# pH of weak bases

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

This module covers calculation of pH for weak base solutions for AP Chemistry, including the base dissociation constant $K_b$, the $K_a$-$K_b$-$K_w$ relationship, approximations, percent ionization, and pH of basic salt solutions.

**Prerequisites:** Autoionization of water and $K_w = [H^+][OH^-]$; pH and pOH definition and conversion; Equilibrium ICE table setup

## Learning objectives

- Calculate $K_b$ from $K_a$ of a conjugate acid
- Calculate pH of a pure weak base solution
- Calculate pH of basic salt solutions
- Calculate percent ionization of weak bases
- Apply the 5% rule for equilibrium approximations

## Base Dissociation Constant ($K_b$) and $K_a$-$K_b$-$K_w$ Relationship

**Base dissociation constant** — Equilibrium constant for the partial ionization of a weak base in water, measures base strength. A larger $K_b$ corresponds to a stronger base.

*Notation:* $K_b$

*Example:* Ammonia, a common weak base, has $K_b = 1.8 \times 10^{-5}$

When a weak base $\text{B}$ dissolves in water, it accepts a proton from water, following the equilibrium:

$$\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq)$$

Water is the pure solvent, so it is excluded from the equilibrium expression (activity = 1 for pure liquids). The $K_b$ expression is:

$$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$$

For any conjugate acid-base pair at 25°C, the product of the acid dissociation constant of the acid and the base dissociation constant of the conjugate base equals $K_w$, the autoionization constant of water:

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

$$\text{p}K_a + \text{p}K_b = 14 \quad (\text{at } 25^\circ C)$$

**Worked example:** The $K_a$ of the ammonium ion ($NH_4^+$, conjugate acid of ammonia $NH_3$) is $5.6 \times 10^{-10}$ at 25°C. Calculate $K_b$ for ammonia.

1. Recall the core relationship for conjugate pairs:

   $$K_a \times K_b = K_w = 1.0 \times 10^{-14}$$
2. Rearrange to isolate $K_b$:

   $$K_b = \frac{K_w}{K_a}$$
3. Substitute the given values:

   $$K_b = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-10}} = 1.8 \times 10^{-5}$$
4. Check for reasonableness: Ammonia is a weak base, so $K_b < 1$, which matches our result.

> **Exam tip:** If a problem gives $\text{p}K_a$ instead of $K_a$, subtract $\text{p}K_a$ from 14 to get $\text{p}K_b$ directly, saving time on multiple-choice questions.

## pH Calculation for a Pure Weak Base Solution

To find the pH of a solution of a pure weak base with known initial concentration and $K_b$, we use an ICE (Initial, Change, Equilibrium) table to find equilibrium $[OH^-]$, then convert to pH. If the initial base concentration is $c$, the ICE table gives equilibrium concentrations: $[B] = c - x$, $[BH^+] = x$, $[OH^-] = x$, where $x = [OH^-]$. Substituting into the $K_b$ expression gives:

$$K_b = \frac{x^2}{c - x}$$

Because $K_b$ is very small for weak bases, $x << c$, so we can approximate $c - x \approx c$, simplifying the expression to:

$$x = [OH^-] \approx \sqrt{K_b \times c}$$

After calculating $x$, we check the 5% rule: if $\frac{x}{c} \times 100\% < 5\%$, the approximation is valid. If not, we solve the quadratic equation for the exact value of $x$. Once we have $[OH^-]$, calculate $\text{pOH} = -\log[OH^-]$, then $\text{pH} = 14 - \text{pOH}$ at 25°C.

**Worked example:** Calculate the pH of a 0.15 M solution of methylamine ($CH_3NH_2$), where $K_b = 4.4 \times 10^{-4}$ at 25°C.

1. Write the equilibrium reaction:

   $$CH_3NH_2(aq) + H_2O(l) \rightleftharpoons CH_3NH_3^+(aq) + OH^-(aq)$$
2. Set up the ICE table: initial $[CH_3NH_2] = 0.15$ M, all other starting concentrations = 0; change: $-x$, $+x$, $+x$; equilibrium: $0.15 - x$, $x$, $x$.
3. Apply the approximation:

   $$x = \sqrt{K_b \times c} = \sqrt{(4.4 \times 10^{-4})(0.15)} = 8.1 \times 10^{-3} \text{ M}$$
4. Check the 5% rule: $\frac{8.1 \times 10^{-3}}{0.15} \times 100\% = 5.4\%$, which is just over 5%, so we solve the quadratic:

   $$x^2 + 4.4 \times 10^{-4}x - 6.6 \times 10^{-5} = 0, \text{ giving } x = 7.9 \times 10^{-3} \text{ M}$$
5. Calculate final pH:

   $$\text{pOH} = -\log(7.9 \times 10^{-3}) = 2.10, \quad \text{pH} = 14 - 2.10 = 11.9$$

> **Exam tip:** AP exam graders accept answers within 0.1 pH unit of the correct value, even if you use the approximation when percent ionization is 5-6%, but always explicitly state whether your approximation is valid to earn full points on FRQ.

## pH of Basic Salts

Basic salts are ionic compounds formed from the neutralization of a strong base and a weak acid. They dissolve completely in water to release a spectator cation (from the strong base, which does not react with water) and an anion (the conjugate base of the weak acid, which acts as a weak base in solution). We calculate pH for basic salts exactly the same way as for any other weak base.

**Worked example:** Calculate the pH of a 0.25 M solution of sodium hypochlorite (NaOCl). The $K_a$ of hypochlorous acid (HOCl) is $3.5 \times 10^{-8}$ at 25°C.

1. Complete dissociation of the salt: $NaOCl(s) \rightarrow Na^+(aq) + OCl^-(aq)$, so $[OCl^-]_{initial} = 0.25$ M, and $Na^+$ is a spectator ion that can be ignored.
2. Write the base equilibrium for $OCl^-$ and calculate $K_b$:

   $$OCl^-(aq) + H_2O(l) \rightleftharpoons HOCl(aq) + OH^-(aq), \quad K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-8}} = 2.9 \times 10^{-7}$$
3. Approximate $[OH^-]$:

   $$[OH^-] = \sqrt{K_b \times c} = \sqrt{(2.9 \times 10^{-7})(0.25)} = 2.7 \times 10^{-4} \text{ M}$$
4. Check the 5% rule: $\frac{2.7 \times 10^{-4}}{0.25} \times 100\% = 0.11\% < 5\%$, so the approximation is valid.
5. Calculate final pH:

   $$\text{pOH} = -\log(2.7 \times 10^{-4}) = 3.57, \quad \text{pH} = 14 - 3.57 = 10.4$$

> **Exam tip:** Always identify spectator ions first when solving basic salt pH problems: all group 1 and heavy group 2 metal cations from strong bases do not affect pH, so you only need to focus on the conjugate base anion.

## Percent Ionization of Weak Bases

Percent ionization is the percentage of the initial weak base that has ionized to produce $OH^-$ at equilibrium. It is calculated as:

$$\text{Percent ionization} = \frac{[OH^-]_{equilibrium}}{[B]_{initial}} \times 100\%$$

Percent ionization correlates with both base strength and solution dilution. For a given weak base, percent ionization increases as the solution becomes more dilute. This follows Le Chatelier's principle: increasing the volume (diluting) shifts equilibrium toward the side with more moles of solute (1 mole of base produces 2 moles of ions), so more base ionizes.

**Worked example:** A 0.10 M solution of an unknown weak base has a pH of 10.5 at 25°C. Calculate the percent ionization of the base.

1. Calculate pOH from pH:

   $$\text{pOH} = 14 - 10.5 = 3.5$$
2. Calculate $[OH^-]$ from pOH:

   $$[OH^-] = 10^{-\text{pOH}} = 10^{-3.5} = 3.2 \times 10^{-4} \text{ M}$$
3. Substitute into the percent ionization formula:

   $$\text{Percent ionization} = \frac{3.2 \times 10^{-4}}{0.10} \times 100\% = 0.32\%$$
4. Check reasonableness: A percent ionization of 0.32% is well below 5%, which confirms the base is weak, matching the problem description.

> **Exam tip:** If you are asked to calculate $K_b$ from percent ionization, rearrange the formula to get $x = [OH^-] = \left(\frac{\text{percent}}{100}\right) \times c$, then plug $x$ and $c$ into $K_b = \frac{x^2}{c-x}$ to solve directly for $K_b$.

## Common pitfalls

- **Wrong:** Using $[OH^-] = c$ (equal to initial base concentration) for weak bases, like you do for strong bases
  - Why it fails: Students confuse the 100% dissociation rule for strong bases with partial dissociation for weak bases, and skip the required equilibrium calculation
  - Correct: Always confirm if the base is weak or strong first; if weak, always use $K_b$ and ICE to calculate $[OH^-]$
- **Wrong:** Solving directly for $[H^+]$ instead of $[OH^-]$ when setting up the equilibrium for weak bases
  - Why it fails: Students memorize weak acid pH calculation and replicate it incorrectly, leading to wrong exponents and a final pH that is far too low
  - Correct: For any weak base equilibrium, always set up the ICE table to solve for $[OH^-]$ first, then convert to pH via pOH
- **Wrong:** Forgetting that the anion of a weak acid acts as a weak base when calculating pH of basic salts, and assuming the salt is neutral
  - Why it fails: Students forget only salts from strong acid-strong base neutralization are neutral; conjugate bases of weak acids hydrolyze to produce $OH^-$
  - Correct: For any salt, split into cation and anion; if the anion is the conjugate base of a weak acid, treat it as a weak base for pH calculation
- **Wrong:** Misremembering the $K_a$-$K_b$ relationship, and using $K_b = \frac{K_a}{K_w}$ instead of $K_b = \frac{K_w}{K_a}$
  - Why it fails: Students skip writing the full relationship and flip the fraction from memory
  - Correct: Always write the full relationship $K_a K_b = K_w$ first before rearranging, every time
- **Wrong:** Including liquid water in the $K_b$ equilibrium expression
  - Why it fails: Students include all reactants out of habit, forgetting pure solvent activity is 1
  - Correct: Always omit pure liquid water from any $K_a$ or $K_b$ expression for aqueous equilibria

## Cheatsheet

| Category | Formula/Rule | Notes |
| --- | --- | --- |
| $K_b$ definition | $K_b = \frac{[\text{BH}^+][OH^-]}{[\text{B}]}$ | Water excluded; larger $K_b$ = stronger base |
| Conjugate pair relationship | $K_a \times K_b = 1.0 \times 10^{-14}$; $\text{p}K_a + \text{p}K_b = 14$ | Valid at 25°C only |
| Approximate $[OH^-]$ | $[OH^-] \approx \sqrt{K_b \times c}$ | Valid if percent ionization < 5%; $c$ = initial base concentration |
| pH conversion | $\text{pH} = 14 - \text{pOH}$ | Always solve for $[OH^-]$ first for weak bases |
| Percent ionization | $\% \text{ionization} = \frac{[OH^-]_{eq}}{[B]_{initial}} \times 100\%$ | Increases as weak base concentration decreases |
| 5% rule | $\frac{[OH^-]}{[B]_{initial}} \times 100\% < 5\%$ | If >5%, solve quadratic for exact $[OH^-]$ |
| Basic salt pH | Treat conjugate base anion as weak base; cation is spectator | Applies to salts of strong base + weak acid |
| Quadratic solution | $x = \frac{-K_b + \sqrt{K_b^2 + 4 K_b c}}{2}$ | Use only the positive root for concentration |

## What's next

Mastering pH of weak bases is a critical foundation for the remaining topics in AP Chemistry Unit 8: Acids and Bases, and supports key equilibrium concepts from earlier units. When calculating pH at the equivalence point of a strong acid-weak base titration, you will rely on the $K_a$-$K_b$ relationship from this module to find the pH of the conjugate acid product. For buffer solutions made from a weak base and its conjugate salt, you will use $K_b$ directly to calculate buffer pH, so mastering weak base pH calculation is non-negotiable for these topics. Beyond Unit 8, this topic supports solubility equilibria, where the pH of the solution changes the solubility of ionic compounds with basic anions.

- [pH of weak acids](https://www.owlsprep.com/study/ap-chemistry-u8-ph-of-weak-acids/)
- [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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