Solution Chemistry Calculator: Molarity, Dilution and pH

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buffer calculator henderson-hasselbalch

Buffer calculator: pH from the Henderson-Hasselbalch equation and how to design one

A buffer is a weak acid with its conjugate base, or the reverse, in comparable amounts. The tool computes the pH from the Henderson-Hasselbalch equation, checks it against the exact balance and designs the concentrations for a target pH.

Henderson-Hasselbalch, the useful range and when it fails

From the Ka expression one gets pH = pKa + log([A⁻] ÷ [HA]). With equal amounts the logarithm is zero and pH = pKa: 4.76 for the acetic acid and acetate pair, 9.26 for the ammonia and ammonium pair (from the ammonia pKb of 4.74). With 0.1 mol/L acetic acid and 0.05 mol/L acetate, the ratio is 0.5 and pH = 4.757 − 0.301 = 4.46. Each factor of 10 in the ratio shifts the pH by one unit, which is why the useful range is pKa ± 1, between ratios 1:10 and 10:1.
To design one, invert the equation: ratio = 10 to the power (target pH − pKa). An acetate buffer at pH 5.00 needs a ratio of 10^(5.00 − 4.757) = 1.75; with 0.1 mol/L in total, that is 0.0364 mol/L of acid and 0.0636 mol/L of acetate. For phosphate, the step that matters in biology is the second, H2PO4⁻ and HPO4²⁻ (pKa2 = 7.21): for pH 7.40 the base/acid ratio is 1.56. Picking a pair with a pKa within one unit of the target pH is the rule of thumb; the tool warns when the pair is a poor choice.
The equation assumes equilibrium concentrations equal the prepared ones. That fails when the buffer is very dilute or the pH is far from the pKa: with acetate and acetic acid at 1 × 10⁻⁴ mol/L each, the equation gives 4.76 but the exact balance gives 4.87. At 0.01 mol/L the difference drops to 0.001. That is why the tool shows both values and warns when they differ by more than 0.05. Buffer capacity, the amount of acid or base the buffer can take, grows with total concentration and peaks at a 1:1 ratio.

Buffer pH values and design ratios

Acetate 1:1

Input: ácido acético 0.1 + acetato 0.1
Expected output: pH 4.76

pKa = 4.757 (Ka = 1.75e-5).

Excess acid

Input: ácido acético 0.1 + acetato 0.05
Expected output: pH 4.46

Ratio 0.5: log = −0.301. Exact: 4.456.

Phosphate, step 2, 1:1

Input: H2PO4⁻ 0.05 + HPO4²⁻ 0.05
Expected output: pH 7.21

pKa2 = 7.208.

Ratio for pH 7.40 in phosphate

Input: pH 7.40; pKa2 = 7.21
Expected output: base/ácido 1.56

10^(7.40 − 7.208).

Ammonia and ammonium 1:1

Input: NH3 0.1 + NH4Cl 0.1; Kb = 1.8e-5
Expected output: pH 9.26

pOH = pKb = 4.745.

Design: acetate pH 5.00, 0.1 mol/L

Input: pH 5.00; pKa 4.757; total 0.1
Expected output: 0.0364 ácido + 0.0636 acetato

Base/acid ratio 1.75.

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

Can I mix the acid with its salt, or do I need acid and strong base?

Both routes work. Acetic acid with sodium acetate gives the buffer directly; acetic acid with half the amount of NaOH forms the acetate in solution. The tool works with the final concentrations of each form.

Why does the phosphate buffer have three pKa values?

Because phosphoric acid loses three protons in steps (pKa 2.16; 7.21; 12.32). Each step is a different buffering pair; the second, near physiological pH, is the most used.

Does a buffer pH change on dilution?

Little, because the base/acid ratio is kept, but buffer capacity falls. In extreme dilutions (below roughly 1 × 10⁻³ mol/L) water and ionization start to matter and the pH departs from Henderson-Hasselbalch.

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

Henderson-Hasselbalch, the useful range and when it fails

From the Ka expression one gets pH = pKa + log([A⁻] ÷ [HA]). With equal amounts the logarithm is zero and pH = pKa: 4.76 for the acetic acid and acetate pair, 9.26 for the ammonia and ammonium pair (from the ammonia pKb of 4.74). With 0.1 mol/L acetic acid and 0.05 mol/L acetate, the ratio is 0.5 and pH = 4.757 − 0.301 = 4.46. Each factor of 10 in the ratio shifts the pH by one unit, which is why the useful range is pKa ± 1, between ratios 1:10 and 10:1.

To design one, invert the equation: ratio = 10 to the power (target pH − pKa). An acetate buffer at pH 5.00 needs a ratio of 10^(5.00 − 4.757) = 1.75; with 0.1 mol/L in total, that is 0.0364 mol/L of acid and 0.0636 mol/L of acetate. For phosphate, the step that matters in biology is the second, H2PO4⁻ and HPO4²⁻ (pKa2 = 7.21): for pH 7.40 the base/acid ratio is 1.56. Picking a pair with a pKa within one unit of the target pH is the rule of thumb; the tool warns when the pair is a poor choice.

The equation assumes equilibrium concentrations equal the prepared ones. That fails when the buffer is very dilute or the pH is far from the pKa: with acetate and acetic acid at 1 × 10⁻⁴ mol/L each, the equation gives 4.76 but the exact balance gives 4.87. At 0.01 mol/L the difference drops to 0.001. That is why the tool shows both values and warns when they differ by more than 0.05. Buffer capacity, the amount of acid or base the buffer can take, grows with total concentration and peaks at a 1:1 ratio.

Buffer pH values and design ratios

Acetate 1:1

Input: ácido acético 0.1 + acetato 0.1
Expected output: pH 4.76

pKa = 4.757 (Ka = 1.75e-5).

Excess acid

Input: ácido acético 0.1 + acetato 0.05
Expected output: pH 4.46

Ratio 0.5: log = −0.301. Exact: 4.456.

Phosphate, step 2, 1:1

Input: H2PO4⁻ 0.05 + HPO4²⁻ 0.05
Expected output: pH 7.21

pKa2 = 7.208.

Ratio for pH 7.40 in phosphate

Input: pH 7.40; pKa2 = 7.21
Expected output: base/ácido 1.56

10^(7.40 − 7.208).

Ammonia and ammonium 1:1

Input: NH3 0.1 + NH4Cl 0.1; Kb = 1.8e-5
Expected output: pH 9.26

pOH = pKb = 4.745.

Design: acetate pH 5.00, 0.1 mol/L

Input: pH 5.00; pKa 4.757; total 0.1
Expected output: 0.0364 ácido + 0.0636 acetato

Base/acid ratio 1.75.

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

Can I mix the acid with its salt, or do I need acid and strong base?

Why does the phosphate buffer have three pKa values?

Because phosphoric acid loses three protons in steps (pKa 2.16; 7.21; 12.32). Each step is a different buffering pair; the second, near physiological pH, is the most used.

Does a buffer pH change on dilution?

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