Equilibrium · Lesson 10 of 14

Weak Bases and Conjugate Constants

“Solve base-ionization problems and use water’s ionic product to connect a base with its conjugate acid.”

Learning Objectives

• Write Kb for a weak base reacting with water. • Calculate hydroxide concentration, pOH and pH using a checked equilibrium model. • Derive the conjugate-pair relationship KaKb = Kw. • Use ionization constants consistently when a conjugate ion is initially present.

A base can generate hydroxide by accepting a proton

Ammonia is basic even though it contains no hydroxide group. Its nitrogen lone pair accepts a proton from water, leaving hydroxide behind. The reverse reaction returns a proton from ammonium to hydroxide. The resulting equilibrium is closely parallel to weak-acid ionization, except that the directly calculated ion is usually hydroxide.

Weak-base ionizationLaTeX

B + H_{2} O ⇌ B H^{+} + O H^{-}, K_{b} \approx \frac{[ B H ^{+} ] [ O H ^{-} ]}{[ B ]}, p K_{b} = - lo g_{10} K_{b}

B is the unprotonated base and BH+ its conjugate acid. Water activity is incorporated; concentrations approximate activities under dilute conditions.

A larger Kb means a stronger proton-accepting base and a smaller pKb. Aqueous ammonia is sometimes written as NH4OH; writing NH3 + H2O makes the proton-transfer process clearer. Amines such as methylamine and dimethylamine are related to ammonia, with hydrogen atoms replaced by carbon-containing groups. Their available lone pairs can also accept protons. Different amines have different strengths; not all have equally small Kb values.

Weak base	Illustrative Kb at 298 K
Dimethylamine	$5.4 \times 1 0^{- 4}$
Triethylamine	$6.45 \times 1 0^{- 5}$
Ammonia	$1.77 \times 1 0^{- 5}$
Quinine	$1.10 \times 1 0^{- 6}$
Pyridine	$1.77 \times 1 0^{- 9}$
Aniline	$4.27 \times 1 0^{- 10}$
Urea	$1.3 \times 1 0^{- 14}$

The same change-table method applies

For base concentration c and no initially added conjugate acid, let x be the hydroxide produced. Then BH+ also has concentration x and unprotonated B has concentration c − x. The resulting equation is Kb = x squared divided by c − x. If ionization is small, x is approximately the square root of Kb times c. Convert the resulting hydroxide concentration to pOH before using pKw to find pH.

Simple weak-base modelLaTeX

K_{b} = \frac{x ^{2}}{c - x} = \frac{c α ^{2}}{1 - α}, [O H^{-}] \approx K_{b} c

The square-root estimate assumes small ionization, no important initial common ion, and a base contribution that dominates water autoionization.

Example — Hydrazine from its measured pH

Problem
A 0.0040 M hydrazine solution has pH 9.70 at 298 K. Determine its Kb and pKb in the dilute model.

1.Calculate pOH first: $14.00 - 9.70 = 4.30$ , so $x = [O H^{-}] = 1 0^{- 4.30} = 5.012 \times 1 0^{- 5} M$ .
2.Hydrazine accepts a proton: $N_{2} H_{4} + H_{2} O ⇌ N_{2} H_{5}^{+} + O H^{-}$ . Thus hydrazinium is approximately x and hydrazine remaining is $0.0040 - x$ .
3.Use $K_{b} = x^{2} / (0.0040 - x) \approx 6.36 \times 1 0^{- 7}$ , and pKb approximately 6.20.
4.As a check, hydrogen-ion concentration is approximately $1.995 \times 1 0^{- 10}$ M. Its product with the calculated hydroxide concentration is $1 0^{- 14}$ , as required at 298 K.

Example — Ammonia ionization

Problem
For 0.050 M NH3 at 298 K, use $K_{b} = 1.77 \times 1 0^{- 5}$ to estimate degree of ionization and pH.

1.Estimate $x = (1.77 \times 1 0^{- 5}) (0.050) = 9.407 \times 1 0^{- 4} M$ .
2.The fraction is $x /0.050 = 0.01881$ , or 1.88%, small enough for the stated approximation.
3.Then pOH is about 3.027 and pH about 10.973. The exact quadratic gives a slightly lower hydroxide concentration and pH about 10.97, consistent with the estimate.

A conjugate acid and base share one equilibrium link

Consider ammonium losing a proton to water and ammonia accepting one from water. Add their balanced reactions: the ammonia and ammonium cancel, leaving water autoionization. When reactions are added, their equilibrium constants multiply because the activity factors for cancelled species also cancel. This general rule is consistent with standard Gibbs energies adding.

Multiplying constants when reactions addLaTeX

K_{net} = K_{1} K_{2} \dots

The reactions must be added with the coefficients and directions corresponding to their quoted constants. Reverse or scale constants first if the reactions are changed.

A conjugate-pair relationshipLaTeX

K_{a} (B H^{+}) K_{b} (B) = K_{w}, p K_{a} + p K_{b} = p K_{w}

Ka must belong to the conjugate acid of the exact base used for Kb. At 298 K the pK sum is approximately 14.

The relationship is reciprocal: a base with a large Kb has a conjugate acid with a small Ka. It does not apply to two unrelated substances merely because one is labelled acid and the other base. For ammonia with Kb = $1.77 \times 1 0^{- 5}$ , ammonium has Ka approximately $5.65 \times 1 0^{- 10}$ at 298 K, and its pKa is about 9.25.

Why conjugate constants multiply to the water product— Each reaction arrow represents a reversible equilibrium. Adding the reactions cancels ammonia and ammonium, leaving water autoionization.

Example — Inferring a conjugate acid’s constant

Problem
Find the ionization constant and pKa of the conjugate acid of ammonia when $K_{b} = 1.77 \times 1 0^{- 5}$ and $K_{w} = 1 0^{- 14}$ .

1.The conjugate acid is NH4+, so use $K_{a} (N H_{4}^{+}) = K_{w} / K_{b} (N H_{3})$ .
2.Calculate $K_{a} = 1 0^{- 14} / (1.77 \times 1 0^{- 5}) = 5.65 \times 1 0^{- 10}$ .
3.Then pKa is 9.248. Check by adding it to pKb = 4.748: their sum is 14.000.

An initially present conjugate ion changes the calculation

An ammonia solution containing ammonium chloride already has NH4+ before further ammonia ionization occurs. We must include that starting concentration. It suppresses the additional proton acceptance by ammonia, illustrating the common-ion effect developed in the next lesson. The constant is unchanged; the equilibrium concentrations satisfying it are different.

Example — Ammonia with ammonium chloride

Problem
A solution contains 0.100 M NH3 and 0.200 M NH4Cl. Its ammonia pKb is 4.75. Estimate pH at 298 K.

1.Convert $K_{b} = 1 0^{- 4.75} = 1.778 \times 1 0^{- 5}$ . Let x be the extra hydroxide from base ionization.
2.The concentration equation is $K_{b} = (0.200 + x) x / (0.100 - x)$ . Since x is small compared with both initial concentrations, $x \approx K_{b} (0.100/0.200) = 8.891 \times 1 0^{- 6} M$ .
3.Then pOH is approximately 5.051 and pH is 8.949. The extra ionization is tiny, supporting the approximation. Ignoring the initial ammonium would give a much more basic prediction.

Common mistake

Keep the directly calculated ion clear: weak-acid calculations usually give hydronium, while weak-base calculations usually give hydroxide. Also, KaKb = Kw is restricted to a conjugate pair at the same temperature and compatible conventions.

Quick check

Which expression describes NH3 ionization in water?

Quick check

For a base at 298 K with pKb = 5, its conjugate acid has pKa

Quick check

Why do constants multiply when reactions are added?

Quick check

A solution at 298 K has pH = 9.70. Its pOH is

Quick check

For ammonia initially mixed with NH4Cl, the ionization table must include

Quick check

Which pair can be used together in KaKb = Kw?

Practice Problems

Find pH and degree of ionization for 0.0010 M aniline using Kb = $4.27 \times 1 0^{- 10}$ , and check the relevance of water ionization.
A 0.0050 M codeine solution has pH 9.95 at 298 K. Find Kb and pKb.
Find the conjugate-acid Ka of a base whose Kb is $2.0 \times 1 0^{- 6}$ .
Explain why stronger bases have weaker conjugate acids.
Derive the multiplication rule by writing the acid and base equilibria of a conjugate pair and cancelling species.
Compare the hydroxide concentrations expected from 0.10 M ammonia alone and from 0.10 M ammonia with 0.20 M ammonium chloride.

Key Takeaways

Weak bases accept protons from water, producing conjugate acids and hydroxide. Calculate pOH from hydroxide and then use the temperature-appropriate pKw. Conjugate acid and base constants multiply to Kw, and initially present conjugate ions must be included in the equilibrium model.

Previous · Lesson 9

Weak Acids and Their Ionization

Next · Lesson 11

Polyprotic Acids, Acid Strength, and Common Ions

A base can generate hydroxide by accepting a proton

The same change-table method applies

A conjugate acid and base share one equilibrium link

An initially present conjugate ion changes the calculation

Quiz

Practice Problems

Key Takeaways