Grade by Grade
E. Chemical Systems and Equilibrium · Acid-Base and Solubility Equilibria

Weak Bases, Kb, and the Ka × Kb = Kw Link

E2.4 E3.7 E3.4 E3.5About 70 minutes
Learning goals
  • I can use an ICE table and Kb to calculate [OH⁻], pOH and pH of a weak base.
  • I can calculate Kb from the pH of a weak base solution.
  • I can derive and use Ka × Kb = Kw (and pKa + pKb = 14.00) for a conjugate pair.
  • I can find Kb for the conjugate base of any acid in the Ka table.

Words to know

weak base
A base that accepts protons from water only partly: B + H₂O ⇌ BH⁺ + OH⁻.
amine
An organic compound related to ammonia in which one or more hydrogens are replaced by carbon groups (CH₃NH₂, methylamine). Amines are weak bases.
pKb
pKb = −log Kb. A smaller pKb means a stronger base.
Ka × Kb = Kw
For any conjugate acid-base pair in water at 25 °C, the product of the acid's Ka and the base's Kb is 1.0 × 10⁻¹⁴.

The smell of fish, the sting of window cleaner

The fishy smell of old fish comes partly from amines, molecules related to ammonia. Ammonia-based glass cleaners and the amines in fish share something: they are weak bases. They pull protons off water molecules, but only a small fraction of them do. A squeeze of lemon on fish is an acid-base reaction: the acid converts the volatile amines into ions, which stay in the food instead of reaching your nose.

In this lesson you will calculate the pH of weak bases with Kb, just as you did for weak acids with Ka. Then you will discover a simple link between the two constants that lets you find Kb for any base whose conjugate acid is in your Ka table.

Weak-base calculations

For a weak base B with initial concentration C:

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq) Kb = [BH⁺][OH⁻] ÷ [B]

The ICE table is the same shape as for a weak acid, but x is now [OH⁻]:

  • Equilibrium: [B] = C − x, [BH⁺] = x, [OH⁻] = x
  • Kb = x² ÷ (C − x) ≈ x² ÷ C, so x = √(Kb × C) when the 5% check passes.

The last step is the one students forget: x gives pOH, and you must convert to pH with pH = 14.00 − pOH.

Worked example

Example 1: pH of household-strength ammonia

Find the pH and percentage ionization of 0.10 mol/L NH₃(aq). Kb = 1.8 × 10⁻⁵.

  1. NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. ICE equilibrium: [NH₃] = 0.10 − x, [NH₄⁺] = [OH⁻] = x.
  2. C ÷ Kb ≈ 5600 ≥ 500, so x = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ mol/L = [OH⁻].
  3. Check: 1.34 × 10⁻³ ÷ 0.10 × 100% = 1.3% ✓
  4. pOH = −log(1.34 × 10⁻³) = 2.87; pH = 14.00 − 2.87 = 11.13.

pH = 11.13, 1.3% ionized. (The numbers mirror Example 1 of the weak-acid lesson because Kb of NH₃ equals Ka of acetic acid.)

Worked example

Example 2: Kb from a measured pH

A 0.15 mol/L solution of an amine has a pH of 11.80. Find Kb.

  1. pOH = 14.00 − 11.80 = 2.20, so [OH⁻] = 10⁻²·²⁰ = 6.31 × 10⁻³ mol/L = x.
  2. [BH⁺] = 6.31 × 10⁻³ mol/L, and [B] = 0.15 − 0.00631 = 0.1437 mol/L.
  3. Kb = (6.31 × 10⁻³)² ÷ 0.1437 = 2.8 × 10⁻⁴.

Kb = 2.8 × 10⁻⁴, a fairly strong weak base (about 15 times stronger than ammonia).

Ka × Kb = Kw

Write the Ka reaction for an acid (NH₄⁺) and the Kb reaction for its conjugate base (NH₃), then add them. From the lesson Kp, Combining Equilibrium Constants and the K Family (unit Chemical Equilibrium): when you add two equations, you multiply their equilibrium constants. Everything cancels except water's autoionization.

KaNH₄⁺+H₂ONH₃+H₃O⁺KbNH₃+H₂ONH₄⁺+OH⁻Kw2 H₂OH₃O⁺+OH⁻Add the equations: NH₄⁺ and NH₃ cancel.Adding equations multiplies their K values: Ka × Kb = Kw
Adding the acid reaction of NH₄⁺ and the base reaction of NH₃ leaves only the autoionization of water. So Ka(NH₄⁺) × Kb(NH₃) = Kw.

You can check it with the expressions themselves:

Ka × Kb = ([NH₃][H₃O⁺] ÷ [NH₄⁺]) × ([NH₄⁺][OH⁻] ÷ [NH₃]) = [H₃O⁺][OH⁻] = Kw

So for any conjugate pair at 25 °C:

Ka × Kb = Kw = 1.0 × 10⁻¹⁴ and pKa + pKb = 14.00

This is the exact version of the rule from the conjugate-pairs lesson: a large Ka forces a small Kb. It also means you need only one table. To find Kb of F⁻, look up Ka of HF and divide: Kb = Kw ÷ Ka.

HF / F⁻pKa 3.18pKb 10.82CH₃COOH / CH₃COO⁻pKa 4.74pKb 9.26NH₄⁺ / NH₃pKa 9.25pKb 4.75HCO₃⁻ / CO₃²⁻pKa 10.33pKb 3.6702468101214Every bar has total length 14: pKa + pKb = 14.00
For each conjugate pair, pKa and pKb add up to 14.00. A strong acid (small pKa, short left bar) has a weak conjugate base (large pKb, long right bar).
Worked example

Example 3: Kb of a conjugate base

Find Kb for the fluoride ion, F⁻, and for the carbonate ion, CO₃²⁻. Which is the stronger base?

  1. F⁻ is the conjugate base of HF (Ka = 6.6 × 10⁻⁴). Kb = 1.0 × 10⁻¹⁴ ÷ 6.6 × 10⁻⁴ = 1.5 × 10⁻¹¹.
  2. CO₃²⁻ is the conjugate base of HCO₃⁻ (Ka = 4.7 × 10⁻¹¹), not of H₂CO₃. Kb = 1.0 × 10⁻¹⁴ ÷ 4.7 × 10⁻¹¹ = 2.1 × 10⁻⁴.
  3. Compare: 2.1 × 10⁻⁴ is far larger than 1.5 × 10⁻¹¹.

Kb(F⁻) = 1.5 × 10⁻¹¹; Kb(CO₃²⁻) = 2.1 × 10⁻⁴. Carbonate is the much stronger base, which is why washing soda solutions are strongly basic.

Worked example

Example 4: pH of a solution of an anion

Find the pH of 0.10 mol/L sodium hypochlorite, NaClO (the active ingredient in household bleach, which is several times more concentrated than this). Ka(HClO) = 2.9 × 10⁻⁸.

  1. Na⁺ does not affect pH. ClO⁻ is a weak base: ClO⁻ + H₂O ⇌ HClO + OH⁻.
  2. Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 2.9 × 10⁻⁸ = 3.4 × 10⁻⁷.
  3. x = √(3.4 × 10⁻⁷ × 0.10) = 1.86 × 10⁻⁴ mol/L = [OH⁻]. Check: 0.19% ✓
  4. pOH = 3.73; pH = 14.00 − 3.73 = 10.27.

pH ≈ 10.27. (Commercial bleach is more basic than this because manufacturers add some sodium hydroxide. Never mix bleach with acids or ammonia: dangerous gases form.)

Just for fun

Ammonium and ammonia are the ideal couple: whatever one gives up, the other makes up for. Their Ka and Kb always multiply to exactly Kw, no matter how much they argue about who has the proton.

Hands-on activity

Antacid investigation: how much base is in a tablet?

You need: red cabbage indicator, white vinegar, 2 or 3 calcium carbonate antacid tablets (check the label: "calcium carbonate"), a spoon to crush them, a clear glass, a teaspoon and a tablespoon.

Grown-up note: an adult should supervise. The tablets are medicine: use them only for the experiment, store the rest safely, and do not eat any. Do not drink the mixtures.

  1. Put 1 tablespoon of vinegar and 2 tablespoons of cabbage indicator in the glass. It should be red or pink.
  2. Crush one tablet into a powder. Add it a little at a time (about a quarter at a time), stirring, and watch for fizzing and colour changes.
  3. Record how much tablet is needed before the colour shifts toward purple. If one tablet is not enough, crush a second.
  4. Write the proton-transfer equations: CO₃²⁻ + CH₃COOH ⇌ HCO₃⁻ + CH₃COO⁻, then HCO₃⁻ + CH₃COOH ⇌ H₂CO₃ + CH₃COO⁻ (and H₂CO₃ → CO₂ + H₂O).
  5. Use Ka values to explain why both steps favour products.

What to look for: fizzing (CO₂) and a gradual colour change from red toward purple as acetic acid is used up. Both steps favour products because CH₃COOH (Ka 1.8 × 10⁻⁵) is a stronger acid than HCO₃⁻ (4.7 × 10⁻¹¹) and than H₂CO₃ (4.5 × 10⁻⁷). A typical tablet with 500 mg of CaCO₃ contains 0.0050 mol of carbonate, which can react with 0.010 mol of acetic acid, about 12 mL of 5% vinegar. Your 1 tablespoon (about 15 mL) of vinegar therefore needs a little more than one tablet, which is why you have a second one ready.

Your turn

1

Find the pH of 0.25 mol/L NH₃ (Kb = 1.8 × 10⁻⁵).

2

Find Kb for the acetate ion. (Ka of CH₃COOH = 1.8 × 10⁻⁵)

3

The pKa of an acid is 3.18. What is the pKb of its conjugate base?

4

Find the pH of 0.20 mol/L pyridine, C₅H₅N (Kb = 1.7 × 10⁻⁹).

5

Find the pH of 0.050 mol/L methylamine (Kb = 4.4 × 10⁻⁴). Check the approximation.

6

Which is the stronger base: CN⁻ (Ka of HCN = 6.2 × 10⁻¹⁰) or F⁻ (Ka of HF = 6.6 × 10⁻⁴)? Find both Kb values.

More practice: warm-up

1

For a weak base, x in the ICE table equals which concentration?

2

Ka × Kb for a conjugate pair = ?

3

pKa + pKb = ?

4

If pOH = 3.00, what is the pH?

5

Write the Kb expression for CO₃²⁻.

More practice: on your own

1

Find the pH of 0.050 mol/L NaCN.

2

A solution of NH₃ has pH 11.27. Find its concentration.

3

Find the pH of 0.0050 mol/L NH₃. Is the approximation valid?

4

Explain, using Ka × Kb = Kw, why the conjugate base of a strong acid has no measurable effect on pH.

5

Which 0.10 mol/L solution has the higher pH: sodium acetate or sodium fluoride? Explain without full calculations.

Challenge

1

Find the pH of 0.10 mol/L Na₂CO₃. Treat CO₃²⁻ as a weak base (ignore the second step).

2

Show that for a weak base where the approximation holds, pOH = ½(pKb − log C). Use it to check Example 1.

3

The amine in Example 2 has Kb = 2.8 × 10⁻⁴. Find Ka and pKa of its conjugate acid, BH⁺, and decide whether BH⁺ is a stronger or weaker acid than NH₄⁺ (Ka 5.6 × 10⁻¹⁰).

Check your understanding

Quiz

1.

What is Kb for the conjugate base of an acid with Ka = 2.0 × 10⁻⁶?

2.

A 0.10 mol/L weak base has [OH⁻] = 1.0 × 10⁻³ mol/L. What is its pH?

3.

For which pair is pKa + pKb = 14.00 true at 25 °C?

4.

Which ion is the strongest base?

5.

To find Kb for CO₃²⁻, which Ka should you use?

6.

In a weak-base problem, a student finds x = 2.0 × 10⁻³ mol/L and reports pH = 2.70. What went wrong?

For parents and teachers

Big idea (E2.4, E3.7, E3.5): weak bases are solved like weak acids, but x is [OH⁻], so you find pOH first and then pH = 14.00 − pOH. For any conjugate pair, Ka × Kb = Kw (pKa + pKb = 14.00), so one table of Ka values gives every Kb.

How to help: After each base problem, ask, "Is your answer above 7?" A base must give a pH above 7. Ask your teen to show you how adding the two equations in the figure gives Kw. If you try the antacid activity, keep the tablets for the experiment only.

Common difficulties:

  • Reporting pOH as the pH.
  • Using the wrong Ka for an anion: for CO₃²⁻ use Ka of HCO₃⁻, not of H₂CO₃.
  • Forgetting that the cation of a strong base (Na⁺, K⁺) has no effect on pH.
  • Skipping the 5% check for stronger weak bases such as methylamine.