The ionization constant of propanoic acid is 1.32 × 10⁻⁵. Calculate the degree of ionization of the acid in it

Exercises · 6.59

Q. The ionization constant of propanoic acid is 1.32 \times 10⁻⁵. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

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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, the degree of ionization $\alpha$ is found from $K_a = C\alpha^2/(1-\alpha)$ ; in pure 0.05 M solution, $\alpha \approx 1.62 \times 10^{-2}$ and pH ≈ 3.09. In the presence of 0.01 M HCl, the common ion effect suppresses ionization drastically, giving $\alpha' \approx 1.32 \times 10^{-3}$ .

Concept First: Weak Acid Ionization

Propanoic acid ( $\mathrm{CH_3CH_2COOH}$ ) is a weak monoprotic acid. Its ionization in water is an equilibrium:

\mathrm{CH_3CH_2COOH \rightleftharpoons CH_3CH_2COO^- + H^+}

The equilibrium constant $K_a = 1.32 \times 10^{-5}$ tells us the acid is weak — only a tiny fraction of molecules actually donate a proton. The degree of ionization $\alpha$ is the fraction of acid molecules that have ionized. If the initial concentration is $C$ , then at equilibrium:

$[\mathrm{HA}] = C(1-\alpha)$
$[\mathrm{A^-}] = C\alpha$
$[\mathrm{H^+}] = C\alpha$ (from the acid alone)

The key formula is:

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

When $\alpha$ is very small (typically $\alpha < 0.05$ ), we can approximate $1-\alpha \approx 1$ , giving $\alpha \approx \sqrt{K_a/C}$ . But we must check the approximation — if it fails, we solve the quadratic.

1. Pure 0.05 M solution — find $\alpha$

Given $C = 0.05$ M, $K_a = 1.32 \times 10^{-5}$ .

First, test the approximation: $\alpha \approx \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = \sqrt{2.64 \times 10^{-4}} = 1.62 \times 10^{-2}$ .

Is this small enough? $1.62\%$ is borderline — the approximation $1-\alpha \approx 1$ introduces about $1.6\%$ error. For exam accuracy (usually 2–3 significant figures), this is acceptable. But let's be thorough and solve exactly.

Write the exact equation:

\frac{0.05\,\alpha^2}{1-\alpha} = 1.32 \times 10^{-5}

Multiply through:

0.05\,\alpha^2 = 1.32 \times 10^{-5} (1-\alpha)

0.05\,\alpha^2 = 1.32 \times 10^{-5} - 1.32 \times 10^{-5}\,\alpha

Bring all terms to one side:

0.05\,\alpha^2 + 1.32 \times 10^{-5}\,\alpha - 1.32 \times 10^{-5} = 0

This is a quadratic in $\alpha$ . Using the quadratic formula:

\alpha = \frac{-1.32 \times 10^{-5} \pm \sqrt{(1.32 \times 10^{-5})^2 + 4(0.05)(1.32 \times 10^{-5})}}{2 \times 0.05}

The term $(1.32 \times 10^{-5})^2 = 1.74 \times 10^{-10}$ is negligible compared to $4(0.05)(1.32 \times 10^{-5}) = 2.64 \times 10^{-6}$ . So:

\alpha \approx \frac{-1.32 \times 10^{-5} + \sqrt{2.64 \times 10^{-6}}}{0.1}

\sqrt{2.64 \times 10^{-6}} = 1.625 \times 10^{-3}

\alpha \approx \frac{1.625 \times 10^{-3} - 1.32 \times 10^{-5}}{0.1} = \frac{1.6118 \times 10^{-3}}{0.1} = 1.612 \times 10^{-2}

Tip

The approximation gave $1.62 \times 10^{-2}$ — identical to three significant figures. For weak acids where $K_a/C < 10^{-3}$ , the approximation is safe. Here $K_a/C = 2.64 \times 10^{-4}$ , so it's fine.

So $\alpha = 1.62 \times 10^{-2}$ (or $1.62\%$ ).

2. pH of the pure solution

$[\mathrm{H^+}] = C\alpha = 0.05 \times 1.62 \times 10^{-2} = 8.10 \times 10^{-4}$ M.

\mathrm{pH} = -\log_{10}(8.10 \times 10^{-4}) = 3.09

(Check: $\log 8.1 \approx 0.908$ , so $4 - 0.908 = 3.092$ .)

3. In presence of 0.01 M HCl — common ion effect

Now the solution already contains $\mathrm{H^+}$ from a strong acid (HCl) at 0.01 M. This shifts the weak acid equilibrium to the left — ionization is suppressed.

Let the new degree of ionization be $\alpha'$ . The initial concentration of propanoic acid is still $C = 0.05$ M. At equilibrium: …

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Q.The ionization constant of propanoic acid is 1.32 × 10⁻⁵. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Concept First: Weak Acid Ionization

1. Pure 0.05 M solution — find α\alphaα

2. pH of the pure solution

3. In presence of 0.01 M HCl — common ion effect

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1. Pure 0.05 M solution — find $\alpha$

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