pH of weak acids — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Dissociation equilibrium of monoprotic weak acids, acid dissociation constant $K_a$, approximation method for $[H_3{O}^{+}]$, 5% validation rule, quadratic solution for non-approximable cases, and percent dissociation calculations for varying concentrations.

You should already know: Definition of pH and pOH. Equilibrium constant expressions for homogeneous reactions. The difference between strong and weak acid dissociation.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is pH of weak acids?

Unlike strong acids, which dissociate completely in dilute aqueous solution, weak acids only partially dissociate, so equilibrium $[H_3{O}^{+}]$ cannot be directly equated to the initial weak acid concentration. This requires equilibrium-based calculations to find pH, which is one of the core skills tested in AP Chemistry Unit 8: Acids and Bases. 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, often as part of larger equilibrium or titration questions.

Standard notation conventions used on the AP exam: $[HA{]}_0$ = initial concentration of monoprotic weak acid before dissociation, $K_a$ = acid dissociation constant, $x$ = moles per liter of HA that dissociate at equilibrium, so $[H_3{O}^{+}{]}_{eq}=[A^{-}{]}_{eq}=x$ for pure weak acid solutions. A key conceptual point always tested: 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

For any monoprotic weak acid $HA$, dissociation in water follows the equilibrium: $$HA(aq)+H_{2}O(l)\rightleftharpoons H_3{O}^{+}(aq)+A^{-}(aq)$$ Liquid water is omitted from the equilibrium expression because its concentration is nearly constant in dilute solutions, so it is absorbed into the equilibrium constant. The acid dissociation constant $K_a$ is defined as: $$K_a=\frac{[H_3{O}^{+}{]}_{eq}[A^{-}{]}_{eq}}{[HA{]}_{eq}}$$ From the stoichiometry of dissociation, for a solution of pure weak acid, $[H_3{O}^{+}{]}_{eq}=[A^{-}{]}_{eq}=x$, and $[HA{]}_{eq}=[HA{]}_0-x$. This relationship is derived directly from ICE (Initial-Change-Equilibrium) tables, the standard tool for equilibrium problems you already know. $K_a$ values are always small for weak acids (typically $<1$), 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 $[H_3{O}^{+}]$: $[H_3{O}^{+}]=10^{-pH}=10^{-2.87}=1.35\times 10^{-3}M$. This equals $x=[A^{-}]$, from dissociation stoichiometry.
2. Calculate equilibrium $[HA]$: $[HA{]}_{eq}=[HA{]}_0-x=0.12-0.00135=0.11865M$.
3. Substitute into the $K_a$ expression: $$K_a=\frac{(1.35\times 10^{-3})(1.35\times 10^{-3})}{0.11865}\approx 1.5\times 10^{-5}$$
4. Confirm $K_a$ is unitless (standard for dilute solutions, per AP convention).

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.

3. Approximation Method and 5% Validation Rule

Most weak acids have very small $K_a$ values, so $x=[H_3{O}^{+}]$ is much smaller than $[HA{]}_0$. This means $[HA{]}_0-x\approx [HA{]}_0$, which simplifies the $K_a$ expression to: $$K_a\approx \frac{x^2}{[HA{]}_0}⟹x=[H_3{O}^{+}]=\sqrt{K_a\times [HA{]}_0}$$ This approximation drastically reduces calculation time, which is valuable for both MCQ and FRQ sections. However, the approximation is only valid if the error introduced by ignoring $x$ is small enough to be acceptable. The AP Chemistry standard for validation is the 5% rule: if $\frac{x}{[HA{]}_0}\times 100\%\le 5\%$, the approximation is acceptable. If the percent is 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 $[HA]=0.45M$, $[H_3{O}^{+}]=[A^{-}]=0$. Change: $-x,+x,+x$. Equilibrium: $[HA]=0.45-x$, $[H_3{O}^{+}]=[A^{-}]=x$.
2. Apply the approximation: assume $0.45-x\approx 0.45$, so $K_a=\frac{x^2}{0.45}=6.3\times 10^{-5}$.
3. Solve for $x$: $x^2=0.45\times 6.3\times 10^{-5}=2.835\times 10^{-5}$, so $x=5.33\times 10^{-3}M$.
4. Validate with the 5% rule: $\frac{5.33\times 10^{-3}}{0.45}\times 100\%=1.18\%<5\%$, so the approximation is valid.
5. Calculate pH: $pH=-{\log }_{10}(5.33\times 10^{-3})=2.27$.

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

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 $a{x}^2+bx+c=0$: $$x^2+K_{a}x-K_a[HA{]}_0=0$$ 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: $\frac{x^2}{0.15-x}=0.012$.
2. Rearrange to quadratic form: $x^2+0.012x-(0.012\times 0.15)=0⟹x^2+0.012x-0.0018=0$.
3. Identify coefficients: $a=1$, $b=0.012$, $c=-0.0018$.
4. Solve the quadratic: $$x=\frac{-0.012\pm \sqrt{(0.012{)}^2-4(1)(-0.0018)}}{2(1)}=\frac{-0.012\pm 0.0854}{2}$$
5. Take the positive root: $x=\frac{0.0734}{2}=0.0367M$. Check 5% rule: $\frac{0.0367}{0.15}\times 100\%=24.5\%$, so approximation would have given a large error.
6. Calculate pH: $pH=-\log (0.0367)=1.43$.

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

Percent dissociation (or percent ionization) is the percentage of the original weak acid that has dissociated at equilibrium, defined as: $$\%dissociation=\frac{[HA{]}_{dissociated}}{[HA{]}_0}\times 100\%=\frac{x}{[HA{]}_0}\times 100\%$$ 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 the equilibrium shifts right to produce more moles of dissolved ions, increasing the fraction of dissociated acid. Unlike strong acids (which always dissociate 100%, so 10x dilution increases pH by 1 unit), 10x dilution of a weak acid increases pH by less than 1 unit, because of the 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: $x=[H_3{O}^{+}]=10^{-2.72}=1.91\times 10^{-3}M$. Percent dissociation: $\frac{1.91\times 10^{-3}}{0.20}\times 100\%=0.95\%$.
2. For 0.020 M: Use approximation $x=\sqrt{K_a[HA{]}_0}=\sqrt{1.8\times 10^{-5}\times 0.020}=6.0\times 10^{-4}M$.
3. Validate 5% rule: $\frac{6.0\times 10^{-4}}{0.020}\times 100\%=3\%<5\%$, so approximation is valid.
4. 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 (and how to avoid them)

• Wrong move: Using the $[HA{]}_0$ approximation without 5% validation, even when $x$ is more than 5% of $[HA{]}_0$. Why: Students memorize the shortcut and forget to check if it applies, especially on time-pressured MCQ. Correct move: Always calculate $\frac{x}{[HA{]}_0}\times 100\%$ 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 a negative concentration and 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, $K_{a1}$ is thousands of times larger than $K_{a2}$. Correct move: Calculate pH only from the first dissociation for polyprotic weak acids, unless explicitly told to include subsequent steps.

7. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

All of the following solutions are 0.10 M at 25°C. Which solution has the highest pH? $K_a$ values: hydrocyanic acid (HCN) $=6.2\times 10^{-10}$, formic acid $=1.8\times 10^{-4}$, lactic acid $=1.4\times 10^{-4}$, nitrous acid $=4.0\times 10^{-4}$. A) HCN B) Formic acid C) Lactic acid D) Nitrous acid

Worked Solution: Higher pH corresponds to lower $[H_3{O}^{+}]$, which for equal concentrations of weak acids corresponds to a smaller $K_a$ (weaker acid). Ranking $K_a$ from smallest to largest: HCN $(6.2\times 10^{-10})<$ lactic acid $(1.4\times 10^{-4})<$ formic acid $(1.8\times 10^{-4})<$ nitrous acid $(4.0\times 10^{-4})$. The smallest $K_a$ gives the lowest $[H_3{O}^{+}]$ and highest pH. The correct answer is A.

Question 2 (Free Response)

Propanoic acid is a monoprotic weak acid with $K_a=1.5\times 10^{-5}$ at 25°C. (a) Calculate the pH of a 0.25 M solution of propanoic acid. Validate any approximations you use. (b) Calculate the percent dissociation of propanoic acid in this solution. (c) If the solution is diluted by a factor of 10 to 0.025 M, will the percent dissociation increase, decrease, or stay the same? Justify your answer using Le Chatelier's principle.

Worked Solution: (a) Set up equilibrium: $K_a=\frac{x^2}{0.25-x}\approx \frac{x^2}{0.25}=1.5\times 10^{-5}$. Solve for $x$: $x^2=0.25\times 1.5\times 10^{-5}=3.75\times 10^{-6}$, so $x=1.94\times 10^{-3}M$. Validate: $\frac{1.94\times 10^{-3}}{0.25}\times 100\%=0.78\%<5\%$, approximation is valid. Calculate pH: $pH=-\log (1.94\times 10^{-3})=2.71$. (b) Percent dissociation is already calculated in validation: $\boxed{0.78\%}$. (c) Percent dissociation will increase. Dilution reduces the concentration of all species by the same factor, so $Q<K$, shifting the dissociation equilibrium $HA(aq)\rightleftharpoons H^{+}(aq)+A^{-}(aq)$ to the right to re-establish equilibrium. This increases the fraction of dissociated propanoic acid.

Question 3 (Application / Real-World Style)

Citric acid is a triprotic weak acid that gives citrus fruit its sour taste. The first dissociation constant of citric acid is $K_{a1}=7.4\times 10^{-4}$, and $K_{a2}$ and $K_{a3}$ are more than 1000x smaller than $K_{a1}$. A fresh squeezed lemon juice has an initial citric acid concentration of 0.30 M. Calculate the pH of lemon juice, and compare its acidity to pure water (pH 7.0) and 0.1 M HCl (pH 1.0).

Worked Solution:

1. Use approximation and ignore second and third dissociations per the problem statement: $x=\sqrt{K_{a1}[HA{]}_0}=\sqrt{7.4\times 10^{-4}\times 0.30}=\sqrt{2.22\times 10^{-4}}=0.0149M$.
2. Validate 5% rule: $\frac{0.0149}{0.30}\times 100\%=5.0\%$, which meets the AP 5% rule threshold for approximation.
3. Calculate pH: $pH=-\log (0.0149)=1.83$. In context: Lemon juice is far more acidic than pure water, but slightly less acidic than 0.1 M strong hydrochloric acid, matching its role as a mild sour acid in food.

8. Quick Reference Cheatsheet

9. What's Next

This chapter is the foundation for all subsequent acid-base equilibrium topics in AP Chemistry Unit 8. Immediately next, you will apply the same approximation, 5% rule, and quadratic solution techniques to weak base pH calculations, using $K_b$ instead of $K_a$. Mastery of weak acid pH is also required for all buffer calculations, acid-base titration curve analysis, and pH at equivalence point calculations, which make up a large share of Unit 8 exam points. Without mastering the methods in this chapter, you will not be able to solve buffer pH problems or calculate pH before the equivalence point in a weak acid-strong base titration. Follow-on topics to study next: pH of weak bases Buffer pH calculations Acid-base titration curves Acid strength trends