3.1.5—Ion product of water (Kw)

$Kw = [H+][OH−]$

Solution	Ion comparison
acidic	[H+] > [OH−]
neutral	[H+] = [OH−]
basic	[H+] < [OH−]

At 25 °C, Kw = 1.0 × 10⁻¹⁴, so a neutral solution has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³. Neutrality always means equal ion concentrations; neutral pH is not necessarily 7 when temperature changes.

At a fixed temperature, Kw is constant, so [OH-] = Kw/[H+]: a higher [H+] means a lower [OH-]. For example, at pH 9.3 and 25 C, [OH-] = 2.0 x 10^-5 mol dm^-3. Classify a solution from the ion comparison; do not assume neutral pH is 7 at every temperature.

Representative question

[Maximum number: 1]

Calculate the concentration of hydroxide ions in an ammonia solution with pH=9.3. Use sections 1 and 2 of the data booklet.

$\left[\mathrm{OH}^{-}\right]\left\langle\left\langle=\frac{\mathrm{K}_{\mathrm{w}}}{\left[\mathrm{H}^{+}\right]}=\frac{10^{-14}}{10^{-9.3}}=10^{-4.7}\right\rangle\right\rangle=2.0 \times 10^{-5}\left\langle\left\langle\mathrm{~mol} \mathrm{dm}^{-3}\right\rangle\right\rangle$

Retrieve the route: track proton transfer and conjugates, calculate pH and Kw, distinguish strength, balance neutralization, read titration curves, use Ka/Kb and hydrolysis, select indicators, and explain and calculate buffer behaviour.

Check donor versus acceptor, one-proton differences, logarithm direction, ion comparison, strength versus concentration, equivalence versus endpoint, pKa landmarks, conjugate equations and dilution ratios.

3.1.5—Ion product of water (Kw)

The Ion Product of Water

Classifying Solutions with Kw

Assessment in practice

Proton Transfer Reactions Summary

Question 1