Ka + Kb = Kw- The Relationship Explained

What Are Ka, Kb, and Kw?

These three constants are the backbone of acid-base chemistry. If you're studying chemistry, you'll encounter them constantly. Here's what each one means in plain terms.

Ka - The Acid Dissociation Constant

Ka measures how strongly an acid donates protons (H+ ions) in water. The stronger the acid, the higher the Ka value.

For the reaction: HA ⇌ H+ + A-

The Ka expression is: Ka = [H+][A-] / [HA]

Strong acids like HCl have very large Ka values. Weak acids like acetic acid have small Ka values (around 1.8 × 10^-5).

Kb - The Base Dissociation Constant

Kb measures how strongly a base accepts protons in water. The stronger the base, the higher the Kb value.

For the reaction: B + H2O ⇌ BH+ + OH-

The Kb expression is: Kb = [BH+][OH-] / [B]

Kw - The Water Ionization Constant

Water self-ionizes: 2H2O ⇌ H3O+ + OH-

At 25°C, Kw = [H+][OH-] = 1.0 × 10^-14

This constant is the foundation that links Ka and Kb together.

The Relationship: Ka × Kb = Kw

This equation is not arbitrary. It comes directly from how acids and bases behave in water.

Consider a conjugate acid-base pair. If you have the weak acid HA, its conjugate base is A-. The Ka of HA and the Kb of A- are mathematically linked through water's ionization.

Ka × Kb = Kw

This means if you know one constant, you can calculate the other:

• Kb = Kw / Ka
• Ka = Kw / Kb

Why This Relationship Matters

You don't need to measure both constants experimentally. Measure one, calculate the other. This saves time and is especially useful when dealing with conjugate pairs.

For example, if you know the Ka of acetic acid is 1.8 × 10^-5, you can find the Kb of the acetate ion:

Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.6 × 10^-10

The acetate ion is a weak base, and that tiny Kb value confirms it.

Understanding the pKa and pKb Connection

Most tables list pKa values, not Ka values. The same relationship holds for the logarithmic forms:

pKa + pKb = pKw = 14 (at 25°C)

This makes calculations even simpler. Just subtract one p value from 14 to get the other.

Quick Reference Table

Getting Started: How to Use These Relationships

Here's a straightforward method for solving problems involving Ka and Kb:

Step 1: Identify the Conjugate Pair

Find the weak acid and its conjugate base (or weak base and its conjugate acid). They always come in pairs.

Step 2: Check What You Know

Are you given Ka, pKa, Kb, or pKb? You only need one to find the rest.

Step 3: Apply the Formula

Use Kw = 1.0 × 10^-14 (at 25°C) to connect everything:

• Need Kb? → Kb = Kw / Ka
• Need Ka? → Ka = Kw / Kb
• Need pKb? → pKb = 14 - pKa

Step 4: Calculate pH if Needed

If you find [H+] or [OH-] from your constant, use:

pH = -log[H+] or pOH = -log[OH-], then pH + pOH = 14

Common Mistakes to Avoid

Mixing up Ka and Kb for the wrong species. Ka applies to the acid form; Kb applies to the conjugate base form.

Forgetting that Kw changes with temperature. The value 1.0 × 10^-14 is specifically for 25°C. At different temperatures, pKw is different (at 0°C it's about 14.94, at 100°C it's about 13.00).

Assuming strong acids and bases follow this relationship. Strong acids have Ka values so large they're essentially infinite. Weak acids have measurable Ka values you can work with.

The Bottom Line

Ka × Kb = Kw is not a trick equation. It's a direct consequence of how acids and bases behave in water. Memorize it. Understand why it works. You'll use it constantly in acid-base calculations.

If you have Ka, you can find Kb. If you have pKa, you can find pKb. The entire system is interconnected through water's ionization constant.