Chem & Algebra
Algebra · 6. Exponential and Logarithmic Functions

6.3Logarithmic Functions

Section · module m49363Read §6.3 at OpenStax ↗

Study this section

Written from the section, each card quoting the span it came from. The ones you miss come back sooner.

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.