pH and pOH calculations for bases

The acid-side pH calculation extends to bases via pOH, and to reversing a pH back to a concentration — both confirmed exam tasks.

The two extra calculation directions

FINDING [H₃O⁺] FROM A GIVEN pH (the reverse direction): [H₃O⁺] = 10^(−pH) — undo the log rather than apply it.
FINDING THE pH OF A STRONG BASE: find [OH⁻] from the base's concentration (accounting for how many OH⁻ ions each formula unit releases), calculate pOH = −log₁₀[OH⁻], then use pH + pOH = 14 to find pH.
For a strong monobasic base like LiOH or NaOH, [OH⁻] equals the base concentration directly (complete ionisation, one OH⁻ per formula unit).

Worked example Finding [H₃O⁺] from a given pH, and the pH of a strong base A HCl solution has pH 2.4. Find [H₃O⁺]. Separately, find the pH of 0.450 mol L⁻¹ LiOH. PART 1: [H₃O⁺] = 10^(−pH) = 10^(−2.4) ≈ 4.0 × 10⁻³ mol L⁻¹. PART 2 — FIND [OH⁻]: LiOH is a strong base, fully ionising with one OH⁻ per formula unit, so [OH⁻] = 0.450 mol L⁻¹. CALCULATE pOH: pOH = −log₁₀(0.450) ≈ 0.35. CONVERT TO pH: pH = 14 − pOH = 14 − 0.35 = 13.65. ✓ Always check which direction the question is asking — pH-to-concentration uses 10^(−pH); concentration-to-pH for a BASE requires the extra pOH step that an acid calculation does not need. ✕

Using pH = −log[OH⁻] directly for a base

The formula pH = −log₁₀[H₃O⁺] uses the HYDRONIUM concentration, not hydroxide. For a base, you must find [OH⁻] first, convert to pOH = −log₁₀[OH⁻], and then use pH + pOH = 14 to reach the actual pH — skipping straight to −log[OH⁻] and calling it "pH" gives the pOH value by mistake.

Given	Find
pH	[H₃O⁺] = 10^(−pH)
[OH⁻] of a strong base	pOH = −log[OH⁻], then pH = 14 − pOH
pOH	pH = 14 − pOH

A fast check before submitting a base pH calculation

Have I found [OH⁻] first, accounting for how many hydroxide ions each formula unit releases?
Have I converted pOH to pH using pH + pOH = 14, rather than reporting pOH as if it were pH?

Practice for this standard covers the whole topic — flashcards and quiz.