Weak Base Equilibrium and Kb

Introduction

A weak base can accept a proton from water, producing hydroxide while some base remains unprotonated. Kb measures the equilibrium balance. As with weak acids, the starting molarity is not automatically equal to the ion concentration, and a mass-balance calculation is needed for pH.

Core explanation

For a weak base B, write B + H₂O ⇌ BH⁺ + OH⁻. The pure water activity is absorbed into Kb; in a dilute approximation, Kb ≈ [BH⁺][OH⁻]/[B]. Ammonia follows NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. Its formula lacks an OH group, but proton transfer from water creates hydroxide in solution.

Begin with C₀ of B and negligible BH⁺. If x reacts, [B]eq = C₀ − x, [BH⁺]eq = x, and [OH⁻] ≈ x when water contribution is small. Then Kb ≈ x²/(C₀ − x). After finding x, pOH ≈ −log₁₀x and pH = pKw − pOH at the stated temperature. A small Kb can support a small-x approximation, but test x/C₀ afterward.

The protonated product BH⁺ is B's conjugate acid and can donate its proton back to water. Adding a salt containing BH⁺ gives a common ion and alters the weak-base equilibrium. Such a mixture may be a buffer when enough B and BH⁺ coexist. The simple no-initial-product ICE setup then needs revision.

Kb is a temperature-dependent constant associated with a written base reaction. It is not a measure of the amount of base added to a beaker. A concentrated weak-base solution can be strongly basic in pH terms despite incomplete protonation, while a very dilute strong base may produce less hydroxide.

At high ionic strength, activities rather than raw concentrations govern thermodynamic Kb. At extreme dilution, water's own OH⁻ contribution matters. The introductory equation x²/(C₀ − x) is useful within its stated dilute, dominant-solute regime.

Step-by-step reasoning

1. Write B's proton transfer with water and name BH⁺. 2. Build Kb from equilibrium activities or dilute concentrations. 3. Use an ICE table to solve hydroxide amount x. 4. Calculate pOH and pH with the temperature-specific pKw.

Visual explanation

Draw an NH₃ molecule receiving a proton from H₂O, leaving NH₄⁺ and OH⁻. Show many NH₃ units and fewer products in a weak-base equilibrium sketch.

Real-world analogy

A receiver accepts a package from a donor, leaving the donor with one fewer item. A base accepts a proton from water, leaving hydroxide as water's conjugate base.

Real-world example

Ammonia solutions used in cleaning are basic because NH₃ reacts partly with water to form NH₄⁺ and OH⁻. Their pH requires equilibrium, not complete-conversion stoichiometry.

Why?

Why can NH₃ raise pH without containing OH⁻ in its formula? It removes a proton from water, generating hydroxide through the base-ionization reaction in solution.

Common misconception

“Every base releases a pre-existing OH group.” Brønsted bases accept protons; ammonia forms OH⁻ by reacting with water despite having no OH in its formula.

Worked example

A 0.100 M weak base B has Kb = 1.0 × 10⁻⁵ in the simple concentration model at 25 °C. Approximate x = √(KbC₀) = 0.0010 M OH⁻. The fraction reacted is 1%, so the small-x approximation is reasonable. pOH ≈ 3.00 and pH ≈ 11.00. The solution is basic but far less than 0.100 M in OH⁻.

Quick check

1. What conjugate acid forms when NH₃ accepts a proton? Answer: NH₄⁺, the ammonium ion.

Exam focus

Use hydroxide from equilibrium x, then pOH and pH. Do not equate weak-base analytical concentration with [OH⁻].

Advanced insight

The Ka of BH⁺ and Kb of B are linked through water's ion product under consistent standard states. This allows acid-base calculations to shift between conjugate-pair viewpoints.

Summary

Kb describes partial proton uptake by a weak base in water. The resulting hydroxide concentration depends on both Kb and starting composition, and pH follows through pKw.

Practice questions

1. Write the simple Kb expression for NH₃. Answer: [NH₄⁺][OH⁻]/[NH₃] at equilibrium in a dilute concentration model. 2. Is 0.20 M NH₃ automatically 0.20 M OH⁻? Answer: No. Only some NH₃ reacts with water at equilibrium. 3. At 25 °C, what is pH if pOH = 4.00? Answer: Approximately 10.00 under pKw = 14.00.