Ka, Kb, pKa, and pKb Values Explained

The acid dissociation constant Ka and the base dissociation constant Kb measure the strength of weak acids and bases: Ka = [H+][A−] ÷ [HA], and pKa = −log Ka; a larger Ka (smaller pKa) means a stronger acid. In brief: acetic acid has Ka = 1.8 × 10−5 (pKa = 4.76), hydrofluoric acid has Ka = 6.8 × 10−4 (pKa = 3.17), so HF is the stronger acid; for a conjugate pair, Ka × Kb = Kw and pKa + pKb = 14.00.

These constants allow quantitative predictions of pH, of the direction of reactions, and of the behavior of buffers. This article defines the constants, shows how to convert between K and pK, lists values for common acids and bases, shows how to use an ICE table to find pH and percent ionization, works through examples and gives some history.

What are Ka and Kb?

For a weak acid HA: HA + H2O ⇌ H3O+ + A−, Ka = [H3O+][A−] ÷ [HA]. For a weak base B: B + H2O ⇌ BH+ + OH−, Kb = [BH+][OH−] ÷ [B]. Water is a pure liquid and does not appear in the expressions. The constants depend only on temperature (values here are for 25 °C).

What are pKa and pKb?

pKa = −log Ka and pKb = −log Kb. They convert very small numbers into simple ones: Ka = 10−pKa. A difference of one pK unit is a tenfold difference in strength. Strong acids have negative pKa values.

Values for common acids and bases

Ka and pKa of weak acids at 25 °C
AcidFormulaKapKa
Hydrofluoric acidHF6.8 × 10−43.17
Nitrous acidHNO24.5 × 10−43.35
Formic acidHCOOH1.8 × 10−43.74
Benzoic acidC6H5COOH6.3 × 10−54.20
Acetic acidCH3COOH1.8 × 10−54.76
Hypochlorous acidHOCl3.5 × 10−87.46
Ammonium ionNH4+5.6 × 10−109.25
Hydrocyanic acidHCN6.2 × 10−109.21
PhenolC6H5OH1.3 × 10−109.89
Kb and pKb of weak bases at 25 °C
BaseFormulaKbpKb
MethylamineCH3NH24.4 × 10−43.36
AmmoniaNH31.8 × 10−54.75
Cyanide ionCN−1.6 × 10−54.79
PyridineC5H5N1.7 × 10−98.77
AnilineC6H5NH24.3 × 10−109.37

How do you use an ICE table?

ICE stands for initial, change, equilibrium. For HA with initial concentration c: initial [HA] = c, [H+] ≈ 0, [A−] = 0; change −x, +x, +x; equilibrium c − x, x, x. Then Ka = x2 ÷ (c − x). If the acid comes with its salt (a buffer), the initial [A−] is not zero.

Worked examples

  1. Ka from pH. A 0.100 mol/L solution of a weak acid has pH 3.00. [H+] = [A−] = 1.0 × 10−3; [HA] = 0.100 − 0.001 = 0.099. Ka = (1.0 × 10−3)2 ÷ 0.099 = 1.0 × 10−5; pKa = 5.00.
    HA H+ A− I 0.100 0 0 C −0.001 +0.001 +0.001 E 0.099 1.0×10−3 1.0×10−3 Ka = (1.0×10−3)2 / 0.099 = 1.0×10−5
  2. pH from Ka. 0.0500 mol/L HF, Ka = 6.8 × 10−4: solving x2 + 6.8 × 10−4x − 3.4 × 10−5 = 0 gives x = 5.5 × 10−3 mol/L; pH = 2.26; ionization = 11%.
  3. Kb of a conjugate base. Fluoride: Kb = 1.0 × 10−14 ÷ 6.8 × 10−4 = 1.5 × 10−11.
  4. pK conversion. pKa of NH4+ = 9.25, so pKb of NH3 = 14.00 − 9.25 = 4.75, Kb = 10−4.75 = 1.8 × 10−5.
  5. Ranking. By pKa: HF (3.17) > acetic acid (4.76) > HCN (9.21). HF is the strongest, and the conjugate bases rank in the opposite order: CN− > CH3COO− > F−.
    0 1 2 3 4 5 6 7 8 9 10 pKa HF 3.17 CH3COOH 4.76 HCN 9.21
  6. pH of a weak base. 0.100 mol/L NH3: [OH−] = √(1.8 × 10−5 × 0.100) = 1.34 × 10−3 mol/L; pH = 11.13.

How are Ka and Kb of a pair related?

Adding the equilibria HA ⇌ H+ + A− and A− + H2O ⇌ HA + OH− gives H2O ⇌ H+ + OH−, so Ka × Kb = Kw. A weak acid has a conjugate base of moderate strength, and a very weak acid has a fairly strong conjugate base.

How were the constants developed?

Wilhelm Ostwald derived the dilution law in 1888 and used Ka to compare acids from conductivity data. Tables of pKa values became standard in the 20th century, and are used throughout biochemistry and pharmacology.

Which mistakes are common?

Mistake 1: Thinking that a larger pKa means a stronger acid

It means a weaker acid.

Mistake 2: Including water in the expression

Pure liquid water is omitted.

Mistake 3: Applying Ka to a strong acid

Strong acids ionize completely and have no useful Ka in water.

Mistake 4: Using the approximation when ionization is large

Solve the quadratic when x exceeds 5% of the initial concentration.