pH of weak bases

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

1. Base Dissociation Constant ($K_b$) and $K_a$-$K_b$-$K_w$ Relationship★★☆☆☆⏱ 3 min

📘 Definition

Base dissociation constant

$K_{b}$

Equilibrium constant for the partial ionization of a weak base in water, measures base strength. A larger $K_{b}$ corresponds to a stronger base.

Example:

Ammonia, a common weak base, has $K_{b} = 1.8 \times 1 0^{- 5}$

When a weak base $B$ 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 ( $N H_{4}^{+}$ , conjugate acid of ammonia $N H_{3}$ ) is $5.6 \times 1 0^{- 10}$ 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 $p K_{a}$ instead of $K_{a}$ , subtract $p K_{a}$ from 14 to get $p K_{b}$ directly, saving time on multiple-choice questions.

2. pH Calculation for a Pure Weak Base Solution★★★☆☆⏱ 4 min

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 $[O H^{-}]$ , then convert to pH. If the initial base concentration is $c$ , the ICE table gives equilibrium concentrations: $[B] = c - x$ , $[B H^{+}] = x$ , $[O H^{-}] = x$ , where $x = [O H^{-}]$ . 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 $[O H^{-}]$ , calculate $pOH = - lo g [O H^{-}]$ , then $pH = 14 - pOH$ at 25°C.

📐 Worked Example

Calculate the pH of a 0.15 M solution of methylamine ( $C H_{3} N H_{2}$ ), where $K_{b} = 4.4 \times 1 0^{- 4}$ 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 $[C H_{3} N H_{2}] = 0.15$ 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: $\frac{8.1 \times 1 0 ^{- 3}}{0.15} \times 100% = 5.4%$ , 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.

3. pH of Basic Salts★★★☆☆⏱ 3 min

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 $3.5 \times 1 0^{- 8}$ at 25°C.

1
Complete dissociation of the salt: $N a O C l (s) \to N a^{+} (a q) + O C l^{-} (a q)$ , so $[O C l^{-}]_{ini t ia l} = 0.25$ M, and $N a^{+}$ is a spectator ion that can be ignored.
2
Write the base equilibrium for $O C l^{-}$ 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 $[O H^{-}]$ :
$[OH^-] = \sqrt{K_b \times c} = \sqrt{(2.9 \times 10^{-7})(0.25)} = 2.7 \times 10^{-4} \text{ M}$
4
Check the 5% rule: $\frac{2.7 \times 1 0 ^{- 4}}{0.25} \times 100% = 0.11% < 5%$ , 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.

4. Percent Ionization of Weak Bases★★☆☆☆⏱ 2 min

Percent ionization is the percentage of the initial weak base that has ionized to produce $O H^{-}$ 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 $[O H^{-}]$ 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 $x = [O H^{-}] = (\frac{percent}{100}) \times c$ , then plug $x$ and $c$ into $K_{b} = \frac{x ^{2}}{c - x}$ to solve directly for $K_{b}$ .

5. Common Pitfalls

Wrong move:

Using $[O H^{-}] = c$ (equal to initial base concentration) for weak bases, like you do for strong bases

Why:

Students confuse the 100% dissociation rule for strong bases with partial dissociation for weak bases, and skip the required equilibrium calculation

Correct move:

Always confirm if the base is weak or strong first; if weak, always use $K_{b}$ and ICE to calculate $[O H^{-}]$

Wrong move:

Solving directly for $[H^{+}]$ instead of $[O H^{-}]$ when setting up the equilibrium for weak bases

Why:

Students memorize weak acid pH calculation and replicate it incorrectly, leading to wrong exponents and a final pH that is far too low

Correct move:

For any weak base equilibrium, always set up the ICE table to solve for $[O H^{-}]$ first, then convert to pH via pOH

Wrong move:

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:

Students forget only salts from strong acid-strong base neutralization are neutral; conjugate bases of weak acids hydrolyze to produce $O H^{-}$

Correct move:

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 move:

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:

Students skip writing the full relationship and flip the fraction from memory

Correct move:

Always write the full relationship $K_{a} K_{b} = K_{w}$ first before rearranging, every time

Wrong move:

Including liquid water in the $K_{b}$ equilibrium expression

Why:

Students include all reactants out of habit, forgetting pure solvent activity is 1

Correct move:

Always omit pure liquid water from any $K_{a}$ or $K_{b}$ expression for aqueous equilibria

6. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
$K_{b}$ definition	$K_{b} = \frac{[ BH ^{+} ] [ O H ^{-} ]}{[ B ]}$	Water excluded; larger $K_{b}$ = stronger base
Conjugate pair relationship	$K_{a} \times K_{b} = 1.0 \times 1 0^{- 14}$ ; $p K_{a} + p K_{b} = 14$	Valid at 25°C only
Approximate $[O H^{-}]$	$[O H^{-}] \approx K_{b} \times c$	Valid if percent ionization < 5%; $c$ = initial base concentration
pH conversion	$pH = 14 - pOH$	Always solve for $[O H^{-}]$ first for weak bases
Percent ionization	$% ionization = \frac{[ O H ^{-} ] _{e q}}{[ B ] _{ini t ia l}} \times 100%$	Increases as weak base concentration decreases
5% rule	$\frac{[ O H ^{-} ]}{[ B ] _{ini t ia l}} \times 100% < 5%$	If >5%, solve quadratic for exact $[O H^{-}]$
Basic salt pH	Treat conjugate base anion as weak base; cation is spectator	Applies to salts of strong base + weak acid
Quadratic solution	$x = \frac{- K _{b} + K _{b}^{2} + 4 K _{b} c}{2}$	Use only the positive root for concentration

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 weak base solutions
2022 · FRQ
Calculate pH of basic salt

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.