A weak acid has K_a = 4.0 × 10^-7. Using the approximation, calculate [H_3O^+] in a 0.10 M solution. [2 points]

The K_a of an acid is 1.0 × 10^-4. Calculate the K_b of its conjugate base. [2 points]

United StatesChemistrySyllabus dot point

How do we calculate the pH and percent ionization of a weak acid or base using Ka or Kb?

Topic 8.3 Weak Acid and Base Equilibria: use Ka or Kb with an ICE table to calculate the pH and percent ionization of a weak acid or base, and relate Ka, Kb and Kw.

A focused answer to AP Chemistry Topic 8.3, covering the acid and base ionization constants Ka and Kb, ICE-table calculations of pH and percent ionization for weak acids and bases, and the relationship Ka times Kb equals Kw, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The ionization constants
Calculating pH with an ICE table
Percent ionization and the Ka-Kb-Kw relationship
Try this

What this topic is asking

The College Board (Topic 8.3) wants you to use the acid or base ionization constant ( $K_a$ or $K_b$ ) with an ICE table to calculate the pH and percent ionization of a weak acid or base, and to relate $K_a$ , $K_b$ and $K_w$ . Unlike strong acids, weak acids only partially ionize, so an equilibrium calculation is required.

The ionization constants

These are just equilibrium constants for the ionization reactions. A larger $K_a$ means a stronger acid (more ionized at equilibrium); a smaller $K_a$ means a weaker acid. Most weak acids on the exam have $K_a$ values around $10^{-5}$ , so they ionize only slightly.

Calculating pH with an ICE table

The procedure mirrors the equilibrium calculation of Topic 7.7, applied to acid ionization. Because $K_a$ is usually small, the approximation almost always holds, giving $x = \sqrt{K_a C_0}$ . For a weak base you do the same with $K_b$ to find $[\text{OH}^-]$ , then convert to pOH and pH.

Percent ionization and the Ka-Kb-Kw relationship

The percent ionization measures the fraction of the acid that has ionized:

\text{percent ionisation} = \frac{x}{C_0} \times 100\%

A more dilute weak acid has a higher percent ionization, even though its pH is higher (less concentrated). For a conjugate acid-base pair, the constants are linked by

K_a \times K_b = K_w = 1.0 \times 10^{-14} \text{ at } 25\ ^\circ\text{C}

so a stronger acid has a weaker conjugate base. Taking negative logarithms gives $\text{p}K_a + \text{p}K_b = 14.00$ .

pH of a weak base

Calculate the pH of a $0.100$ M solution of the weak base B, with $K_b = 1.8 \times 10^{-5}$ .

step 1 Set up the ICE expression

$\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-$ ; with $[\text{OH}^-] = [\text{BH}^+] = x$ and $[\text{B}] \approx 0.100$ , $K_b = \dfrac{x^2}{0.100}$ .

step 2 Solve for x with the approximation

$x^2 = (1.8 \times 10^{-5})(0.100) = 1.8 \times 10^{-6}$ , so $x = 1.3 \times 10^{-3}$ M $= [\text{OH}^-]$ .

step 3 Find the pOH

$\text{pOH} = -\log(1.3 \times 10^{-3}) = 2.87$ .

step 4 Convert to pH

$\text{pH} = 14.00 - 2.87 = 11.13$ , confirming a basic solution.

Try this

Q1. A weak acid has $K_a = 4.0 \times 10^{-7}$ . Using the approximation, calculate $[\text{H}_3\text{O}^+]$ in a $0.10$ M solution. [2 points]

Cue. $x = \sqrt{K_a C_0} = \sqrt{(4.0 \times 10^{-7})(0.10)} = 2.0 \times 10^{-4}$ M.

Q2. The $K_a$ of an acid is $1.0 \times 10^{-4}$ . Calculate the $K_b$ of its conjugate base. [2 points]

Cue. $K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-4}} = 1.0 \times 10^{-10}$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A

0.250

M solution of a weak acid HA has

K_a = 1.8 \times 10^{-5}

. (a) Write the ionization equation and

K_a

expression. (b) Using an ICE table and the small-

x

approximation, calculate

[\text{H}_3\text{O}^+]

. (c) Calculate the pH. (d) Calculate the percent ionization of the acid.

Show worked answer →

A 4-point quantitative FRQ on a weak acid.

(a) Equation and expression (1 point): $\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$ ; $K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$ .
(b) Hydronium (1 point): with $[\text{H}_3\text{O}^+] = [\text{A}^-] = x$ and $[\text{HA}] \approx 0.250$ , $K_a = \dfrac{x^2}{0.250} = 1.8 \times 10^{-5}$ , so $x^2 = 4.5 \times 10^{-6}$ and $x = 2.1 \times 10^{-3}$ M.
(c) pH (1 point): pH $= -\log(2.1 \times 10^{-3}) = 2.67$ .
(d) Percent ionization (1 point): $\dfrac{2.1 \times 10^{-3}}{0.250} \times 100\% = 0.85\%$ .

Markers reward the equation and $K_a$ expression, the hydronium concentration via the approximation, the pH, and the percent ionization.

AP 2021 (style)1 marksSection I (multiple choice). For a conjugate acid-base pair, the relationship between

K_a

and

K_b

25\ ^\circ\text{C}

is (A)

K_a = K_b

(B)

K_a + K_b = 14

(C)

K_a \times K_b = K_w = 1.0 \times 10^{-14}

(D)

K_a / K_b = 1

. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (C).

For a conjugate acid-base pair, $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ at $25\ ^\circ\text{C}$ . This means a stronger acid (larger $K_a$ ) has a weaker conjugate base (smaller $K_b$ ). The trap is (B): that relation is for pKa + pKb, not $K_a + K_b$ .

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)