pOH and pH: Converting Between Them

pH gets all the attention, but it has a twin. pOH measures hydroxide ions in exactly the same way pH measures hydrogen ions. Once you see how the two are linked, you can move between any of the four quantities, pH, pOH, [H⁺] and [OH⁻], in one or two steps.

What pOH means

pOH = −log₁₀[OH⁻]

A low pOH means lots of hydroxide ions: a strongly alkaline solution. A high pOH means very little hydroxide: an acidic solution. So pOH runs in the opposite direction to pH.

Solution	pH	pOH
0.1 mol/dm³ HCl	1	13
Pure water (25 °C)	7	7
0.1 mol/dm³ NaOH	13	1

Where pH + pOH = 14 comes from

Water always contains a little H⁺ and OH⁻ from its own self-ionisation:

H₂O ⇌ H⁺ + OH⁻

The product of the two concentrations is a constant called the ionic product of water, Kw:

Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C

Take −log of both sides:

−log[H⁺] + (−log[OH⁻]) = −log(1.0 × 10⁻¹⁴)

pH + pOH = 14.00

That’s all “14” is: the pKw of water at 25 °C. It isn’t a law of nature; it’s a consequence of one particular temperature. More on that below.

The conversion map

Think of the four quantities as the corners of a square:

• [H⁺] ↔ pH: use pH = −log[H⁺] or [H⁺] = 10⁻ᵖᴴ
• [OH⁻] ↔ pOH: use pOH = −log[OH⁻] or [OH⁻] = 10⁻ᵖᴼᴴ
• pH ↔ pOH: add or subtract from 14.00
• [H⁺] ↔ [OH⁻]: divide 1.0 × 10⁻¹⁴ by the one you know

From any corner you can reach any other.

Worked example 1: pH of a strong base

Find the pH of 0.0040 mol/dm³ NaOH.

• NaOH is a strong base, so [OH⁻] = 0.0040 mol/dm³
• pOH = −log(0.0040) = 2.40
• pH = 14.00 − 2.40 = 11.60

Worked example 2: [OH⁻] from pH

A cleaning solution has a pH of 11.8. What is [OH⁻]?

• pOH = 14.00 − 11.8 = 2.2
• [OH⁻] = 10⁻²·² = 6.3 × 10⁻³ mol/dm³

Worked example 3: [H⁺] in an alkaline solution

What is [H⁺] in 0.25 mol/dm³ KOH?

• [OH⁻] = 0.25 mol/dm³
• [H⁺] = Kw ÷ [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 0.25 = 4.0 × 10⁻¹⁴ mol/dm³

Even strongly alkaline solutions still contain some hydrogen ions, just very few.

Worked example 4: [OH⁻] in an acidic solution

Lemon juice has a pH of 2.3. What is [OH⁻]?

• pOH = 14.00 − 2.3 = 11.7
• [OH⁻] = 10⁻¹¹·⁷ = 2.0 × 10⁻¹² mol/dm³

Worked example 5: Group 2 hydroxide

What is the pH of 0.0030 mol/dm³ Sr(OH)₂?

• Each Sr(OH)₂ releases two OH⁻: [OH⁻] = 0.0060 mol/dm³
• pOH = −log(0.0060) = 2.22
• pH = 11.78

Worked example 6: finding a concentration from pH

A solution of barium hydroxide has a pH of 12.30. What is the concentration of Ba(OH)₂?

• pOH = 14.00 − 12.30 = 1.70
• [OH⁻] = 10⁻¹·⁷⁰ = 0.0200 mol/dm³
• Each Ba(OH)₂ gives 2 OH⁻, so [Ba(OH)₂] = 0.0200 ÷ 2 = 0.0100 mol/dm³

Worked example 7: a weak base

Ammonia solution, 0.10 mol/dm³, has Kb = 1.8 × 10⁻⁵. Estimate its pH.

For a weak base, [OH⁻] ≈ √(Kb × c):

• [OH⁻] ≈ √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ mol/dm³
• pOH = 2.87
• pH = 11.13

Compare this with 0.10 mol/dm³ NaOH (pH 13.00). The weak base has about 75 times less hydroxide at the same concentration.

Why chemists bother with pOH

You could always convert [OH⁻] straight to [H⁺] using Kw and never mention pOH. But pOH has real advantages:

1. It keeps base calculations symmetrical with acid ones. A strong base at 0.01 mol/dm³ has pOH = 2, exactly as a strong acid at 0.01 mol/dm³ has pH = 2.
2. It links neatly to Kb and pKb. Weak base calculations mirror weak acid calculations if you work in pOH.
3. It reduces calculator errors. Numbers like 4.0 × 10⁻¹⁴ are easy to mistype.

A strategy for any conversion question

When a question gives you one of the four quantities and asks for another, don’t reach for a formula straight away. Work through these steps instead:

1. Write down what you know and what you want. For example: “know [OH⁻], want pH”.
2. Decide whether you’re crossing between the acid side and the base side. pH and [H⁺] are on one side; pOH and [OH⁻] are on the other.
3. If you need to cross, do it in log form. Subtracting from 14 is much less error-prone than dividing into 1.0 × 10⁻¹⁴.
4. Convert log to concentration (or back) last. That keeps the awkward powers of ten until the very end.
5. Sense-check the direction. Alkaline solutions must have pH above 7 and pOH below 7 at 25 °C.

This route means you only ever use two operations: “−log” and “subtract from 14”. Most students who get these questions wrong have mixed up the order of operations, not the chemistry.

pOH in the real world

You’ll rarely see pOH printed on a product label, but the idea turns up whenever hydroxide concentration is what really matters. Soap makers and paper mills track the strength of their sodium hydroxide solutions. Water treatment engineers dose lime to raise pH and remove dissolved metals as insoluble hydroxides, and whether a metal hydroxide precipitates depends directly on [OH⁻]. In analytical chemistry, many precipitation calculations with solubility products are simpler to set up in terms of hydroxide concentration, so pOH is often the more natural quantity to calculate first.

Temperature changes the “14”

Water’s self-ionisation is endothermic, so Kw grows as temperature rises:

Temperature	Kw	pKw (= pH + pOH)	pH of pure (neutral) water
0 °C	1.1 × 10⁻¹⁵	14.94	7.47
25 °C	1.0 × 10⁻¹⁴	14.00	7.00
37 °C	2.4 × 10⁻¹⁴	13.62	6.81
50 °C	5.5 × 10⁻¹⁴	13.26	6.63
100 °C	5.1 × 10⁻¹³	12.29	6.14

At 50 °C, pure water has a pH of about 6.63, yet it is not acidic, because [H⁺] still equals [OH⁻]. The definition of neutral is [H⁺] = [OH⁻], not pH = 7.

This matters in real life. Blood, at 37 °C, is neutral at pH 6.81, so its normal pH of 7.4 is more alkaline relative to neutral than it first seems.

Example 8. At 50 °C, a solution has pOH = 5.00. What is its pH, and is it acidic or alkaline?

• pH = pKw − pOH = 13.26 − 5.00 = 8.26
• Neutral at this temperature is pH 6.63, so a pH of 8.26 is alkaline.

Common mistakes

• Giving pOH as the final answer when the question asked for pH.
• Using 14 at temperatures other than 25 °C.
• Forgetting the ×2 for Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂.
• Thinking pOH is only for bases. Every aqueous solution has a pOH, even strong acids.

Quick practice

1. pH of 0.080 mol/dm³ LiOH?
2. [OH⁻] in a solution of pH 4.5?
3. pOH of a solution with [H⁺] = 3.0 × 10⁻⁹ mol/dm³?

Answers: (1) pOH = 1.10, pH = 12.90. (2) pOH = 9.5, [OH⁻] = 3.2 × 10⁻¹⁰ mol/dm³. (3) pH = 8.52, pOH = 5.48.

Key takeaways

• pOH = −log[OH⁻]; it runs opposite to pH.
• At 25 °C, pH + pOH = 14.00 because Kw = 1.0 × 10⁻¹⁴.
• Any of pH, pOH, [H⁺] and [OH⁻] can be found from any other.
• Kw rises with temperature, so neutral pH is below 7 when water is warm.
• For more practice, try the pH practice problems or check answers with the pH calculator.