Solution Chemistry Calculator: Molarity, Dilution and pH

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weak acid pH calculator

Calculate the pH of a weak acid and weak base (Ka, Kb, acetic, ammonia, HF)

A weak acid ionizes only a little, which is why the pH is not simply −log c. Enter the concentration and pick the acid from the Ka table (or type your own): the tool solves the equilibrium and compares it with the textbook approximation.

From the equilibrium to the pH, with and without approximation

For HA ⇌ H⁺ + A⁻ with concentration c, the ionization x satisfies Ka = x² ÷ (c − x), or x² + Ka·x − Ka·c = 0. The positive root is x = (−Ka + √(Ka² + 4·Ka·c)) ÷ 2, and pH = −log x. For 0.1 mol/L acetic acid with Ka = 1.75 × 10⁻⁵: x = 1.31 × 10⁻³ mol/L, pH 2.88 and a degree of ionization of 1.3 %. With Ka = 1.8 × 10⁻⁵, common in textbooks, the exact pH is 2.875 and the approximation gives 2.872, which explains slightly different answers between sources.
The approximation x ≈ √(Ka·c) assumes c − x ≈ c and only holds with x up to 5 % of c. In 0.01 mol/L hydrofluoric acid (Ka = 6.3 × 10⁻⁴), x is 22 % of c: the approximation gives 2.60 and the correct value is 2.65. In 1 × 10⁻⁵ mol/L acetic acid, the approximation (4.87) is off by more than 0.25 from the correct 5.15. That is why the tool shows both routes side by side and says whether the 5 % test passes. For weak bases the same applies with Kb and [OH⁻]: 0.1 mol/L ammonia (Kb = 1.8 × 10⁻⁵) has pOH 2.875 and pH 11.12.
In very dilute solutions water cannot be ignored. The full charge balance includes [OH⁻] = Kw ÷ [H⁺] with Kw = 1.0 × 10⁻¹⁴ at 25 °C. For a strong acid, 1 × 10⁻⁸ mol/L HCl gives [H⁺] = 1.05 × 10⁻⁷ mol/L and pH 6.98, never 8, and 5 × 10⁻⁸ mol/L HCl gives 6.89. Polyprotic acids ionize in steps and the tool solves all of them: 0.1 mol/L H2SO4 has a complete first ionization and the second (Ka2 = 1.0 × 10⁻²) takes the pH to 0.96, not 1.00. All values are for 25 °C.

pH from the exact equilibrium

Acetic acid 0.1 mol/L

Input: Ka = 1.75e-5; c = 0.1
Expected output: pH 2.88

Exact 2.8814; approximation 2.8785.

Acetic acid 0.1 mol/L with textbook Ka

Input: Ka = 1.8e-5; c = 0.1
Expected output: pH 2.88

Exact 2.8753; approximation 2.8724 (2.87).

Ammonia 0.1 mol/L

Input: Kb = 1.8e-5; c = 0.1
Expected output: pH 11.12

pOH 2.875.

Hydrofluoric acid 0.01 mol/L

Input: Ka = 6.3e-4; c = 0.01
Expected output: pH 2.65

x = 22 % of c: the approximation (2.60) is not valid.

HCl 1e-8 mol/L

Input: ácido forte; c = 1e-8
Expected output: pH 6.98

With water; without it, it would give 8.

NaOH 0.01 mol/L

Input: base forte; c = 0.01
Expected output: pH 12.00

[OH⁻] = 0.01; pOH 2.

Full tool FAQ

How do I calculate the molarity of a solution?

Molarity is moles of solute per liter of solution: c = m ÷ (M × V). For 5.844 g of NaCl (58.44 g/mol) in 1 L of solution: 5.844 ÷ 58.44 = 0.1000 mol, so 0.1000 mol/L. V is the final solution volume, measured to the mark of the flask, not the volume of water used.

What is the difference between molarity and molality?

How do I convert % w/w to mol/L?

How much should I weigh to prepare a solution, with hydrate and purity?

How do I use C1V1 = C2V2 and what does a 1:10 dilution mean?

How do I calculate the pH of a weak acid and when does the 5 % approximation hold?

Why is the pH of 1e-8 mol/L HCl not 8?

How does a buffer work and when does Henderson-Hasselbalch fail?

How do I read a titration curve and choose the indicator?

What are the limits of the calculations and what should the tool not be used for?

Frequently asked questions

Why does 0.1 mol/L acetic acid not have pH 1?

Because it ionizes only 1.3 %. Only 1.3 × 10⁻³ mol/L of the 0.1 mol/L becomes H⁺, which gives pH 2.88. A strong acid at the same concentration would have pH 1.

Can the pH be negative or above 14?

It can, in concentrated strong acid or base solutions (2 mol/L HCl would have a calculated pH of −0.30). At those concentrations activity differs from concentration, and the measured value departs from the calculated one.

Where do the Ka values come from?

From the LibreTexts reference table, citing the CRC Handbook, at 25 °C. Different books give slightly different values (1.75 or 1.8 × 10⁻⁵ for acetic acid); you can type your own.

Next steps

The same solution in every unit
Unit	Value
Molarity (mol/L)	0.100 mol/L
Millimolarity (mmol/L)	100 mmol/L
Normality (eq/L)	0.100 eq/L
Molality (mol/kg)	0.101 mol/kg
Mole fraction (x)	0.00181
Mass concentration (g/L)	5.84 g/L
Milligrams per liter (mg/L)	5840 mg/L
% mass/volume (g/100 mL)	0.584 % m/v
% volume/volume	needs solute density
% mass/mass	0.584 % m/m
Mass fraction (titer)	0.00584
Parts per million (ppm, mg/kg)	5840 ppm
Parts per billion (ppb, µg/kg)	5.84 × 10⁶ ppb

The same solution in every unit

Unit

Value

Molarity (mol/L)

0.100 mol/L

Millimolarity (mmol/L)

100 mmol/L

Normality (eq/L)

0.100 eq/L

Molality (mol/kg)

0.101 mol/kg

Mole fraction (x)

0.00181

Mass concentration (g/L)

5.84 g/L

Milligrams per liter (mg/L)

5840 mg/L

% mass/volume (g/100 mL)

0.584 % m/v

% volume/volume

needs solute density

% mass/mass

0.584 % m/m

Mass fraction (titer)

0.00584

Parts per million (ppm, mg/kg)

5840 ppm

Parts per billion (ppb, µg/kg)

5.84 × 10⁶ ppb

From the equilibrium to the pH, with and without approximation

For HA ⇌ H⁺ + A⁻ with concentration c, the ionization x satisfies Ka = x² ÷ (c − x), or x² + Ka·x − Ka·c = 0. The positive root is x = (−Ka + √(Ka² + 4·Ka·c)) ÷ 2, and pH = −log x. For 0.1 mol/L acetic acid with Ka = 1.75 × 10⁻⁵: x = 1.31 × 10⁻³ mol/L, pH 2.88 and a degree of ionization of 1.3 %. With Ka = 1.8 × 10⁻⁵, common in textbooks, the exact pH is 2.875 and the approximation gives 2.872, which explains slightly different answers between sources.

The approximation x ≈ √(Ka·c) assumes c − x ≈ c and only holds with x up to 5 % of c. In 0.01 mol/L hydrofluoric acid (Ka = 6.3 × 10⁻⁴), x is 22 % of c: the approximation gives 2.60 and the correct value is 2.65. In 1 × 10⁻⁵ mol/L acetic acid, the approximation (4.87) is off by more than 0.25 from the correct 5.15. That is why the tool shows both routes side by side and says whether the 5 % test passes. For weak bases the same applies with Kb and [OH⁻]: 0.1 mol/L ammonia (Kb = 1.8 × 10⁻⁵) has pOH 2.875 and pH 11.12.

In very dilute solutions water cannot be ignored. The full charge balance includes [OH⁻] = Kw ÷ [H⁺] with Kw = 1.0 × 10⁻¹⁴ at 25 °C. For a strong acid, 1 × 10⁻⁸ mol/L HCl gives [H⁺] = 1.05 × 10⁻⁷ mol/L and pH 6.98, never 8, and 5 × 10⁻⁸ mol/L HCl gives 6.89. Polyprotic acids ionize in steps and the tool solves all of them: 0.1 mol/L H2SO4 has a complete first ionization and the second (Ka2 = 1.0 × 10⁻²) takes the pH to 0.96, not 1.00. All values are for 25 °C.

pH from the exact equilibrium

Acetic acid 0.1 mol/L

Input: Ka = 1.75e-5; c = 0.1
Expected output: pH 2.88

Exact 2.8814; approximation 2.8785.

Acetic acid 0.1 mol/L with textbook Ka

Input: Ka = 1.8e-5; c = 0.1
Expected output: pH 2.88

Exact 2.8753; approximation 2.8724 (2.87).

Ammonia 0.1 mol/L

Input: Kb = 1.8e-5; c = 0.1
Expected output: pH 11.12

pOH 2.875.

Hydrofluoric acid 0.01 mol/L

Input: Ka = 6.3e-4; c = 0.01
Expected output: pH 2.65

x = 22 % of c: the approximation (2.60) is not valid.

HCl 1e-8 mol/L

Input: ácido forte; c = 1e-8
Expected output: pH 6.98

With water; without it, it would give 8.

NaOH 0.01 mol/L

Input: base forte; c = 0.01
Expected output: pH 12.00

[OH⁻] = 0.01; pOH 2.

Full tool FAQ

How do I calculate the molarity of a solution?

What is the difference between molarity and molality?

How do I convert % w/w to mol/L?

How much should I weigh to prepare a solution, with hydrate and purity?

How do I use C1V1 = C2V2 and what does a 1:10 dilution mean?

How do I calculate the pH of a weak acid and when does the 5 % approximation hold?

Why is the pH of 1e-8 mol/L HCl not 8?

How does a buffer work and when does Henderson-Hasselbalch fail?

How do I read a titration curve and choose the indicator?

What are the limits of the calculations and what should the tool not be used for?

Frequently asked questions

Why does 0.1 mol/L acetic acid not have pH 1?

Because it ionizes only 1.3 %. Only 1.3 × 10⁻³ mol/L of the 0.1 mol/L becomes H⁺, which gives pH 2.88. A strong acid at the same concentration would have pH 1.

Can the pH be negative or above 14?

Where do the Ka values come from?

From the LibreTexts reference table, citing the CRC Handbook, at 25 °C. Different books give slightly different values (1.75 or 1.8 × 10⁻⁵ for acetic acid); you can type your own.

The same solution in every unit
Unit	Value
Molarity (mol/L)	0.100 mol/L
Millimolarity (mmol/L)	100 mmol/L
Normality (eq/L)	0.100 eq/L
Molality (mol/kg)	0.101 mol/kg
Mole fraction (x)	0.00181
Mass concentration (g/L)	5.84 g/L
Milligrams per liter (mg/L)	5840 mg/L
% mass/volume (g/100 mL)	0.584 % m/v
% volume/volume	needs solute density
% mass/mass	0.584 % m/m
Mass fraction (titer)	0.00584
Parts per million (ppm, mg/kg)	5840 ppm
Parts per billion (ppb, µg/kg)	5.84 × 10⁶ ppb

The same solution in every unit

Unit

Value

Molarity (mol/L)

0.100 mol/L

Millimolarity (mmol/L)

100 mmol/L

Normality (eq/L)

0.100 eq/L

Molality (mol/kg)

0.101 mol/kg

Mole fraction (x)

0.00181

Mass concentration (g/L)

5.84 g/L

Milligrams per liter (mg/L)

5840 mg/L

% mass/volume (g/100 mL)

0.584 % m/v

% volume/volume

needs solute density

% mass/mass

0.584 % m/m

Mass fraction (titer)

0.00584

Parts per million (ppm, mg/kg)

5840 ppm

Parts per billion (ppb, µg/kg)

5.84 × 10⁶ ppb