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Kw and pH of basesEdexcel A-Level Chemistry: Revision notes

Section 1

The ionic product of water, Kw

Water ionises very slightly: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). Its concentration is effectively constant, so the equilibrium law gives the ionic product of water:

Kw=[H+][OH−]K_w = [H^+][OH^-], with units mol² dm⁻⁶.

At 298 K, Kw=1.00×10−14K_w = 1.00 \times 10^{-14} mol² dm⁻⁶. This holds for any dilute aqueous solution, so if [OH−][OH^-] rises, [H+][H^+] must fall.

Key termsionic product of waterKw
Common mistake

Writing [H2O][H_2O] in the KwK_w expression. It is not included.

Section 2

pKw

pKw is defined as pKw=−lg⁡KwpK_w = -\lg K_w. At 298 K it equals 14.00. Taking logs of Kw=[H+][OH−]K_w=[H^+][OH^-] gives pKw=pH+pOHpK_w = pH + pOH, so pOH = pKw − pH. In pure water [H+]=[OH−][H^+] = [OH^-], so [H+]=Kw[H^+] = \sqrt{K_w} and the neutral pH is pKw ÷ 2.

Key termspKwpOH

Section 3

pH of a strong base

A strong base (NaOH, KOH, Ba(OH)₂) is fully dissociated, so [OH−][OH^-] comes straight from its concentration, multiplied by the number of OH⁻ ions per formula unit. Then:

[H+]=Kw[OH−][H^+] = \frac{K_w}{[OH^-]} and pH = −lg [H+][H^+].

Worked example: 0.0250 mol dm⁻³ NaOH. [H+]=1.00×10−14÷0.0250=4.00×10−13[H^+] = 1.00 \times 10^{-14} \div 0.0250 = 4.00 \times 10^{-13} mol dm⁻³, so pH = 12.40.

Ba(OH)₂ example: 0.0150 mol dm⁻³ gives [OH−][OH^-] = 0.0300, so [H+]=3.33×10−13[H^+] = 3.33 \times 10^{-13} and pH = 12.48.

Key termsstrong base
Common mistake

Using pH = −lg [OH−][OH^-]. That is pOH. Convert using KwK_w (or pH = pKw − pOH).

Section 4

Dilution of a base

Diluting a strong base lowers [OH−][OH^-] in proportion to the volume change. A tenfold dilution cuts [OH−][OH^-] by a factor of ten, so [H+][H^+] rises tenfold and pH falls by exactly 1.

Example: 25.0 cm³ of 0.400 mol dm⁻³ NaOH diluted to 1.00 dm³ gives 0.0100 mol dm⁻³, so pH falls from 13.60 to 12.00.

Key termsdilution

Section 5

Temperature and Kw

The ionisation of water is endothermic, so KwK_w increases as temperature rises (5.48 × 10⁻¹⁴ at 323 K). Pure water still has [H+]=[OH−][H^+] = [OH^-] and is neutral, but its pH is lower: 5.48×10−14=2.34×10−7\sqrt{5.48 \times 10^{-14}} = 2.34 \times 10^{-7} mol dm⁻³ gives pH 6.63. The pH of a strong base also changes with temperature, since [H+]=Kw/[OH−][H^+] = K_w/[OH^-].

Key termsneutral
Exam tip

pH 7 means neutral only at 298 K. Always say [H+]=[OH−][H^+]=[OH^-] for neutral.

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Exam questions on Kw and pH of bases

  1. A technician prepares a 0.0250 mol dm⁻³ solution of sodium hydroxide, a strong base, for a cleaning product. All measurements are at 298 K, where the ionic product of water, KwK_w, is 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
    Calculate the pH of the sodium hydroxide solution.2 marks
  2. A water-treatment plant dissolves barium hydroxide, Ba(OH)₂, a strong base, to make a 0.0150 mol dm⁻³ solution at 298 K, where KwK_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶.
    Calculate the pH of the barium hydroxide solution.2 marks
  3. The ionic product of water varies with temperature. At 298 K, KwK_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. At 323 K (50 °C), KwK_w = 5.48 × 10⁻¹⁴ mol² dm⁻⁶. A student measures the pH of pure water at 323 K with a calibrated meter.
    Calculate the pH of pure water at 323 K.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).