3.1.9 (HL)—pOH scale

Syllabus
First assessment 2025
Objective
3.1.9
Level
HL

pOH and Ion Concentrations

HL only

pOH=log10[OH];[OH]=10(pOH);pH+pOH=14at25°CpOH = −log10[OH−]; [OH−] = 10^(−pOH); pH + pOH = 14 at 25 °C

Move between pH and pOH, then between the logarithm and concentration. Keep the 25 °C condition attached to pH+pOH=14.

A reliable route is [OH⁻] → pOH → pH → [H⁺], or the reverse, with each logarithm shown. Use pH + pOH = pKw; replacing pKw by 14 is valid only at 25 °C.

Worked pOH example: for 0.025moldm30.025\,\mathrm{mol\,dm^{-3}} KOH(aq)\ce{KOH(aq)}, complete dissociation gives [OHX]=0.025moldm3[\ce{OH^-}]=0.025\,\mathrm{mol\,dm^{-3}}. Therefore pOH=log10(0.025)=1.60\mathrm{pOH}=-\log_{10}(0.025)=1.60. The low pOH is consistent with a basic solution; at 298 K, pH=14.001.60=12.40\mathrm{pH}=14.00-1.60=12.40.

Interconverting pH and pOH

HL only

Assessment in practice

Representative question

Question 1

[Maximum number: 1]

What is the concentration of OH(aq)\mathrm{OH}^{-}(\mathrm{aq}), in moldm3\mathrm{mol} \mathrm{dm}^{-3}, in a solution at 298.15 K with a pH of 4.50 ?

pH=log10[H+][H+]=10pHKw=[H+][OH]Kw=1.00×1014 mol2dm6\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \quad\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}} \quad K_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \quad K_{\mathrm{w}}=1.00 \times 10^{-14} \mathrm{~mol}^{2} \mathrm{dm}^{-6}
A

3.16×10103.16 \times 10^{-10}

B

3.16×1093.16 \times 10^{-9}

C

3.16×1053.16 \times 10^{-5}

D

3.16×1043.16 \times 10^{-4}

Proton Transfer Reactions Summary

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.