Key Equations

K_w = [H₃O⁺][OH⁻] = 1.0

\times

10⁻¹⁴ (at 25 °C)

pH = −log [H_{3} O^{+}]

pOH = −log[OH⁻]

[H₃O⁺] = 10^−pH

[OH⁻] = 10^−pOH

pH + pOH = pK_w = 14.00 at 25 °C

K_{a} = \frac{[H_{3} O^{+}] [A^{−}]}{[HA]}

K_{b} = \frac{[{HB}^{+}] [{OH}^{−}]}{[B]}

K_a

\times

K_b = 1.0

\times

10⁻¹⁴ = K_w

Percent ionization = \frac{{[H_{3} O^{+}]}_{eq}}{{[HA]}_{0}} \times 100

pK_a = −log K_a

pK_b = −log K_b

pH = p K_{a} + log \frac{[A^{−}]}{[HA]}