What will be the value of pH of 0.01 mol dm^-3 CH3COOH (Ka = 1.74 × 10^-5)? (i) 3.4 (ii) 3.6 (iii) 3.9 (iv) 3.

NCERT Exemplar · Q13

Q. What will be the value of pH of 0.01 mol dm^-3 CH3COOH (Ka = 1.74 \times 10^-5)?

(i) 3.4

(ii) 3.6

(iii) 3.9

(iv) 3.0

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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 a weak acid, ionization is incomplete and governed by $K_a$ . Using the approximation that $[\mathrm{H}^+] \approx \sqrt{K_a \cdot c}$ when ionization is small, we find $\mathrm{pH} \approx 3.4$ .

Why this approach works

Acetic acid is a weak acid, meaning it does not dissociate completely in water. Instead, an equilibrium is established:

\mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H^+}

The extent of ionization is controlled by the acid dissociation constant $K_a$ . For weak acids with small $K_a$ values (here $1.74 \times 10^{-5}$ ), only a tiny fraction of molecules ionize, so we can simplify the equilibrium calculation with a standard approximation.

Step-by-step solution

1. Write the equilibrium expression

At equilibrium, if $\alpha$ is the degree of ionization and $c = 0.01 \, \mathrm{mol \, dm^{-3}}$ is the initial concentration:

K_a = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c\alpha^2}{1-\alpha}

2. Apply the weak-acid approximation

Since $K_a$ is small, ionization is minimal and $\alpha \ll 1$ . This means $(1 - \alpha) \approx 1$ , simplifying our expression:

K_a \approx c\alpha^2

Tip

The approximation $1 - \alpha \approx 1$ is valid when $\alpha < 0.05$ (less than 5% ionization). We can verify this after solving.

3. Solve for the hydrogen ion concentration

From $K_a = c\alpha^2$ , we get $\alpha = \sqrt{\frac{K_a}{c}}$ .

The hydrogen ion concentration is:

[\mathrm{H^+}] = c\alpha = c \cdot \sqrt{\frac{K_a}{c}} = \sqrt{K_a \cdot c}

Substituting the values:

[\mathrm{H^+}] = \sqrt{1.74 \times 10^{-5} \times 0.01} = \sqrt{1.74 \times 10^{-7}}

$[\mathrm{H^+}] = \sqrt{17.4 \times 10^{-8}} = \sqrt{17.4} \times 10^{-4}$ …

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Q.What will be the value of pH of 0.01 mol dm^-3 CH3COOH (Ka = 1.74 × 10^-5)?

Why this approach works

Step-by-step solution

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