Algebra · 6. Exponential and Logarithmic Functions

6.4Logarithmic Functions

Section · module m49363

What leans on this

Chemistry that uses this algebra — the answer to why you are learning it.

  • The Clausius-Clapeyron equation P = Ae^(−ΔHvap/RT) is rearranged into ‘the linear equation’ ln P = −(ΔHvap/R)(1/T) + ln A by taking a natural log of both sides — a student who does not know what ln undoes cannot follow that step or evaluate ln(P2/P1) in the worked examples.

  • S = k ln W, and ΔS = k ln(Wf/Wi), require evaluating a natural log of a microstate count; the ‘Determination of ΔS’ example computes k·ln(6/1) directly, which is meaningless arithmetic to a student who has not met what ln does to a ratio.

  • pH = −log[H3O+] is a direct logarithm evaluation, and the sign is where it goes wrong — a hydronium concentration of 1.2×10⁻³ M gives pH = −(−2.92) = 2.92, not −2.92, because the outer minus sign has to be applied after taking the log, not folded into it.

  • Every hydrolysis worked example ends by converting the solved-for hydronium concentration to pH via −log, so the same logarithm-evaluation and sign-handling skill from pH/pOH is needed here as a final step, not a novelty.

  • pKa = −log Ka is itself a logarithm evaluation, needed before the Henderson-Hasselbalch equation can be used numerically, e.g. pKa = 6.4 for carbonic acid in the blood-buffer example.

  • Every titration-curve point is a pH = −log[H3O+] or pOH = −log[OH⁻] evaluation from a stoichiometrically computed concentration, repeated at each titrant volume — the section is a sequence of logarithm evaluations chained to a stoichiometry calculation.

  • log(0.077), a number less than 1, evaluates to a negative logarithm that then gets subtracted (as a double negative) from E°cell — a student who has not internalized that log of a fraction is negative will mishandle the sign and move the potential the wrong direction.

  • Finding the potential of Cd | Cd²⁺ (0.10 M) ‖ Ni²⁺ (0.50 M) | Ni cannot be finished without evaluating log(0.2) and keeping the sign straight — a reader who has never evaluated a logarithm of a value less than 1 cannot produce the numeric E this worked example asks for, not just find it uncomfortable; algebra:6.4 has no cell-potential example of its own to have already covered this ground the way algebra:6.8 covers radioactive decay for chemistry:20.4.