3.1.17 (HL)—Buffer pH and composition
- Syllabus
- First assessment 2025
- Objective
- 3.1.17
- Level
- HL
pH≈pKa+log10([A−]/[HA])
Buffer pH depends on pKa and the conjugate-to-parent ratio. Dilution changes both concentrations by the same factor, so the ratio and pH remain approximately constant.
When a small amount of acid is added, A⁻ removes it to form HA; added base is removed by HA to form A⁻. Dilution leaves their ratio nearly unchanged but reduces buffer capacity, so resistance to a large addition is not unchanged.
basicbuffer:pOH≈pKb+log10([BH+]/[B]);thenpH=pKw−pOH
Worked buffer example: an ethanoate buffer contains 0.100moldm−3 CHX3COOH and 0.200moldm−3 CHX3COOX−, with pKa=4.76. Substitution gives pH=pKa+log10([AX−]/[HA])=4.76+log10(0.200/0.100)=5.06. The pH is above pKa because the conjugate base is more concentrated than the acid.
Representative question
This 1.00 moldm−3 solution of nitrous acid was used to prepare a buffer with pH 3.00 .
Calculate the concentration of the conjugate base of nitrous acid required to make this buffer. The pKa of nitrous acid is 3.25.
Concentration of conjugate base:
(a)
(ii)
Ka<10−3.25>=5.62×10−4
« [H+]=10−3.00 » =0.001 «mol dm −3 »
« [A−]=(5.62×10−4×1.00)/0.001»=0.562 «mol dm−3 »
Alternative solution:
pH=pKa+log10[ salt ]/[ acid ] « log10[ salt ]/[ acid ] 》 =−0.25
OR
«[salt]/[acid]» = 0.562、
《 [A−]=0.562/1.00»=0.562 «mol dm−3 » ↓
Award[3]for correct final answer.
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.