Question 1

How are Ka and Kb related in a conjugate acid-base pair?

Accepted Answer

Ka and Kb are inversely proportional, meaning their product always equals Kw (the ion product of water). When you multiply the Ka expression for a weak acid by the Kb expression for its conjugate base, the result is the equilibrium expression for Kw. This relationship allows you to calculate one constant if you know the other, since Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C.

Question 2

Why does a stronger acid have a weaker conjugate base?

Accepted Answer

Because Ka and Kb are inversely proportional through the Kw relationship, a stronger acid with a higher Ka value must have a proportionately weaker conjugate base with a lower Kb value. This inverse relationship is a direct consequence of the mathematical product Ka × Kb = Kw remaining constant. The stronger the acid donates protons, the less likely its conjugate base is to accept them.

Question 3

What do pKa and pKb values tell you about acid and base strength?

Accepted Answer

pKa and pKb are the negative logarithms of Ka and Kb, respectively. Lower pKa values indicate stronger acids, while lower pKb values indicate stronger bases. The relationship pKa + pKb = pKw = 14 at 25°C holds for all conjugate pairs. For example, an acid with pKa of 2.1 is stronger than one with pKa of 4.6.

Question 4

How can you calculate a conjugate acid's Ka if you know the base's Kb?

Accepted Answer

Use the relationship Ka × Kb = Kw (1.0 × 10⁻¹⁴ at 25°C). Rearrange to Ka = Kw / Kb. For example, if a base has Kb = 1 × 10⁻⁶, then its conjugate acid's Ka = (1.0 × 10⁻¹⁴) / (1 × 10⁻⁶) = 1 × 10⁻⁸. This calculation works because the product of acid base strengths and dissociation constants remains constant.

Question 5

What is the mathematical relationship between Ka, Kb, and Kw?

Accepted Answer

The product of Ka for a weak acid and Kb for its conjugate base equals Kw, the ion product of water: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C. This relationship arises from adding the ionization equations for the acid and its conjugate base, which together produce the water autoionization equation. Taking negative logarithms yields pKa + pKb = pKw = 14.

Question 6

How does the strength of an acid compare to the strength of its conjugate base?

Accepted Answer

An acid and its conjugate base have opposite relative strengths due to their inverse proportional relationship through Kw. If an acid is strong (high Ka), its conjugate base is weak (low Kb), and vice versa. This reciprocal relationship means that the stronger a species is as an acid, the weaker it is as a base, maintaining the constant product Ka × Kb = Kw.

Question 7

Can you use pKa and pKb values to predict which conjugate pair is stronger?

Accepted Answer

Yes. Compare pKa values directly: lower pKa means stronger acid. Compare pKb values directly: lower pKb means stronger base. Since pKa + pKb = 14 for conjugate pairs, an acid with pKa = 3 has a conjugate base with pKb = 11, making the acid relatively strong and its conjugate base relatively weak. This relationship helps predict acid-base behavior in solutions.

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