SME
Grade 10 Science · Problem type
A strong base: pOH and pH
Use this for “A solution of 0.010 M NaOH: what are its pOH and pH?”
Diagram
Since pH = −log₁₀[H⁺], here [H⁺] = 10⁻¹² = 1 × 10⁻¹² mol/L.
Above 7: basic. pOH = 14 − pH = 2.
This is an example. Type your own values to start a new problem.
pOH
=
−log₁₀(
[OH⁻]
)
pH
pH
calculated
Formulas
pOH = −log₁₀([OH⁻])
2 = −log₁₀(0.01)
pH + pOH = 14
12 + 2 = 14
Assumptions
•
The water is at 25 °C, where [H⁺][OH⁻] = 1 × 10⁻¹⁴ and pH + pOH = 14.
•
Each pH unit is ten times the [H⁺]. Below 7 is acidic, above 7 basic.
•
NaOH is a strong base: every unit gives one OH⁻, so [OH⁻] is the base’s molarity.
Step-by-step
WE KNOW
[OH⁻] = 0.01 mol/L
FIND
pOH (pOH), pH (pH)
Step 1 · Find pOH (pOH)
Use pOH = −log₁₀([OH⁻])
Each step of 1 on the scale is ten times the [OH⁻]: the log counts the powers of ten.
pOH = −log₁₀(0.01)
pOH = 2
Step 2 · Find pH (pH)
Use pH + pOH = 14
In water at 25 °C, pH and pOH add to 14.
pH = 14 − pOH
pH = 14 − 2
pH = 12
Check
Put the numbers back in. Both sides should match.
2 = −log₁₀(0.01) ✓
12 + 2 = 14 ✓
Topic check
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