Equilibrium
Water, pH and strong acids and bases
Calculate pH and pOH.
Water itself ionises very slightly: H₂O ⇌ H⁺ + OH⁻. The ionic product of water is Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K. In pure water, [H⁺] = [OH⁻], so each is 10⁻⁷ M. Kw increases with temperature, because ionisation is endothermic.
Such small numbers are easier to handle as logarithms. pH = –log[H⁺], and pOH = –log[OH⁻]. At 298 K: pH + pOH = pKw = 14.
- pH < 7: acidic.
- pH = 7: neutral (at 298 K).
- pH > 7: basic.
Strong acids and bases ionise completely, so their concentration gives [H⁺] or [OH⁻] directly:
- 0.01 M HCl: [H⁺] = 10⁻² M, so pH = 2.
- 0.001 M NaOH: [OH⁻] = 10⁻³ M, so pOH = 3 and pH = 14 – 3 = 11.
A trap: in a very dilute acid, the H⁺ from water itself (about 10⁻⁷ M) is no longer negligible and must be added. An acid of 10⁻⁸ M cannot make water basic: its pH comes out at about 6.98, just below 7.
For a mixture of a strong acid and a strong base: find the millimoles of H⁺ and of OH⁻, subtract the smaller from the larger (they neutralise each other), and divide what is left over by the total volume.
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