3.1.9 (HL)—pOH scale
- Syllabus
- First assessment 2025
- Objective
- 3.1.9
- Level
- HL
pOH=−log10[OH−];[OH−]=10(−pOH);pH+pOH=14at25°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 0.025moldm−3 KOH(aq), complete dissociation gives [OHX−]=0.025moldm−3. Therefore pOH=−log10(0.025)=1.60. The low pOH is consistent with a basic solution; at 298 K, pH=14.00−1.60=12.40.
Representative question
What is the concentration of OH−(aq), in moldm−3, in a solution at 298.15 K with a pH of 4.50 ?
3.16×10−10
3.16×10−9
3.16×10−5
3.16×10−4
A
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.