# Neutralization Calculator

URL: https://speedycalculator.com/calculators/chemistry/neutralization/
Last updated: 2026-10-30

> The neutralization calculator finds the volume or concentration of acid or base needed for complete neutralization, using the number of H⁺ or OH⁻ each formula unit provides, gives the normality of a solution from mass and equivalent weight, and predicts the pH after mixing a strong acid with a strong base.

## Inputs

- Calculate
- Acid concentration (mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³)
- Acid volume (mL, L, µL, cm³, m³, fl oz (US), US gal)
- H⁺ per acid formula unit
- Base concentration (mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³)
- Base volume (mL, L, µL, cm³, m³, fl oz (US), US gal)
- OH⁻ per base formula unit
- Mass of acid or base dissolved (g, mg, µg, kg, oz, lb)
- Volume of solution (L, mL, µL, m³, dm³, cm³, US gal, fl oz (US))
- Equivalent weight
- Equivalent weight (g/eq)
- Molar mass (g/mol, kg/mol, Da (u), kDa)
- Valence (H⁺ or OH⁻ per formula unit)

## Results

- Volume of base needed
- Volume of acid needed
- Base concentration
- Acid concentration
- Equivalents neutralized (mol H⁺ = mol OH⁻)
- Normality of the acid
- Normality of the base
- Equivalent weight
- Normality
- Molarity
- pH of the mixture
- Excess acid or base
- Result

## Formula

```
At the equivalence point: n_H·C_a·V_a = n_OH·C_b·V_b · Normality N = m ÷ (EW·V) with EW = M ÷ valence · N = n·C · after mixing: excess = n_H·C_a·V_a − n_OH·C_b·V_b, [H⁺] = d/2 + √(d²/4 + Kw), d = excess ÷ V_total
```

## Explanation

Neutralization is the reaction of an acid with a base to form a salt and, for Arrhenius acids and bases, water: H⁺ + OH⁻ → H₂O. Complete neutralization (the equivalence point) is reached when the moles of H⁺ the acid can donate equal the moles of OH⁻ the base can accept. That is the principle behind every acid–base titration and every calculation of how much lime to add to acidic water or how much baking soda neutralises a spill.

The key is to count equivalents, not just moles. One mole of sulfuric acid, H₂SO₄, provides two moles of H⁺, so it needs two moles of NaOH; one mole of Ca(OH)₂ provides two moles of OH⁻. Multiplying each molarity by the number of H⁺ or OH⁻ per formula unit gives the normality, and at the equivalence point N_acid·V_acid = N_base·V_base. Normality can also be built from a weighed mass and the equivalent weight (molar mass divided by the number of reacting H⁺ or OH⁻).

The last mode mixes known amounts of a strong acid and a strong base and tells you which is in excess and the resulting pH – useful for checking whether a titration has overshot.

### How to use the Neutralization Calculator

- **Calculate**
- **Acid concentration**
- **Acid volume**
- **H⁺ per acid formula unit**
- **Base concentration**
- **Base volume**
- **OH⁻ per base formula unit**
- **Mass of acid or base dissolved**
- **Volume of solution**
- **Equivalent weight**
- **Equivalent weight (g/eq)**
- **Molar mass**
- **Valence (H⁺ or OH⁻ per formula unit)**

Results update as you type, every field has its own unit menu, and everything is calculated in your browser.

### How it works

The calculator multiplies each concentration by its volume and by the number of H⁺ or OH⁻ per formula unit to get equivalents, then solves the equivalence condition for the unknown. For normality it divides the mass by the equivalent weight and the volume. For mixtures it subtracts equivalents and solves the water equilibrium for the excess to get the pH.

### Worked example

How much 0.20 M NaOH neutralizes 25 mL of 0.10 M H₂SO₄? Equivalents of H⁺ = 2 × 0.10 mol/L × 0.025 L = 0.0050 mol. Volume of base = 0.0050 / (1 × 0.20) = 0.025 L = 25 mL.

Normality: 4.9 g of H₂SO₄ (98.079 g/mol, valence 2, equivalent weight 49.04 g/eq) dissolved to 0.5 L gives N = 4.9 / (49.04 × 0.5) = 0.1998 N, which is 0.0999 M.

### Equivalents of common acids and bases

CompoundH⁺ or OH⁻ per formulaMolar mass (g/mol)Equivalent weight (g/eq)HCl136.4636.46HNO₃163.0163.01H₂SO₄298.0849.04H₃PO₄397.9932.66NaOH140.0040.00KOH156.1156.11Ca(OH)₂274.0937.05Ba(OH)₂2171.3485.67

### Tips

- Use the final volume of the solution, not the volume of solvent added, for molarity.
- For dilutions, C₁V₁ = C₂V₂ only holds when both concentrations use the same unit.
- Add acid to water, never water to acid.
- Everything is calculated in your browser.

### Common mistakes to avoid

- Mixing w/w, w/v and v/v percentages.
- Forgetting that pH is logarithmic – one unit is a tenfold change in [H⁺].
- Using pKa where Ka is asked for (or the other way round).

### Inputs explained

FieldWhat to enterCalculateChoose one: Volume of base needed, Volume of acid needed, Concentration of the base (titration), Concentration of the acid (titration) …Acid concentrationUnits: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³Acid volumeUnits: mL, L, µL, cm³, m³, fl oz (US) …H⁺ per acid formula unitChoose one: 1 (HCl, HNO₃, CH₃COOH), 2 (H₂SO₄, H₂C₂O₄), 3 (H₃PO₄, citric acid)Base concentrationUnits: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³Base volumeUnits: mL, L, µL, cm³, m³, fl oz (US) …OH⁻ per base formula unitChoose one: 1 (NaOH, KOH, NH₃), 2 (Ca(OH)₂, Ba(OH)₂, Na₂CO₃), 3 (Al(OH)₃)Mass of acid or base dissolvedUnits: g, mg, µg, kg, oz, lbVolume of solutionUnits: L, mL, µL, m³, dm³, cm³ …Equivalent weightChoose one: I know the equivalent weight, Calculate it from molar mass and valenceEquivalent weight (g/eq)A numberMolar massUnits: g/mol, kg/mol, Da (u), kDaValence (H⁺ or OH⁻ per formula unit)A number

### What each result means

- **Volume of base needed** = [H⁺ per acid formula unit]×[Acid concentration]×[Acid volume]÷([OH⁻ per base formula unit]×[Base concentration])
- **Volume of acid needed** = [OH⁻ per base formula unit]×[Base concentration]×[Base volume]÷([H⁺ per acid formula unit]×[Acid concentration])
- **Base concentration** = [H⁺ per acid formula unit]×[Acid concentration]×[Acid volume]÷([OH⁻ per base formula unit]×[Base volume])
- **Acid concentration** = [OH⁻ per base formula unit]×[Base concentration]×[Base volume]÷([H⁺ per acid formula unit]×[Acid volume])
- **Normality** = [Mass of acid or base dissolved]÷([Equivalent weight]×[Volume of solution])
- **pH of the mixture** = -log10([d]÷2 + sqrt([d]^2÷4 + 1e-14))
- **Excess acid or base** = abs([Excess])

## Worked example

Inputs — Calculate: Volume of base needed; Acid concentration: 0.1 mol/L (M); Acid volume: 25 mL; H⁺ per acid formula unit: 2 (H₂SO₄, H₂C₂O₄); Base concentration: 0.2 mol/L (M); OH⁻ per base formula unit: 1 (NaOH, KOH, NH₃).

Results — **Volume of base needed = 25 mL**; Equivalents neutralized (mol H⁺ = mol OH⁻) = 0.005 eq; Normality of the acid = 0.2 N (eq/L); Normality of the base = 0.2 N (eq/L).

## Key terms

- **Equivalence point**: The point where the acid and base have reacted in exact stoichiometric amounts.
- **Equivalent**: The amount of acid or base that supplies or accepts one mole of H⁺.
- **Normality (N)**: Equivalents per litre of solution.
- **Equivalent weight**: Molar mass divided by the number of H⁺ or OH⁻ per formula unit.
- **Titration**: Measuring the volume of a solution of known concentration needed to react with a sample.

## FAQ

**How do you calculate the volume needed to neutralize an acid?**
Use n_H·C_a·V_a = n_OH·C_b·V_b. For 25 mL of 0.1 M H₂SO₄ with 0.2 M NaOH: V_b = 2 × 0.1 × 25 / (1 × 0.2) = 25 mL.

**How many moles of NaOH neutralize one mole of H₂SO₄?**
Two – sulfuric acid gives two H⁺ per molecule: H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O.

**What is normality?**
Equivalents per litre. For acids and bases N = molarity × number of H⁺ or OH⁻ supplied, so 1 M H₂SO₄ is 2 N and 1 M HCl is 1 N.

**How do I find the equivalent weight?**
Divide the molar mass by the number of H⁺ (acid) or OH⁻ (base) per formula unit. H₂SO₄: 98.08 / 2 = 49.04 g/eq; Ca(OH)₂: 74.09 / 2 = 37.05 g/eq.

**Is the pH 7 at the equivalence point?**
Only for a strong acid with a strong base. A weak acid titrated with a strong base gives a basic equivalence point (about pH 8.7 for 0.1 M acetic acid), and a weak base with a strong acid gives an acidic one.

**What is the pH after mixing 50 mL of 0.1 M HCl with 30 mL of 0.1 M NaOH?**
0.005 − 0.003 = 0.002 mol H⁺ remain in 80 mL, so [H⁺] = 0.025 M and pH = 1.60.

**Does the calculator work for weak acids?**
The volume and concentration modes do, because the stoichiometry is the same. The pH-after-mixing mode assumes a strong acid and a strong base.


## Sources

- [OpenStax – Chemistry 2e](https://openstax.org/details/books/chemistry-2e)
- [IUPAC Compendium of Chemical Terminology (Gold Book)](https://goldbook.iupac.org/)
- [LibreTexts Chemistry](https://chem.libretexts.org/)