Calculate the values of pH and pOH, based on Kw and the concentration of all species present in a neutral solution of water.
The concentrations of hydronium and hydroxide ions are usually reported logarithmically as pH and pOH:
pH = −log[H₃O⁺] and pOH = −log[OH⁻]
The CED notes that "hydrogen ion" and "hydronium ion", and the symbols H⁺(aq) and H₃O⁺(aq), are often used interchangeably. Hydronium and H₃O⁺ are preferred, but H⁺(aq) is also accepted on the AP Exam.
Water autoionizes:
2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq) Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C
This means every aqueous solution contains both ions. Adding acid does not eliminate OH⁻; it suppresses it, because their product is pinned at Kw.
In pure water, pH = pOH defines a neutral solution. At 25 °C, pKw = 14.0, so pH = pOH = 7.0 and
pKw = 14 = pH + pOH at 25 °C
EK 8.1.A.4 is the subtle one, and it is tested: the value of Kw is temperature dependent, so the pH of pure, neutral water deviates from 7.0 at temperatures other than 25 °C. Autoionization is endothermic, so heating water increases Kw, increases both [H₃O⁺] and [OH⁻], and lowers the neutral pH below 7. The water is still neutral — because pH still equals pOH — it just is not at pH 7.
Neutral means [H₃O⁺] = [OH⁻]. It does not mean pH 7.
At 50 °C, Kw = 5.5 × 10⁻¹⁴. (a) Calculate the pH of pure water at 50 °C. (b) Is the water acidic, basic, or neutral? Justify.
(a) In pure water, [H₃O⁺] = [OH⁻] = x.
Kw = x² = 5.5 × 10⁻¹⁴
x = 2.35 × 10⁻⁷ M
pH = −log(2.35 × 10⁻⁷) = 6.63
(b) The water is neutral. Neutrality is defined by [H₃O⁺] = [OH⁻], and that condition holds exactly — both equal 2.35 × 10⁻⁷ M.
The pH is below 7 only because autoionization is endothermic: raising the temperature shifts the autoionization equilibrium toward products, increasing both ion concentrations and therefore raising Kw. Since both rose equally, the water remains neutral. Note also that pH + pOH = 6.63 + 6.63 = 13.26, not 14.00 — the 14.00 relationship is specific to 25 °C.