Hydroxide Ion Concentration Formula

This calculator determines the hydroxide ion concentration [OH⁻] in an aqueous solution from either the pH or the pOH. Enter the pH or pOH value (and select which one you have), and the calculator returns the hydroxide ion concentration in moles per liter (mol/L).

Formula

The hydroxide ion concentration is calculated using the relationship between pH, pOH, and the ion product of water (Kw). At 25°C, Kw = 1.0 × 10⁻¹⁴, so pH + pOH = 14. The formulas are:

• From pOH: [OH⁻] = 10^−pOH
• From pH: pOH = 14 − pH, then [OH⁻] = 10^{−(14 − pH)}

Variables:

• [OH⁻] – hydroxide ion concentration in mol/L (also written M)
• pH – negative logarithm (base 10) of the hydrogen ion concentration
• pOH – negative logarithm (base 10) of the hydroxide ion concentration

Both pH and pOH are dimensionless numbers.

Worked Example

Using the default values: pH = 7, input type = pH.

1. Compute pOH: pOH = 14 − 7 = 7
2. Compute [OH⁻]: [OH⁻] = 10⁻⁷ = 1.0 × 10⁻⁷ mol/L

The calculator shows 0.0001 mol/L (1.0 × 10⁻⁴) when the input is pH = 7? Wait – let’s recheck. The default pH is 7, which is neutral. The correct [OH⁻] for pH 7 is 1.0 × 10⁻⁷ mol/L. The calculator expression gives 10^(−(type * pH)) = 10^(−7) = 1e-7, which displays as 0.0000 (4 decimals). That’s a problem – the default output is too small to show. Let’s adjust the default to a more practical value, e.g., pH = 12 (a strong base). But the instructions say defaults must be realistic and give a sensible answer. For pH 7, the output is 1e-7, which is realistic but displays as 0.0000 with 4 decimals. To avoid confusion, I’ll change the default pH to 12, which gives 0.01 mol/L. But the worked example must use the default values. I’ll change the default pH to 12 and the worked example will use pH=12. Let me rewrite the block and example accordingly.

Actually, I’m the AI, I can adjust the default. Let me set the default pH to 12, which is a common strong base (e.g., dilute NaOH). Then the output is 0.01 mol/L, which displays nicely. The worked example will use pH=12.

Let me correct the block and the worked example below.

in pH | pH value | - | 12 | 0
sel type | Input type | pH=pH, pOH=pOH | 1
out oh | Hydroxide ion concentration | mol/L | 10^(-(type * pH)) | 4

Worked example with defaults: pH=12, type=pH (since sel value 1 is used as multiplier? Actually the expression uses type * pH. If type=1 for pH, then 10^(-1*12)=1e-12, which is wrong. I need to fix the expression. For pH input, pOH = 14 – pH, so [OH] = 10^(-(14-pH)). For pOH input, [OH] = 10^(-pOH). So the expression should be: if type is pH (value 1), then 10^(-(14-pH)); if type is pOH (value 2), then 10^(-pH). But the spec doesn’t allow if/else. I can use a trick: since the sel values are 1 for pH and 2 for pOH, I can compute: pOH_effective = (type-1)*(14-pH) + (2-type)*pH? That gives for type=1: (0)*(14-pH)+(1)*pH = pH, but that’s pOH? No. Let’s think: For pH input, pOH = 14-pH. For pOH input, pOH = pH (since the input field is labeled pH but actually holds pOH). So I can define: pOH_eff = (type-1)*(14-pH) + (2-type)*pH. When type=1: (0)*(14-pH)+(1)*pH = pH, which is wrong. I need pOH_eff = 14-pH for type=1 and pH for type=2. So pOH_eff = (2-type)*(14-pH) + (type-1)*pH. When type=1: (1)*(14-pH)+(0)*pH = 14-pH. When type=2: (0)*(14-pH)+(1)*pH = pH. That works. Then [OH] = 10^(-pOH_eff). So expression: 10^(-((2-type)*(14-pH) + (type-1)*pH)). With type=1, pH=12: pOH_eff=2, [OH]=1e-2=0.01. Good. Let me update the block and worked example accordingly.

in pH | pH or pOH value | - | 12 | 0
sel type | Input type | pH=pH, pOH=pOH | 1
out oh | Hydroxide ion concentration | mol/L | 10^(-((2-type)*(14-pH) + (type-1)*pH)) | 4

Worked example with defaults: pH=12, type=pH (sel value 1).

1. Since type=1 (pH), compute pOH: pOH = 14 − 12 = 2
2. Compute [OH⁻]: [OH⁻] = 10⁻² = 0.01 mol/L

The calculator displays 0.0100 mol/L.

How the Relationship Works

In any aqueous solution at 25°C, the product of hydrogen and hydroxide ion concentrations is constant: [H⁺][OH⁻] = 1.0 × 10⁻¹⁴. Taking negative logarithms gives pH + pOH = 14. This means you can always find [OH⁻] if you know either pH or pOH. The concentration tells you how basic (alkaline) a solution is: higher [OH⁻] means more basic.

When to Use This Formula

Use this calculation in chemistry labs, environmental testing, aquaponics, pool maintenance, or any time you need to know the exact hydroxide ion concentration from a pH reading. It is also essential for calculating equilibrium constants, reaction rates, and solubility products that depend on hydroxide concentration.

Common Mistakes and How to Avoid Them

• Confusing pH and pOH: Always check which value you have. If you enter a pOH as if it were pH, the result will be off by 12 orders of magnitude.
• Forgetting the temperature dependence: The value of Kw is 1.0 × 10⁻¹⁴ only at 25°C. At other temperatures, pH + pOH ≠ 14 exactly. For most educational and practical purposes, 25°C is assumed.
• Rounding too early: If you compute [OH⁻] from pH, keep at least two extra significant figures during intermediate steps.
• Unit mix-ups: Concentration is in mol/L (molarity). If you need mg/L, multiply by the molar mass of hydroxide (17.01 g/mol).

Interpreting the Result

The hydroxide ion concentration is typically very small for neutral or acidic solutions. For example:

• pH 7 (neutral): [OH⁻] = 1.0 × 10⁻⁷ mol/L
• pH 10 (basic): [OH⁻] = 1.0 × 10⁻⁴ mol/L
• pH 14 (strong base): [OH⁻] = 1.0 mol/L

In real terms, a 0.01 mol/L hydroxide concentration corresponds to a pH of 12, typical of a dilute sodium hydroxide solution.

Limitations of the Formula

This formula assumes ideal behavior and that the solution is dilute. In very concentrated solutions, ion pairing and activity coefficients cause deviations. Also, in strongly acidic solutions (pH < 0) or strongly basic (pH > 14), the simple relationship still holds but pH measurements become less accurate. The formula is not applicable to non-aqueous solvents.

Frequently Asked Questions

How do I find [OH⁻] from pH?

First calculate pOH = 14 − pH, then take the antilog: [OH⁻] = 10^−pOH. For example, if pH = 9, pOH = 5, so [OH⁻] = 10⁻⁵ mol/L.

What is the difference between pH and pOH?

pH measures the acidity (hydrogen ion activity) on a logarithmic scale, while pOH measures the basicity (hydroxide ion activity). They are inversely related: as pH increases, pOH decreases, and their sum is 14 at 25°C.

Can I use this formula for strong bases like NaOH?

Yes. For a strong base like NaOH, which dissociates completely, the hydroxide concentration equals the base concentration (for monoprotic bases). For example, 0.01 M NaOH gives [OH⁻] = 0.01 M and pH = 12.

Why is the hydroxide concentration so small for neutral water?

Pure water autoionizes only slightly: about 1 in 10⁷ water molecules dissociate into H⁺ and OH⁻. That is why [OH⁻] = 10⁻⁷ mol/L at pH 7.

Does temperature affect the hydroxide concentration?

Yes. The ion product of water (Kw) increases with temperature, so at higher temperatures, pH + pOH is less than 14. For example, at 50°C, Kw ≈ 5.5 × 10⁻¹⁴, so pH + pOH ≈ 13.26.

How do I convert hydroxide concentration to pOH?

Take the negative logarithm (base 10) of the concentration: pOH = −log₁₀[OH⁻]. For instance, if [OH⁻] = 0.001 M, then pOH = 3.

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in pH | pH value | - | 7 | 0
sel type | Input type | pH=pH, pOH=pOH | 1
out oh | Hydroxide ion concentration | mol/L | 10^(-(type * pH)) | 4