14.3pH and pOH
Section · module m68804
What this needs
Learn these first — this section will not make sense without them.
- Needs3.8Inverse Functions
Going from pH back to [H3O⁺] means recognizing 10^x as the inverse of log₁₀ — the worked example literally calls 10^(−7.3) the antilog, or the ‘inverse’ log, of −7.3 — and a student who has never met f⁻¹(f(x)) = x as a general idea has no reason to believe exponentiating undoes a logarithm rather than, say, squaring it.
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.
pKw = pH + pOH is derived by taking −log of both sides of Kw = [H3O+][OH⁻] and splitting −log(AB) into −log A + (−log B) — the log-of-a-product rule is what turns a multiplicative relationship between concentrations into an additive one between p-values.
Going from pH back to [H3O+] requires solving pH = −log[H3O+] for the concentration, i.e. exponentiating both sides to get [H3O+] = 10^(−pH) — the blood-pH example calls this out explicitly as taking the antilog, or the inverse log.
What this touches
Algebra that turns up here. Not a blocker, but this is where you will see it used.
No algebra turns up here. Most of the descriptive sections need none.