pH of weak acids

AP Chemistry· AP Chemistry CED — Acids and Bases· 14 min read

1. Core Concepts of Weak Acid Dissociation★★☆☆☆⏱ 2 min

Unlike strong acids that dissociate completely in dilute solution, weak acids only partially dissociate, so equilibrium $[H_{3} O^{+}]$ 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: $[H A]_{0}$ = 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, $[H_{3} O^{+}]_{e q} = [A^{-}]_{e q} = x$ . 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.

2. Acid Dissociation Constant ($K_a$) Expression★★☆☆☆⏱ 3 min

For any monoprotic weak acid $H A$ , 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: $[H_{3} O^{+}]_{e q} = [A^{-}]_{e q} = x$ , and $[H A]_{e q} = [H A]_{0} - x$ . $K_{a}$ 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 $[H_{3} O^{+}]$ :
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 $[H A]$ :
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 $H A$ , not just the initial concentration. Only simplify to initial concentration after confirming $x$ is negligible.

3. Approximation Method and 5% Validation Rule★★★☆☆⏱ 4 min

Most weak acids have very small $K_{a}$ values, so $x = [H_{3} O^{+}]$ is much smaller than $[H A]_{0}$ . This means $[H A]_{0} - x \approx [H A]_{0}$ , 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 $\frac{x}{[ H A ] _{0}} \times 100% \leq 5%$ , 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, $K_{a} = 6.3 \times 1 0^{- 5}$ , at 25°C.

1
Set up the ICE table: Initial $[H A] = 0.45 M$ , $[H_{3} O^{+}] = [A^{-}] = 0$ . Change: $- x, + x, + x$ . Equilibrium: $[H A] = 0.45 - x$ , $[H_{3} O^{+}] = [A^{-}] = x$ .
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.

4. Quadratic Solution for Non-Approximable Weak Acids★★★★☆⏱ 3 min

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 $a x^{2} + b x + c = 0$ :

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

Here, $a = 1$ , $b = K_{a}$ , $c = - K_{a} [H A]_{0}$ . 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, $K_{a} = 1.2 \times 1 0^{- 2}$ , 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.

5. Percent Dissociation of Weak Acids★★★☆☆⏱ 3 min

📘 Definition

Percent dissociation

% dissociation

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

Example:

$% d i ssoc ia t i o n = \frac{x}{[ H A ] _{0}} \times 100%$

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 ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) 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.

6. Common Pitfalls

Wrong move:

Using the $[H A]_{0}$ approximation without 5% validation, even when $x$ is more than 5% of $[H A]_{0}$

Why:

Students memorize the shortcut and forget to check if it applies, especially on time-pressured MCQ

Correct move:

Always calculate $\frac{x}{[ H A ] _{0}} \times 100%$ after approximating; switch to quadratic if the result is over 5%

Wrong move:

Equating $[H^{+}]$ to the initial concentration of weak acid, the same as strong acids

Why:

Students confuse strong vs weak acid behavior, especially for weak acids with large $K_{a}$ values

Correct move:

Always start with the $K_{a}$ equilibrium expression for any acid explicitly labeled "weak"

Wrong move:

Using the negative root from the quadratic equation, leading to negative concentration and invalid negative pH

Why:

Students rush through calculation and forget concentration cannot be negative

Correct move:

Discard the negative root immediately after solving the quadratic; it has no physical meaning

Wrong move:

Assuming percent dissociation stays constant when a weak acid is diluted

Why:

Students apply strong acid dilution rules (100% dissociation always) to weak acids

Correct move:

Recalculate $x$ for the new concentration, and remember percent dissociation always increases with dilution

Wrong move:

Including liquid water in the $K_{a}$ expression, adding an extra $[H_{2} O]$ term to the denominator

Why:

Students confuse general equilibrium expressions with acid dissociation constants that omit pure solvents

Correct move:

Always omit liquid water from the $K_{a}$ expression for aqueous weak acid dissociation

Wrong move:

Overcomplicating polyprotic weak acid pH by including $H^{+}$ from the second dissociation

Why:

Students forget that for most polyprotic acids, $K_{a 1}$ is thousands of times larger than $K_{a 2}$

Correct move:

Calculate pH only from the first dissociation for polyprotic weak acids, unless explicitly told to include subsequent steps

7. Quick Reference Cheatsheet

Category	Formula	Notes
$K_{a}$ expression (monoprotic HA)	$K_{a} = \frac{[ H _{3} O ^{+} ] [ A ^{-} ]}{[ H A ]}$	Omit liquid water; $[H_{3} O^{+}] = [A^{-}] = x$ for pure weak acid
Approximation for $[H_{3} O^{+}]$	$[H_{3} O^{+}] = K_{a} [H A]_{0}$	Only valid if 5% rule is satisfied
5% Validation Rule	$% = \frac{x}{[ H A ] _{0}} \times 100% \leq 5%$	Explicit validation required for full credit on FRQ
Quadratic Equation (exact solution)	$x^{2} + K_{a} x - K_{a} [H A]_{0} = 0$	Take only the positive root for $[H_{3} O^{+}]$
Percent Dissociation	$% d i ssoc ia t i o n = \frac{x}{[ H A ] _{0}} \times 100%$	Increases with dilution for the same weak acid
pH Calculation	$p H = - lo g_{10} [H_{3} O^{+}]$	Apply after finding equilibrium $[H_{3} O^{+}]$
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 $K_{a 1} > 1000 \times K_{a 2}$ , which is almost always true

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Compare pH of equal concentration acids
2022 · FRQ
Calculate pH of weak acid, validate approximation
2021 · MCQ
Percent dissociation change with dilution

Going deeper

SyllabusOfficial AP Chemistry CED Unit 8College Board official syllabus

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.