pOH Calculators

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pOH is the negative base-10 logarithm of the hydroxide ion concentration: pOH = −log₁₀[OH⁻]. It is the base-equivalent of pH and measures the basicity of a solution. At 25°C, pH and pOH are related by: pH + pOH = pKw = 14. In pure water at 25°C: [OH⁻] = 10⁻⁷ M; pOH = 7. For basic solutions: [OH⁻] > 10⁻⁷ M → pOH < 7. For acidic solutions: [OH⁻] < 10⁻⁷ M → pOH > 7. pOH is a convenient intermediate for calculating the pH of base solutions.

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pOH Formula

pOH = −log₁₀[OH⁻]

pH + pOH = pKw = 14 (at 25°C)

[OH⁻] = 10^(−pOH)

Example: 0.02 M NaOH → [OH⁻] = 0.02 M → pOH = −log(0.02) = 1.70 → pH = 14 − 1.70 = 12.30.

pOH Scale at 25°C

  • pOH = 0: [OH⁻] = 1 M (very strongly basic; pH = 14)
  • pOH = 7: [OH⁻] = 10⁻⁷ M (neutral; pH = 7)
  • pOH = 14: [OH⁻] = 10⁻¹⁴ M (very strongly acidic; pH = 0)

Calculations with Strong Bases

For strong bases (complete dissociation): Ba(OH)₂ → Ba²⁺ + 2OH⁻. 0.05 M Ba(OH)₂: [OH⁻] = 0.10 M. pOH = −log(0.10) = 1.00. pH = 14 − 1.00 = 13.00.

Temperature Dependence

pH + pOH = pKw, which changes with temperature: At 25°C: pKw = 14.00 → pH + pOH = 14. At 37°C: pKw ≈ 13.62 → pH + pOH ≈ 13.62. Neutral pOH = pKw/2 = 6.81 at 37°C. Always specify temperature for precise pOH calculations.

Glossary

pOH
pOH = −log₁₀[OH⁻]; measures hydroxide concentration; [OH⁻] = 10^(−pOH); at 25°C: pH + pOH = 14; pOH < 7 = basic; pOH > 7 = acidic.
pKw
pKw = −log(Kw) = 14.00 at 25°C; defines the relationship pH + pOH = pKw; changes with temperature (pKw ≈ 13.62 at 37°C); neutral pOH = pKw/2.
Strong Base pOH Calculation
For strong bases (NaOH, KOH, Ca(OH)₂, Ba(OH)₂): [OH⁻] = base concentration × number of OH⁻ per formula unit; pOH = −log[OH⁻]; pH = pKw − pOH.

Frequently Asked Questions

pOH = −log₁₀[OH⁻]. It measures hydroxide ion concentration on a logarithmic scale. [OH⁻] = 10^(−pOH). From pH: pOH = 14 − pH (at 25°C). Examples: 0.001 M KOH → [OH⁻] = 0.001 M → pOH = −log(0.001) = 3.00 → pH = 14 − 3 = 11. Blood pH 7.4 → pOH = 14 − 7.4 = 6.6 → [OH⁻] = 10⁻⁶·⁶ = 2.5 × 10⁻⁷ M. For strong bases, [OH⁻] equals the base concentration (complete dissociation). For weak bases, [OH⁻] is calculated from Kb equilibrium.

At 25°C: pH + pOH = pKw = 14.00. This relationship comes from water autoionization: Kw = [H⁺][OH⁻] = 10⁻¹⁴; −log(Kw) = −log([H⁺]) − log([OH⁻]) = pH + pOH = pKw = 14. Useful conversions: Given pH → pOH = 14 − pH → [OH⁻] = 10^(−pOH). Given [OH⁻] → pOH = −log[OH⁻] → pH = 14 − pOH. Given [H⁺] → pH = −log[H⁺] → pOH = 14 − pH. The relationship pH + pOH = 14 is temperature-dependent — valid at 25°C only; at 37°C, pKw ≈ 13.62, so pH + pOH ≈ 13.62.

Step 1: Determine [OH⁻] from the strong base concentration (complete dissociation assumed). Step 2: pOH = −log[OH⁻]. Step 3: pH = 14 − pOH. Examples: 0.05 M NaOH → [OH⁻] = 0.05 M → pOH = −log(0.05) = 1.301 → pH = 14 − 1.301 = 12.70. 0.025 M Ca(OH)₂ → [OH⁻] = 2 × 0.025 = 0.05 M → pOH = 1.301 → pH = 12.70. Note: Ca(OH)₂ provides 2 OH⁻ per formula unit. 0.003 M Ba(OH)₂ → [OH⁻] = 0.006 M → pOH = 2.222 → pH = 11.78.

Pure water at 25°C: Kw = [H⁺][OH⁻] = 10⁻¹⁴. Since H⁺ and OH⁻ are produced equally in water autoionization: [OH⁻] = [H⁺] = √(10⁻¹⁴) = 10⁻⁷ M. pOH = −log(10⁻⁷) = 7.00. pH = −log(10⁻⁷) = 7.00. pH + pOH = 7 + 7 = 14 ✓. At 37°C (body temperature): Kw ≈ 2.42 × 10⁻¹⁴ → pKw ≈ 13.62 → [OH⁻] = [H⁺] = √(2.42 × 10⁻¹⁴) = 1.556 × 10⁻⁷ M → pOH = pH = 6.81. So neutral water at body temperature has pOH = pH = 6.81, not 7.00.