so $1-\alpha\approx1$ ), $K_a \approx C\alpha^2$ , giving $\alpha = \sqrt{K_a/C}$ (Ostwald's dilution law).

Step-by-Step Solution

$K_a = 1.8\times10^{-5}$ , $C=0.02\,\mathrm{M}$ .

$\alpha = \sqrt{\dfrac{1.8\times10^{-5}}{0.02}} = \sqrt{9\times10^{-4}} = 3\times10^{-2} = 0.03$ . …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$x$
›Reveal solutionSolution
Back out each solution's initial concentration from its % ionization via $K_a=C\alpha^2$ , then average the two moles over the combined 2 L volume — giving 0.015 M.

Concept and Intuition

For a weak acid at low degree of dissociation $\alpha$ , the Ostwald dilution law simplifies to $K_a \approx C\alpha^2$ (since $1-\alpha \approx 1$ ). Given $K_a$ and $\alpha$ for each solution, we can solve backward for each solution's original concentration $C$ . When two solutions of a weak acid (not just its ions) are mixed, the total moles of acetic acid (dissociated + undissociated) simply add, and the new concentration is total moles over total volume.

Step-by-Step Solution

Solution A: $\alpha_A = 4.242\% = 0.04242$ . Using $K_a = C\alpha^2$ : $C_A = \dfrac{1.8\times10^{-5}}{(0.04242)^2} = \dfrac{1.8\times10^{-5}}{1.8\times10^{-3}} = 0.01\ \mathrm{M}$ .

Solution B: $\alpha_B = 3\% = 0.03$ . $C_B = \dfrac{1.8\times10^{-5}}{(0.03)^2} = \dfrac{1.8\times10^{-5}}{9\times10^{-4}} = 0.02\ \mathrm{M}$ .

Moles of acetic acid in 1 L of A = 0.01 mol; moles in 1 L of B = 0.02 mol. …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$\mathrm{H_2A}$
›Reveal solutionSolution
Multiplying the two stepwise dissociation constants gives the overall $\mathrm{H_2A \rightleftharpoons 2H^+ + A^{2-}}$ equilibrium constant; dividing by $[H^+]^2$ (from the given pH) yields $[A^{2-}]/[H_2A] = 0.036$ .

Concept and Intuition

For a diprotic acid, the two stepwise equilibria are:
$\mathrm{H_2A \rightleftharpoons H^+ + HA^-}, \quad K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}$ $\mathrm{HA^- \rightleftharpoons H^+ + A^{2-}}, \quad K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]}$
Multiplying these two equilibrium expressions together, the $[HA^-]$ terms cancel, giving the overall two-proton dissociation:
$K_{a1}K_{a2} = \frac{[H^+]^2[A^{2-}]}{[H_2A]}$
Step-by-Step Solution

$K_{a1}K_{a2} = (6\times10^{-2})(6\times10^{-5}) = 36\times10^{-7} = 3.6\times10^{-6}$ .

pH = 2.0 $\Rightarrow$ $[H^+] = 10^{-2}$ M, so $[H^+]^2 = 10^{-4}$ . …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$\mathrm{CN^-}$
›Reveal solutionSolution
For a conjugate acid-base pair, $K_aK_b=K_w$ ; using the given $K_b$ of $\mathrm{CN^-}$ gives $K_a(\mathrm{HCN})\approx4.8\times10^{-9}$ .

Concept and Intuition

$\mathrm{CN^-}$ is the conjugate base of the weak acid HCN. Whenever a conjugate acid-base pair is involved, their ionization constants are linked through the autoionization constant of water: $K_a\times K_b = K_w$ . This lets you get one from the other without doing a fresh equilibrium calculation.

Step-by-Step Solution

Given: ionization constant of $\mathrm{CN^-}$ (as a base, hydrolyzing water) $K_b = 2.08\times10^{-6}$ .

Relation: $K_a(\text{conjugate acid}) \times K_b(\text{base}) = K_w$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$27^0$
›Reveal solutionSolution
Standard weak-acid dissociation approximation $K_a \approx c\alpha^2$ gives $K_a = 5\times10^{-5}$ — option (B).

Concept and Intuition

For a weak acid HA with small degree of dissociation $\alpha$ , the equilibrium expression $K_a = \dfrac{c\alpha^2}{1-\alpha}$ simplifies to $K_a \approx c\alpha^2$ because $\alpha \ll 1$ makes $(1-\alpha) \approx 1$ . This is the standard shortcut used whenever $\alpha$ is a percent-level quantity like 1%.

Step-by-Step Solution

Given: $c = 0.5$ M, $\alpha = 1\% = 0.01$ .

$K_a = \dfrac{c\alpha^2}{1-\alpha} \approx c\alpha^2$ (valid since $\alpha = 0.01 \ll 1$ ).

$K_a = 0.5 \times (0.01)^2 = 0.5 \times 0.0001 = 5\times10^{-5}$ .

Common Mistakes …
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AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$\mathrm{H_3O^+}$
›Reveal solutionSolution
Using the degree-of-dissociation ( $\alpha$ ) formulas for a weak acid at equilibrium gives $[\mathrm{H_3O^+}]=[\mathrm{A^-}]=0.005\ \mathrm{M}$ and $[\mathrm{HA}]=0.495\ \mathrm{M}$ .

Concept and Intuition

For a weak monoprotic acid $\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-}$ starting at concentration $C$ , if $\alpha$ is the fraction that dissociates at equilibrium, then the amount dissociated is $C\alpha$ (this becomes both $[\mathrm{H_3O^+}]$ and $[\mathrm{A^-}]$ in a 1:1 stoichiometry), and the amount remaining undissociated is $C(1-\alpha)$ .

Step-by-Step Solution

Given: $C = 0.5\ \mathrm{M}$ , $\alpha = 1\% = 0.01$ .

$[\mathrm{H_3O^+}] = C\alpha = 0.5 \times 0.01 = 0.005\ \mathrm{M}$ .

$[\mathrm{A^-}] = C\alpha = 0.005\ \mathrm{M}$ (same as H $_3$ O $^+$ since 1 mole of each is produced per mole of HA dissociated). …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$K_a(A)$
›Reveal solutionSolution
A larger dissociation constant $K_a$ means a stronger weak acid and hence a lower pH; ranking the given $K_a$ values in decreasing order (A > B > C) gives the pH order $C>B>A$ .

Concept and Intuition

For a weak acid, $[H^+]=\sqrt{K_aC}$ at equal concentration $C$ . Since $[H^+]$ increases with $K_a$ , pH (which is inversely related to $[H^+]$ ) decreases as $K_a$ increases. So the acid with the largest $K_a$ has the smallest pH, and vice versa.

Step-by-Step Solution

List the given values: $K_a(A)=3.5\times10^{-4}$ , $K_a(B)=1\times10^{-5}$ , $K_a(C)=5\times10^{-10}$ .

Rank by acid strength (largest $K_a$ = strongest acid): $A > B > C$ .

Since stronger acids dissociate more, giving higher $[H^+]$ and hence lower pH, the pH ranking is the exact reverse of the acid-strength ranking. …
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Q.The ionisation constant of an acid, Ka, is the measure of strength of an acid. The Ka values of acetic acid, hypochlorous acid and formic acid are 1.74 × 10^-5, 3.0 × 10^-8 and 1.8 × 10^-4 respectively. Which of the following orders of pH of 0.1 mol dm^-3 solutions of these acids is correct?

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