pKb Value (Base Constant)
The pKb value measures the strength of a base via the base constant Kb of its protolysis with water: the smaller the pKb, the stronger the base. For conjugate pairs pKa + pKb = 14 (25 °C).
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Formula
pK_B = -\lg K_B = -\lg \frac{[\text{BH}^+] \cdot [\text{OH}^-]}{[\text{B}]}Variables & units – pKb Value (Base Constant)
| Symbol | Meaning | Unit |
|---|---|---|
| pKb | Base exponent (negative logarithm of Kb) | dimensionless |
| Kb | Base constant of the protolysis B + H₂O ⇌ BH⁺ + OH⁻ | mol/L |
| [BH⁺] | Concentration of the conjugate acid | mol/L |
| [OH⁻] | Hydroxide concentration at equilibrium | mol/L |
| [B] | Concentration of the base | mol/L |
Derivation & background – pKb Value (Base Constant)
The basis is the protolysis B + H₂O ⇌ BH⁺ + OH⁻ after Brønsted; the practically constant water concentration is absorbed into Kb. Multiplying Ka of the conjugate acid by Kb of the base gives the ion product of water: Ka·Kb = Kw, in logarithmic form pKa + pKb = pKw = 14 at 25 °C. For weak bases the approximation pOH = ½(pKb − lg c₀) holds. Examples: ammonia pKb = 4.75, methylamine pKb ≈ 3.4, aniline pKb ≈ 9.4.
Exam blueprint
Validity range
Applies to weak bases in dilute aqueous solution; pKa + pKb = 14 holds exactly only for conjugate acid-base pairs at 25 °C.
Derivation steps
The mass action law of the protolysis B + H₂O ⇌ BH⁺ + OH⁻ gives Kb; the link to Ka runs through the ion product of water.
- 1Kb = [BH⁺][OH⁻]/[B]; the constant water concentration is absorbed into Kb.
- 2Ka·Kb = Kw = 10⁻¹⁴ (25 °C); taking logarithms gives pKa + pKb = 14.
Rearrangements
pKb from the pKa of the conjugate acid
Holds for conjugate pairs at 25 °C.
pH of a weak base
Approximation for weak bases with c₀ well above Kb.
Kb from Ka
A strong acid means a weak conjugate base and vice versa.
Task variant
pKa(NH₄⁺) = 9.25: what is the pKb of ammonia?
pKb = 14 − pKa = 14 − 9.25 = 4.75. Ammonia is thus a typical weak base.
Calculate the pH of 0.01 mol/L ammonia (pKb = 4.75).
pOH = ½·(pKb − lg c₀) = ½·(4.75 + 2) = 3.38 → pH = 14 − 3.38 = 10.62. The solution is clearly basic.
Common mistakes
Reporting pOH as the final answer.
Finish with pH = 14 − pOH; almost always the pH is asked for.
Applying pKa + pKb = 14 to arbitrary pairs.
The relation holds only for conjugate pairs such as NH₄⁺/NH₃ and only at 25 °C.
Interpreting a small pKb as a weak base.
The smaller the pKb, the stronger the base, exactly as with pKa.
Using the approximation for strong bases.
NaOH dissociates completely: there pOH = −lg c holds directly.
Exam context
- pH calculation of ammonia and amine solutions, conjugate pairs in titration curves and buffer choice on the base side.
These mistakes cost points in real exams. The set drills them until they stick.
Formula cluster
Acid-base systems
Mirror image of pKa: base strength, pOH and conjugate pairs.
Worked example
Ammonia: Kb = 1.78×10⁻⁵ mol/L → pKb = 4.75; check: pKa(NH₄⁺) = 9.25 and 9.25 + 4.75 = 14. pH of 0.1 mol/L NH₃: pOH = ½·(4.75 + 1) = 2.88 → pH = 14 − 2.88 = 11.12.
Applications
Comparing base strengths, pH calculation of ammonia and amine solutions, buffer selection on the base side, pharmacology (basic drugs), titration curves
Quanta exam set
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Which formula describes pKb Value (Base Constant)?
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How do you rearrange pKb = −lg(Kb) for pKb from the pKa of the conjugate acid?
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Which common mistake happens with pKb Value (Base Constant)?
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Frequently asked questions about pKb Value (Base Constant)
How do you calculate the pKb value?+
The pKb value is the negative base-10 logarithm of the base constant Kb: pKb = −lg Kb. The base constant describes the equilibrium of the protolysis of a base with water, B + H₂O ⇌ BH⁺ + OH⁻, and reads Kb = [BH⁺]·[OH⁻]/[B]. Example ammonia: Kb = 1.78×10⁻⁵ mol/L gives pKb = −lg(1.78×10⁻⁵) = 4.75. The smaller the pKb, the stronger the base. Often, however, one does not know Kb itself but the pKa of the conjugate acid; then one uses the relation pKb = 14 − pKa, which holds at 25 °C for conjugate pairs. This gives the pKb of a base directly from the tabulated pKa of its corresponding acid.
Why does pKa + pKb = 14 hold?+
The relation follows from the ion product of water. For a conjugate acid-base pair HA/A⁻ the acid constant Ka of the acid and the base constant Kb of the conjugate base multiply exactly to the ion product of water: Ka·Kb = Kw. At 25 °C, Kw = 10⁻¹⁴. Taking the logarithm of this equation and multiplying by minus one turns the product into a sum: pKa + pKb = pKw = 14. Intuitively this means: the stronger an acid, the weaker its conjugate base, and vice versa. It is important that the 14 holds only at 25 °C, because Kw is temperature-dependent, and only for a genuinely conjugate pair, not for arbitrary acids and bases.
How do you calculate the pH of a weak base?+
For a weak base you first calculate the pOH and from it the pH. The approximation is pOH = ½·(pKb − lg c₀), where c₀ is the initial concentration of the base. It holds when c₀ is much larger than Kb, so the protolysis proceeds only to a small extent. Then you use pH = 14 − pOH. Example: 0.1 mol/L ammonia with pKb = 4.75 gives pOH = ½·(4.75 − lg 0.1) = ½·(4.75 + 1) = 2.88 and thus pH = 14 − 2.88 = 11.12. A common mistake is to leave the pOH as the final answer; almost always the pH is asked for. For strong bases this approximation does not hold; there you calculate pOH = −lg c directly.
What is the difference between pKb and pKa?+
The pKa measures the strength of an acid, the pKb the strength of a base. The pKa is the negative logarithm of the acid constant Ka, which describes the equilibrium HA + H₂O ⇌ H₃O⁺ + A⁻; a small pKa means a strong acid. The pKb is the negative logarithm of the base constant Kb, which describes the equilibrium B + H₂O ⇌ BH⁺ + OH⁻; a small pKb means a strong base. Both quantities are linked via the conjugate pair: pKa + pKb = 14 at 25 °C. In practice tables usually list only the pKa; the pKb of the corresponding base is calculated from it. Both share the same logarithmic logic: one unit of difference corresponds to a factor of ten in the constant.
Why is ammonia a weak base?+
Ammonia NH₃ is a weak base because it protolyses only incompletely in water. In the equilibrium reaction NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ the equilibrium lies predominantly on the left side; only a small fraction of the ammonia molecules actually take up a proton. This is expressed in the small base constant Kb = 1.78×10⁻⁵ mol/L and the pKb of 4.75. For comparison: a strong base like sodium hydroxide dissociates completely. Nevertheless an ammonia solution reacts distinctly basic, because even the small amount of hydroxide ions formed raises the pH above seven; a 0.1-molar solution reaches about pH 11. Weak thus refers to the degree of protolysis, not to the solution being only slightly basic.
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How do you calculate with pKb Value (Base Constant)?
Here is how to work through a typical pKb Value (Base Constant) (pKb = −lg(Kb)) task step by step:
- 1
Task
pKa(NH₄⁺) = 9.25: what is the pKb of ammonia?
Solution path
pKb = 14 − pKa = 14 − 9.25 = 4.75. Ammonia is thus a typical weak base.
- 2
Task
Calculate the pH of 0.01 mol/L ammonia (pKb = 4.75).
Solution path
pOH = ½·(pKb − lg c₀) = ½·(4.75 + 2) = 3.38 → pH = 14 − 3.38 = 10.62. The solution is clearly basic.