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Chemistry · Glossary

What is Acidity constant, pKa?

Also known as: acidity constant · pKa

Definition 44.12 School Chemistry — Grades 1 to 12 · Chapter 44 — Acids and Bases: Ka and pKa

The acidity constant KaK_a of a couple HA\ce{HA}/AX−\ce{A-} is the equilibrium constant of the reaction of the acid with water, HA+HX2O⇌AX−+HX3OX+\ce{HA + H2O <=> A- + H3O+}:

Ka=[AX−] [HX3OX+][HA] c∘.K_a = \frac{[\ce{A-}]\,[\ce{H3O+}]}{[\ce{HA}]\,c^\circ} .

The pKa of the couple is pKa=−log⁡KapK_a = -\log K_a.

Some couples placed by their pK_a at 25\, C, acids on the left, their bases on the right. The lower a couple on the scale, the stronger its acid; the higher, the stronger its base.
Some couples placed by their pKapK_a at 25 ∘C25\,{}^{\circ}\mathrm{C}, acids on the left, their bases on the right. The lower a couple on the scale, the stronger its acid; the higher, the stronger its base.

Examples

Example 44.15 (Ethanoic acid at 0.010 mol/L0.010\,\mathrm{mol}/\mathrm{L})

pKa=4.76pK_a = 4.76, so Ka=10−4.76=1.74×10−5K_a = 10^{-4.76} = 1.74 \times 10^{-5}. Then x2+1.74×10−5 x−1.74×10−7=0x^2 + 1.74 \times 10^{-5}\,x - 1.74 \times 10^{-7} = 0 gives x=4.1×10−4 mol/Lx = 4.1 \times 10^{-4}\,\mathrm{mol}/\mathrm{L} and pH 3.39. Only 4.1×10−4/0.010=4.1 %4.1 \times 10^{-4}/0.010 = 4.1\,\% of the acid has reacted with water; hydrochloric acid at the same concentration, pH 2.0, has reacted entirely.

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