The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its d

Exercises · 6.60

Q. The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

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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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A weak acid partially ionizes; given pH, we find $[\text{H}^+]$ and use it to calculate both $K_a$ (from the equilibrium expression) and $\alpha$ (the fraction ionized). For 0.1 M HCNO with pH 2.34: $K_a = 2.19 \times 10^{-4}$ and $\alpha = 4.57\%$ .

Cyanic acid is a weak acid, meaning it doesn't fully dissociate in water. Instead, it establishes an equilibrium between the undissociated molecule and its ions. The pH tells us the hydrogen ion concentration at equilibrium, which is the key to unlocking both the ionization constant $K_a$ (a measure of acid strength) and the degree of ionization $\alpha$ (the fraction of molecules that actually ionized).

The equilibrium we're dealing with is:

\text{HCNO} \rightleftharpoons \text{H}^+ + \text{CNO}^-

The ionization constant is defined as:

K_a = \frac{[\text{H}^+][\text{CNO}^-]}{[\text{HCNO}]}

Let's work through this systematically.

Step-by-step solution

1. Extract the hydrogen ion concentration from pH

The definition of pH gives us:

\text{pH} = -\log[\text{H}^+]

So:

[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.34} = 4.57 \times 10^{-3} \text{ M}

This is the equilibrium concentration of hydrogen ions.

2. Recognize the stoichiometry at equilibrium

For every HCNO molecule that ionizes, we get one $\text{H}^+$ and one $\text{CNO}^-$ . Since the acid is the only source of both ions:

[\text{H}^+] = [\text{CNO}^-] = 4.57 \times 10^{-3} \text{ M}

3. Find the equilibrium concentration of undissociated acid

We started with 0.1 M HCNO. Some of it ionized (the amount that ionized equals $[\text{H}^+]$ ), so:

[\text{HCNO}]_{\text{equilibrium}} = 0.1 - 4.57 \times 10^{-3} = 0.09543 \text{ M}

Tip

For very weak acids where $[\text{H}^+] \ll c_0$ , you can approximate $[\text{HCNO}] \approx c_0$ . Here, $4.57 \times 10^{-3}$ is about 4.6% of 0.1, so the approximation would introduce a small error but is often acceptable for quick estimates.

4. Calculate the ionization constant $K_a$

Substitute the equilibrium concentrations into the expression:

K_a = \frac{[\text{H}^+][\text{CNO}^-]}{[\text{HCNO}]} = \frac{(4.57 \times 10^{-3})(4.57 \times 10^{-3})}{0.09543}

K_a = \frac{2.088 \times 10^{-5}}{0.09543} = 2.19 \times 10^{-4}

So $K_a = 2.19 \times 10^{-4}$ (using the exact equilibrium concentration $0.1 - 0.00457 = 0.0954$ M; the quick approximation with $c \approx 0.1$ gives $2.09 \times 10^{-4}$ ). …

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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 pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Step-by-step solution

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