Henderson-Hasselbalch Calculator

Calculate buffer pH from acid dissociation constant and molar concentrations.

Bidirectional solving with M, mM, and µM concentration units.

Updated September 2, 2026
Frank Zhao - Creator
CreatorFrank Zhao
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Understanding the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is one of the most widely used relationships in acid–base chemistry. It links the pH of a buffer solution to the pKa of the weak acid and the molar concentrations of the conjugate base and acid.

What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation describes the relationship between pH and the ratio of conjugate base to weak acid in a buffer solution. It is derived from the acid dissociation equilibrium expression and the definition of pH:

pH=pKa+log10 ⁣([A][HA])\mathrm{pH} = \mathrm{pK_a} + \log_{10}\!\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)

Enter any three of the five variables — pH, pKa, Ka, [A], [HA] — and the calculator instantly derives the remaining two, with automatic M / mM / µM unit conversion.

Biochemists

Enzyme assays & protein studies

Clinical chemists

Blood gas panels & diagnostics

Pharmacologists

Drug formulations & delivery

Students

Acid–base equilibrium courses

The equation is bidirectional: you can compute pH from the concentrations, or work backwards to find the required concentration ratio for a target pH.

How to use this calculator

Enter any three of the five variables and the calculator solves for the remaining two.

1

Choose your input mode

Find pH: enter Ka (or pKa), [A], and [HA]. Find concentration: enter pH, pKa, and one concentration. Find pKa: enter pH and both concentrations.

2

Select concentration units

Use the unit dropdown next to each field. You can mix units — for example, enter [A] in mM and [HA] in M. The calculator normalizes everything internally.

3

Read the results

Computed fields appear in blue. You can overwrite any blue field by typing a new value — the calculator recomputes accordingly. The Ka and pKa fields are always linked.

Worked example

Problem — acetic acid / sodium acetate buffer

Ka of acetic acid = 1.8×1051.8 \times 10^{-5}

A buffer is prepared by mixing 0.10 M sodium acetate (NaCH₃COO, the conjugate base) and 0.05 M acetic acid (CH₃COOH). What is the pH of this buffer?

1

Convert Ka to pKa

Step 1

pKa=log10(1.8×105)\mathrm{pK_a} = -\log_{10}(1.8 \times 10^{-5})==4.744.74

Enter Ka = 1.8e-5 — the calculator automatically computes pKa ≈ 4.74.

2

Apply the Henderson-Hasselbalch equation

Step 2

pH=4.74+log10 ⁣(0.100.05)\mathrm{pH} = 4.74 + \log_{10}\!\left(\frac{0.10}{0.05}\right)

Simplify

=4.74+log10(2)= 4.74 + \log_{10}(2)==4.74+0.3014.74 + 0.301==5.045.04

Enter [A] = 0.10 and [HA] = 0.05. The calculator returns pH ≈ 5.04.

Result

The buffer pH is 5.04, which is slightly above the pKa of 4.74 — consistent with having more conjugate base than acid (ratio = 2:1). This makes the buffer slightly more effective at neutralizing added acid than added base.

Formulas and variables

Buffer equation

pH=pKa+log10 ⁣([A][HA])\mathrm{pH} = \mathrm{pK_a} + \log_{10}\!\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)

Given any three of pH, pKa, [A], [HA], the calculator solves for the remaining one.

Dissociation constant

pKa=log10(Ka)\mathrm{pK_a} = -\log_{10}(\mathrm{K_a})

Converts between Ka and pKa. Entering one automatically derives the other.

Variable definitions

pHpH

Solution pH

−log₁₀[H⁺]

pKa\mathrm{pK_a}

pKₐ

−log₁₀(Kₐ)

Ka\mathrm{K_a}

Kₐ

dissociation constant

[A][\mathrm{A^-}]

[A⁻]

conjugate base (M)

[HA][\mathrm{HA}]

[HA]

weak acid (M)

This calculator uses a constraint-based solver. The Ka↔pKa pair and the Henderson-Hasselbalch equation are solved simultaneously — entering Ka, [A], and [HA] yields both pKa and pH in one computation. The solver works in any direction.

Tips and common mistakes

Do not confuse Kₐ with pKₐ

Kₐ = 1.8×10⁻⁵ and pKₐ = 4.74 are the same information in different scales. Entering 4.74 where Kₐ is expected (or vice versa) gives wildly wrong results. Use the labeled fields.

Mix units freely — the calculator converts

You can enter [A⁻] in mM and [HA] in M. The calculator converts both to M internally. Just make sure the unit dropdown matches the number you typed.

pH = pKₐ at the midpoint

When [A⁻] = [HA], the log ratio is zero and pH equals pKₐ. This is the point of maximum buffering capacity — deviations from this midpoint reduce the buffer's ability to resist pH changes.

Use scientific notation for small Kₐ values

Most weak acid Kₐ values are tiny (e.g., 1.8×10⁻⁵). Enter them in scientific notation like 1.8e-5. Do not type the leading zeros.

Overwriting a blue field triggers recomputation

Blue fields are derived values. Typing a new number into one converts it to a user input and forces the calculator to re-solve the remaining fields. Use this to explore "what if" scenarios.

Frequently asked questions

Q.What is a buffer solution?

A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid (HA) and its conjugate base (A⁻), or a weak base (B) and its conjugate acid (HB⁺). Buffers are essential in biological systems (blood pH ≈ 7.4), chemical synthesis, and analytical chemistry.

Q.What is the difference between Ka and pKa?

Ka is the acid dissociation constant — a measure of how strongly an acid dissociates in water. pKa is simply −log₁₀(Ka). A smaller Ka (larger pKa) means a weaker acid. For example, acetic acid has Ka ≈ 1.8×10⁻⁵ and pKa ≈ 4.74. The pKa scale is more intuitive: acids with pKa < 0 are strong acids, while those with pKa > 14 are extremely weak.

Q.When is the Henderson-Hasselbalch equation most accurate?

The equation works best when (1) the concentrations of HA and A⁻ are much larger than [H⁺] or [OH⁻], (2) neither HA nor A⁻ is extremely dilute, and (3) the solution is dilute enough that activity coefficients are close to 1. In practice, it is accurate to within about 0.01 pH units for most buffer systems at concentrations between 0.01 M and 1 M.

Q.Can pKa be negative?

Yes. For very strong acids with Ka > 1 (e.g., HCl with Ka ≈ 10⁷), the pKa is negative. The calculator accepts negative pKa values without restriction.

Q.What does the ratio [A⁻]/[HA] tell me?

When [A⁻]/[HA] = 1, the pH equals the pKa — this is the midpoint of the buffer region where buffering capacity is highest. When the ratio > 1, there is more conjugate base than acid, and pH > pKa. When the ratio < 1, there is more acid, and pH < pKa.

Limitations

  • Ideal solution assumption. The equation assumes activity coefficients are 1. At high ionic strength (> 0.1 M), activity corrections become significant and the calculated pH may deviate from the actual value.
  • Dilute solutions. At very low concentrations (< 0.001 M), water autoionization contributes significantly to [H⁺] and the equation becomes inaccurate.
  • Temperature dependence. Ka (and therefore pKa) changes with temperature. The values used should match the temperature of the buffer solution. For precise work at non-standard temperatures, use temperature-corrected Ka tables.
  • Single acid–base pair. The equation applies to buffers containing a single weak acid and its conjugate base. For polyprotic acids or mixed buffer systems, more complex equilibrium calculations are needed.
Henderson-Hasselbalch Calculator: Buffer pH from Ka and Concentrations