It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of t

Exercises · 6.47

Q. It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK a.

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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 anion concentration equals the hydronium ion concentration from pH, giving $[A^-] = 7.08 \times 10^{-5} \text{ M}$ . Using the ionization equilibrium, $K_a = 5.01 \times 10^{-7}$ and $\text{p}K_a = 6.30$ .

The key here is that an organic acid is almost always a weak acid — it doesn't fully dissociate in water. When you're given the pH of a weak acid solution, that pH directly tells you the hydronium ion concentration, and for a monoprotic acid HA, that's also the concentration of the anion (the conjugate base) produced, assuming no other sources of H⁺ or A⁻ matter.

Let's walk through this carefully.

Find $[H_3O^+]$ from the given pH. The definition of pH is $\text{pH} = -\log_{10}[H_3O^+]$ . So:

[H_3O^+] = 10^{-\text{pH}} = 10^{-4.15}

To evaluate this, recall that $10^{-4.15} = 10^{-4} \times 10^{-0.15}$ . You can compute $10^{-0.15}$ using a calculator or approximate: $10^{-0.15} \approx 0.708$ . So:

[H_3O^+] = 7.08 \times 10^{-5} \text{ M}

Recognize what this tells us about the anion. For a weak monoprotic acid HA, the dissociation is:

\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-

Every HA molecule that dissociates produces one H₃O⁺ and one A⁻. In pure water (no added acid or base), the only source of H₃O⁺ is the acid itself, so:

[A^-] = [H_3O^+] = 7.08 \times 10^{-5} \text{ M}

Watch out

A common mistake is to forget that for a weak acid in water, $[H_3O^+]$ from water autoionization ( $10^{-7}$ M) is negligible compared to $7.08 \times 10^{-5}$ M, so we safely ignore it. But if the pH were near 7, you'd need to account for water's contribution.

Set up the equilibrium expression for $K_a$ . The acid ionization constant is:

K_a = \frac{[H_3O^+][A^-]}{[HA]}

We have $[H_3O^+] = [A^-] = 7.08 \times 10^{-5}$ M. What about $[HA]$ at equilibrium?

Initially, the acid concentration was 0.01 M. A tiny amount, $x = 7.08 \times 10^{-5}$ M, dissociated. So the equilibrium concentration of undissociated HA is:

$[HA] = 0.01 - 7.08 \times 10^{-5} \approx 0.01 - 0.0000708 = 0.0099292 \text{ M}$ …

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Q.It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK a.

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