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Generated on: July 20, 2026 at 09:33 PM

pH and pOH Converter

Inputs

Known quantitypH
pH7
pOH7
Hydrogen-Ion Concentration [H⁺]0.1 µM
Hydroxide-Ion Concentration [OH⁻]0.1 µM
Ion product of water (Kw)1e-14

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Chemistry

pH and pOH Converter

Convert between pH, pOH, [H⁺], and [OH⁻]. Enter any one of the four quantities and get the other three, with an adjustable Kw for non-standard temperatures.

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Results

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This solution is neutral — pH equals pOH, with equal H⁺ and OH⁻ concentrations.

Details

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µM

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Four ways to describe the same solution

The acidity of an aqueous solution can be expressed four equivalent ways: pH, pOH, the hydrogen-ion concentration [H⁺], and the hydroxide-ion concentration [OH⁻]. Knowing any one of them fixes the other three. This converter takes whichever quantity is on hand and returns the rest.

The defining relationships

pH and pOH are negative base-10 logarithms of the two ion concentrations:

pH=−log⁡10[H+]pOH=−log⁡10[OH−]\text{pH} = -\log_{10}[\text{H}^+] \qquad \text{pOH} = -\log_{10}[\text{OH}^-]
SymbolQuantityUnit
pHAcidity measureunitless
pOHBasicity measureunitless
[H⁺]Hydrogen-ion concentrationmol/L (M)
[OH⁻]Hydroxide-ion concentrationmol/L (M)

Each relationship reverses to recover a concentration:

[H+]=10−pH[OH−]=10−pOH[\text{H}^+] = 10^{-\text{pH}} \qquad [\text{OH}^-] = 10^{-\text{pOH}}

Water self-ionizes slightly, and the product of the two ion concentrations is a constant called the ion product of water, Kw:

[H+][OH−]=Kw[\text{H}^+][\text{OH}^-] = K_w

Taking negative logarithms gives the companion sum, where pKw = −log₁₀(Kw):

pH+pOH=pKw\text{pH} + \text{pOH} = pK_w

At 25 °C, Kw = 1.0×10⁻¹⁴ and pKw = 14, so pH + pOH = 14. A solution of pH 4 then has pOH 10, and one of pH 11 has pOH 3.

Worked example

A solution has pH 3.5. Converting to the other three quantities at 25 °C:

pOH=14−3.5=10.5\text{pOH} = 14 - 3.5 = 10.5 [H+]=10−3.5≈3.16×10−4 M[\text{H}^+] = 10^{-3.5} \approx 3.16\times10^{-4}\ \text{M} [OH−]=10−10.5≈3.16×10−11 M[\text{OH}^-] = 10^{-10.5} \approx 3.16\times10^{-11}\ \text{M}

Because pH is below pOH, the solution is acidic, and the two concentrations multiply back to Kw = 1.0×10⁻¹⁴.

Why Kw is adjustable

Kw rises with temperature because the self-ionization of water is endothermic. Near 50 °C it is roughly 5.5×10⁻¹⁴ (pKw ≈ 13.26), and at 100 °C it is larger still. Since pH + pOH = pKw, the neutral point — where pH equals pOH — is not exactly 7 in warm water. Setting Kw to its value at the working temperature keeps the conversion correct; leave it at 1.0×10⁻¹⁴ for standard room-temperature problems.

pH of everyday substances

SubstanceApprox. pHType
Lemon juice2Acidic
Black coffee5Acidic
Pure water7Neutral
Sea water8Basic
Baking soda9Basic
Household bleach13Basic

Frequently Asked Questions (FAQ)

How are pH and pOH related?

At 25 °C, pH + pOH = 14. This comes from the self-ionization of water, where [H⁺] × [OH⁻] = Kw = 1×10⁻¹⁴. Taking the negative logarithm of both sides gives pH + pOH = pKw = 14. So a solution with pH 4 has pOH 10, and one with pH 11 has pOH 3. The sum equals pKw, which is 14 only at 25 °C and shifts at other temperatures.

How do I convert pH to pOH?

Subtract the pH from pKw: pOH = pKw − pH. At 25 °C, pKw = 14, so pOH = 14 − pH. For example, a solution with pH 3.5 has pOH = 14 − 3.5 = 10.5. To go the other way, pH = 14 − pOH. This converter does the subtraction for you and also returns both ion concentrations.

How do I get [H⁺] and [OH⁻] from pH?

Raise 10 to the negative power: [H⁺] = 10^(−pH) and [OH⁻] = 10^(−pOH). For pH 3, [H⁺] = 10⁻³ = 0.001 M; with pOH 11, [OH⁻] = 10⁻¹¹ M. The two concentrations always multiply to Kw, so [H⁺] × [OH⁻] = 1×10⁻¹⁴ at 25 °C.

Why can I change Kw?

The ion product of water, Kw, depends on temperature. It is 1.0×10⁻¹⁴ at 25 °C but rises as water warms — about 1×10⁻¹³·² near 50 °C. Because pH + pOH = pKw, neutral pH is not exactly 7 at other temperatures. Adjusting Kw lets you model these conditions; leave it at 1.0×10⁻¹⁴ for standard room-temperature work.

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