The ionization constant of nitrous acid is 4.5 × 10⁻⁴. Calculate the pH of 0.04 M sodium nitrite solution and

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Q. The ionization constant of nitrous acid is 4.5 \times 10⁻⁴. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

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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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$\text{NaNO}_2$ is the salt of a weak acid ( $\text{HNO}_2$ ) and a strong base, so the nitrite ion hydrolyses and the solution is basic. For $0.04\ \text{M}$ $\text{NaNO}_2$ : degree of hydrolysis $h = 2.36 \times 10^{-5}$ and $\text{pH} = 7.97$ .

Concept

Sodium nitrite dissociates fully into $\text{Na}^+$ and $\text{NO}_2^-$ . The nitrite ion is the conjugate base of the weak acid $\text{HNO}_2$ , so it reacts with water:

\text{NO}_2^- + \text{H}_2\text{O} \rightleftharpoons \text{HNO}_2 + \text{OH}^-

This releases $\text{OH}^-$ , making the solution basic. The hydrolysis constant is $K_h = K_w / K_a$ .

Solution

Given: $K_a(\text{HNO}_2) = 4.5 \times 10^{-4}$ , $c = 0.04\ \text{M}$ , $K_w = 1.0 \times 10^{-14}$ .

Step 1 - Hydrolysis constant

K_h = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}} = 2.22 \times 10^{-11}

Step 2 - Degree of hydrolysis …

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

COMEDK 2025Set 2025-A1 markMCQ
$3.5 \times 10^{-4}$
›Reveal solutionSolution

The hydrolysis constant $K_{eq}=K_w/K_a = 2.86\times10^{-11}$ uniquely fixes option (C).

Hydrolysis of the conjugate base $\text{F}^-$ :
$K_{eq}=\frac{K_w}{K_a}=\frac{1.0\times10^{-14}}{3.5\times10^{-4}}=2.86\times10^{-11}.$
pH of the weak acid ( $0.1$ M HF):
$[\text{H}^+]=\sqrt{K_a\,C}=\sqrt{3.5\times10^{-4}\times 0.1}=5.9\times10^{-3}\ \text{M},$
$\text{pH}=-\log(5.9\times10^{-3})\approx 2.23.$ …
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COMEDK 2023Set 2023-M1 markMCQ
$0.2 \mathrm{~M}$
›Reveal solutionSolution

For a weak acid, $K_a = C\alpha^2/(1-\alpha) \approx C\alpha^2$ . Substituting $C=0.2\,\mathrm{M}$ and $\alpha=0.069$ gives $K_a \approx 9.5\times10^{-4}$ .

Lactic acid is $6.9\%$ dissociated, so degree of dissociation $\alpha = 0.069$ , and $C = 0.2\,\mathrm{M}$ .

Using Ostwald's dilution law:
$K_a = \frac{C\alpha^2}{1-\alpha}$
With the standard approximation ( $\alpha \ll 1$ ): …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. The degree of dissociation of 0.5 M NH3 at 25°C in a solution of pH = 12. (A) 12% (B) 1% (C) 4% (D) 2%$
›Reveal solutionSolution

$[\text{OH}^-]=0.01$ M from pH 12; dividing by 0.5 M gives $\alpha = 0.02 = 2\%$ .

From pH:
$\text{pOH} = 14-12 = 2 \Rightarrow [\text{OH}^-] = 10^{-2} = 0.01\ \text{M}$
Degree of dissociation for $\text{NH}_3 \rightleftharpoons \text{NH}_4^+ + \text{OH}^-$ : …
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KCET 2020Set A-11 markMCQ
$T_1$
›Reveal solutionSolution
For a given amount of heat absorbed, the entropy change is larger at the lower temperature because entropy is inversely proportional to temperature: $\Delta S = Q/T$ . Hence $\Delta S_2 > \Delta S_1$ when $T_1 > T_2$ .

The core idea here is the definition of entropy change for a reversible process: $\Delta S = \frac{Q_{\text{rev}}}{T}$ . When a system absorbs a quantity of heat $Q$ at a constant temperature $T$ , the entropy change is simply $Q/T$ . This tells us that the same heat causes a bigger entropy increase at a lower temperature.

Why? Think of entropy as a measure of disorder or the number of ways energy can be distributed. At a lower temperature, the system already has less thermal motion, so adding the same amount of heat creates a proportionally larger relative increase in the number of accessible microstates — hence a larger entropy change.

Let’s work through the reasoning step by step.

Write the expression for entropy change For a system that absorbs heat $Q$ reversibly at a constant temperature $T$ ,

$\Delta S = \frac{Q}{T}$
This is the defining relation from thermodynamics.

Apply it to the two cases

At temperature $T_1$ : $\Delta S_1 = \frac{Q}{T_1}$

At temperature $T_2$ : $\Delta S_2 = \frac{Q}{T_2}$

The same $Q$ is absorbed in both cases.

Compare the denominators

We are given $T_1 > T_2$ . Since the numerator $Q$ is the same, the fraction with the smaller denominator gives the larger value.

$\frac{Q}{T_2} > \frac{Q}{T_1}$
Therefore $\Delta S_2 > \Delta S_1$ .

Watch out

A common mistake is to think that a higher temperature means a larger entropy change — but the opposite is true. Entropy change is inversely proportional to temperature for a fixed heat transfer. Don’t confuse $\Delta S$ with the absolute entropy $S$ , which does increase with temperature.

Examine the options …
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Q.The ionization constant of nitrous acid is 4.5 × 10⁻⁴. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Concept

Solution

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