# pH of weak acids

> AP Chemistry · Unit 8: Acids and Bases
> Source: https://www.owlsprep.com/study/ap-chemistry-u8-ph-of-weak-acids/

This module covers equilibrium calculations for pH of monoprotic weak acids, including the $K_a$ expression, approximation method, 5% validation rule, quadratic solutions, and percent dissociation calculations.

**Prerequisites:** Definition of pH and strong/weak acid difference; [Equilibrium constant expressions and ICE tables](https://www.owlsprep.com/study/ap-chemistry-u8-introduction-to-equilibrium/)

## Learning objectives

- Calculate Ka of a weak acid from measured pH
- Calculate pH of weak acids using approximation and quadratic methods
- Apply the 5% rule to validate approximations
- Calculate percent dissociation of weak acids and relate it to dilution
- Apply Le Chatelier's principle to percent dissociation changes

## Core Concepts of Weak Acid Dissociation

Unlike strong acids that dissociate completely in dilute solution, weak acids only partially dissociate, so equilibrium $[H_3O^+]$ cannot be directly equated to the initial weak acid concentration. This topic accounts for approximately 7-9% of total AP Chemistry exam points, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections.

Standard AP exam notation: $[HA]_0$ = initial concentration of monoprotic weak acid before dissociation, $K_a$ = acid dissociation constant, and $x$ = equilibrium concentration of dissociated HA. For pure weak acid solutions, $[H_3O^+]_{eq} = [A^-]_{eq} = x$. A core tested concept: the pH of a weak acid is always higher than the pH of an equal concentration of strong acid, because less hydronium is produced from partial dissociation.

## Acid Dissociation Constant ($K_a$) Expression

For any monoprotic weak acid $HA$, dissociation in water follows the equilibrium:

$$HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)$$

Liquid water is omitted from the equilibrium expression because its concentration is nearly constant in dilute solutions, and is absorbed into the equilibrium constant. The $K_a$ expression is defined as:

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

From dissociation stoichiometry and ICE tables, for a solution of pure weak acid: $[H_3O^+]_{eq} = [A^-]_{eq} = x$, and $[HA]_{eq} = [HA]_0 - x$. $K_a$ values are always small ($< 1$) for weak acids, with smaller $K_a$ corresponding to weaker acids.

**Worked example:** A 0.12 M solution of butanoic acid has a measured pH of 2.87 at 25°C. Calculate the $K_a$ of butanoic acid.

1. Convert measured pH to equilibrium $[H_3O^+]$:
2. $$[H_3O^+] = 10^{-pH} = 10^{-2.87} = 1.35 \times 10^{-3}\ M$$
3. This equals $x = [A^-]$, from dissociation stoichiometry.
4. Calculate equilibrium $[HA]$:
5. $$[HA]_{eq} = [HA]_0 - x = 0.12 - 0.00135 = 0.11865\ M$$
6. Substitute into the $K_a$ expression:
7. $$K_a = \frac{(1.35 \times 10^{-3})(1.35 \times 10^{-3})}{0.11865} \approx 1.5 \times 10^{-5}$$
8. Confirm $K_a$ is unitless per AP convention for dilute solutions.

> **Exam tip:** When calculating $K_a$ from pH, always use the equilibrium concentration of $HA$, not just the initial concentration. Only simplify to initial concentration after confirming $x$ is negligible.

## Approximation Method and 5% Validation Rule

Most weak acids have very small $K_a$ values, so $x = [H_3O^+]$ is much smaller than $[HA]_0$. This means $[HA]_0 - x \approx [HA]_0$, which simplifies the $K_a$ expression to:

$$K_a \approx \frac{x^2}{[HA]_0} \implies x = [H_3O^+] = \sqrt{K_a \times [HA]_0}$$

This approximation drastically reduces calculation time, valuable for both MCQ and FRQ. The AP Chemistry standard for validation is the 5% rule: if $\frac{x}{[HA]_0} \times 100\% \leq 5\%$, the approximation is acceptable. If greater than 5%, you must solve the full quadratic equation for an accurate result.

**Worked example:** Calculate the pH of a 0.45 M solution of benzoic acid, $K_a = 6.3 \times 10^{-5}$, at 25°C.

1. Set up the ICE table: Initial $[HA] = 0.45\ M$, $[H_3O^+] = [A^-] = 0$. Change: $-x, +x, +x$. Equilibrium: $[HA] = 0.45 - x$, $[H_3O^+] = [A^-] = x$.
2. Apply the approximation, assume $0.45 - x \approx 0.45$:
3. $$K_a = \frac{x^2}{0.45} = 6.3 \times 10^{-5}$$
4. Solve for $x$:
5. $$x^2 = 0.45 \times 6.3 \times 10^{-5} = 2.835 \times 10^{-5}, \quad x = 5.33 \times 10^{-3}\ M$$
6. Validate with the 5% rule:
7. $$\frac{5.33 \times 10^{-3}}{0.45} \times 100\% = 1.18\% < 5\%$$
8. The approximation is valid. Calculate pH:
9. $$pH = -\log_{10}(5.33 \times 10^{-3}) = 2.27$$

> **Exam tip:** AP FRQ graders require explicit 5% rule validation when you use the approximation method. Always write out the validation step to earn full credit, even if the approximation is obviously valid.

## Quadratic Solution for Non-Approximable Weak Acids

When the 5% rule fails (usually when the weak acid has a relatively large $K_a$, or is very dilute), you must solve the exact form of the $K_a$ expression. Starting from the original relationship:

$$K_a = \frac{x^2}{[HA]_0 - x}$$

Rearrange this into standard quadratic form $ax^2 + bx + c = 0$:

$$x^2 + K_a x - K_a [HA]_0 = 0$$

Here, $a = 1$, $b = K_a$, $c = -K_a [HA]_0$. Solve using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Only the positive root is physically meaningful, since concentration cannot be negative. This method gives an exact value of $x$ with no approximation error.

**Worked example:** Calculate the pH of a 0.15 M solution of chlorous acid, $K_a = 1.2 \times 10^{-2}$, at 25°C.

1. Substitute into the $K_a$ expression:
2. $$\frac{x^2}{0.15 - x} = 0.012$$
3. Rearrange to quadratic form:
4. $$x^2 + 0.012x - (0.012 \times 0.15) = 0 \implies x^2 + 0.012x - 0.0018 = 0$$
5. Identify coefficients: $a=1$, $b=0.012$, $c=-0.0018$.
6. Solve the quadratic:
7. $$x = \frac{-0.012 \pm \sqrt{(0.012)^2 - 4(1)(-0.0018)}}{2(1)} = \frac{-0.012 \pm 0.0854}{2}$$
8. Take the positive root:
9. $$x = \frac{0.0734}{2} = 0.0367\ M$$
10. Checking the 5% rule gives 24.5%, so approximation would introduce large error. Calculate pH:
11. $$pH = -\log(0.0367) = 1.43$$

> **Exam tip:** Double-check the sign of the constant term $c$ when writing the quadratic; it is always negative for weak acid dissociation, which guarantees one positive and one negative root.

## Percent Dissociation of Weak Acids

**Percent dissociation** — The percentage of the original weak acid that has dissociated at equilibrium, calculated as:

*Notation:* % dissociation

*Example:* $\% \ dissociation = \frac{x}{[HA]_0} \times 100\%$

A key conceptual relationship frequently tested on the AP exam: for the same weak acid at the same temperature, percent dissociation increases as the acid is diluted. This follows Le Chatelier's principle: adding water (diluting) reduces the concentration of all species, so equilibrium shifts right to produce more moles of dissolved ions, increasing the fraction of dissociated acid. Unlike strong acids (100% dissociation, 10x dilution increases pH by 1 unit), 10x dilution of a weak acid increases pH by less than 1 unit because of increased percent dissociation.

**Worked example:** A 0.20 M solution of acetic acid ($K_a = 1.8 \times 10^{-5}$) has a pH of 2.72. Calculate the percent dissociation, then calculate the percent dissociation when the solution is diluted to 0.020 M.

1. For 0.20 M solution:
2. $$x = [H_3O^+] = 10^{-2.72} = 1.91 \times 10^{-3}\ M$$
3. Calculate percent dissociation:
4. $$\frac{1.91 \times 10^{-3}}{0.20} \times 100\% = 0.95\%$$
5. For 0.020 M diluted solution, use the approximation:
6. $$x = \sqrt{K_a [HA]_0} = \sqrt{1.8 \times 10^{-5} \times 0.020} = 6.0 \times 10^{-4}\ M$$
7. Validate with 5% rule:
8. $$\frac{6.0 \times 10^{-4}}{0.020} \times 100\% = 3\% < 5\%$$
9. Approximation is valid. Percent dissociation for 0.020 M is 3%, which is triple the original value, matching the dilution rule.

> **Exam tip:** For conceptual MCQ questions asking how percent dissociation changes with dilution, you do not need to calculate: remember *more dilute = higher percent dissociation* to answer instantly.

## Common pitfalls

- **Wrong:** Using the $[HA]_0$ approximation without 5% validation, even when $x$ is more than 5% of $[HA]_0$
  - Why it fails: Students memorize the shortcut and forget to check if it applies, especially on time-pressured MCQ
  - Correct: Always calculate $\frac{x}{[HA]_0} \times 100\%$ after approximating; switch to quadratic if the result is over 5%
- **Wrong:** Equating $[H^+]$ to the initial concentration of weak acid, the same as strong acids
  - Why it fails: Students confuse strong vs weak acid behavior, especially for weak acids with large $K_a$ values
  - Correct: Always start with the $K_a$ equilibrium expression for any acid explicitly labeled "weak"
- **Wrong:** Using the negative root from the quadratic equation, leading to negative concentration and invalid negative pH
  - Why it fails: Students rush through calculation and forget concentration cannot be negative
  - Correct: Discard the negative root immediately after solving the quadratic; it has no physical meaning
- **Wrong:** Assuming percent dissociation stays constant when a weak acid is diluted
  - Why it fails: Students apply strong acid dilution rules (100% dissociation always) to weak acids
  - Correct: Recalculate $x$ for the new concentration, and remember percent dissociation always increases with dilution
- **Wrong:** Including liquid water in the $K_a$ expression, adding an extra $[H_2O]$ term to the denominator
  - Why it fails: Students confuse general equilibrium expressions with acid dissociation constants that omit pure solvents
  - Correct: Always omit liquid water from the $K_a$ expression for aqueous weak acid dissociation
- **Wrong:** Overcomplicating polyprotic weak acid pH by including $H^+$ from the second dissociation
  - Why it fails: Students forget that for most polyprotic acids, $K_{a1}$ is thousands of times larger than $K_{a2}$
  - Correct: Calculate pH only from the first dissociation for polyprotic weak acids, unless explicitly told to include subsequent steps

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| $K_a$ expression (monoprotic HA) | $K_a = \frac{[H_3O^+][A^-]}{[HA]}$ | Omit liquid water; $[H_3O^+] = [A^-] = x$ for pure weak acid |
| Approximation for $[H_3O^+]$ | $[H_3O^+] = \sqrt{K_a [HA]_0}$ | Only valid if 5% rule is satisfied |
| 5% Validation Rule | $\% = \frac{x}{[HA]_0} \times 100\% \leq 5\%$ | Explicit validation required for full credit on FRQ |
| Quadratic Equation (exact solution) | $x^2 + K_a x - K_a [HA]_0 = 0$ | Take only the positive root for $[H_3O^+]$ |
| Percent Dissociation | $\% \ dissociation = \frac{x}{[HA]_0} \times 100\%$ | Increases with dilution for the same weak acid |
| pH Calculation | $pH = -\log_{10}[H_3O^+]$ | Apply after finding equilibrium $[H_3O^+]$ |
| Dilution Rule | 10x dilution → pH increases by <1 unit | Strong acids increase pH by 1 unit for 10x dilution |
| Polyprotic Weak Acid pH | pH = calculated from first $K_a$ | Valid if $K_{a1} > 1000 \times K_{a2}$, which is almost always true |

## What's next

Mastery of weak acid pH calculations is the foundation for all subsequent acid-base equilibrium topics in AP Chemistry Unit 8. The same core methods: ICE tables, approximation, 5% validation, and quadratic solving apply directly to weak base pH calculations, just using $K_b$ instead of $K_a$. Weak acid pH skills are also required for all buffer pH calculations, acid-base titration curve analysis, and pH at equivalence point calculations, which make up a large share of Unit 8 exam points. Without a solid grasp of these methods, you will struggle to solve more complex acid-base problems.

- [pH of weak bases](https://www.owlsprep.com/study/ap-chemistry-u8-ph-of-weak-bases/)
- [Molecular structure of acids and bases](https://www.owlsprep.com/study/ap-chemistry-u8-molecular-structure-of-acids-and/)
- [pH and pKa](https://www.owlsprep.com/study/ap-chemistry-u8-ph-and-pka/)

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