Buffer pH calculator

How do I calculate the pH of a buffer?

With the Henderson–Hasselbalch equation: pH = pKa + log₁₀([A⁻] ÷ [HA]), where [A⁻] is the conjugate base and [HA] the weak acid.

What if the acid and base are equal?

Then the log term is 0 and pH = pKa. For acetic acid and acetate, pKa is 4.76 at 25 °C (IUPAC data, via PubChem), so an equal mixture has pH 4.76.

What happens if I double the base?

The pH rises by log₁₀ 2 ≈ 0.30: a 2 : 1 acetate-to-acetic-acid buffer has pH 5.06.

When does the equation work well?

For buffers where both forms are present in similar amounts — roughly a ratio between 0.1 and 10, so pH within about 1 unit of pKa — and the approximations used to derive it hold.

Why do sources give different pKa values?

Tables round differently: OpenStax lists Ka = 1.8 × 10⁻⁵ for acetic acid (pKa 4.74), while the IUPAC value is 4.76. Use the pKa your course or lab specifies.

How do I make a buffer with a particular pH?

Pick a weak acid whose pKa is within about one unit of the pH you want, then rearrange the equation: [A⁻] ÷ [HA] = 10^(pH − pKa). For pH 5.00 with acetic acid (pKa 4.76) the ratio is 10^0.24 = 1.74, so mix about 1.74 mol of acetate with each mole of acetic acid; entering that pair in the calculator gives pH 5.00. Weak acids and their salts work best below pH 7, weak bases and their salts above 7 (OpenStax Chemistry 2e, section 14.6).

Does diluting a buffer change its pH?

The equation depends only on the ratio [A⁻] ÷ [HA], so the calculated pH does not change when both are diluted equally. What shrinks is the buffer capacity, the amount of strong acid or base it can absorb before the pH moves by about a unit. OpenStax compares 1.0 M acetic acid with 1.0 M acetate against 0.10 M of each: same pH, but the stronger mixture has more capacity.

How the equation works

Start from the acid’s ionization constant, Ka = [H⁺][A⁻] ÷ [HA]. Take the negative logarithm of both sides and rearrange, and the result is pH = pKa + log₁₀([A⁻] ÷ [HA]), with pKa = −log₁₀ Ka. The inputs are the equilibrium concentrations of the conjugate base and the weak acid, which for a buffer made by mixing them are very close to what you put in. OpenStax Chemistry 2e (section 14.6) notes that the equation rests on the assumption that the amount of acid that ionizes is small compared with what is present, so it is only valid when that assumption holds. The pH is fixed by the ratio and the pKa, not by the total amount.

Worked example

OpenStax Example 14.20 uses 100 mL of a buffer that is 0.10 M in acetic acid and 0.10 M in sodium acetate. With pKa = 4.74 the ratio is 1, the logarithm is 0, and the pH is 4.74. Now add 1.0 mL of 0.10 M NaOH. That is 0.10 mmol of hydroxide, which converts the same amount of acetic acid into acetate: acetate goes from 10.0 to 10.10 mmol and acetic acid from 10.0 to 9.90 mmol. The new pH is 4.74 + log₁₀(10.10 ÷ 9.90) = 4.75, a change of only 0.01 units. The same dose of NaOH in unbuffered water at pH 4.74 raises the pH to 10.99 in the textbook example. Moles can replace concentrations here because both species share one volume.

Common mistakes

The most common slip is entering Ka where the calculator wants pKa: convert first with pKa = −log₁₀ Ka. Another is swapping the two concentrations, which flips the sign of the correction: a buffer with more base than acid has a pH above the pKa, not below it. A third is using the equation for a mixture of a strong acid and its salt, or for a polyprotic acid with the wrong pKa; pick the pKa of the dissociation step that actually sets the pH. Finally, remember that the equation gives the pH of the buffer as prepared; if you also add strong acid or base, first work out how much of each component is left, as in the example above.

Sources: OpenStax Chemistry 2e, section 14.6 “Buffers” (Henderson–Hasselbalch equation, buffer capacity, Example 14.20) and section 14.3 (Ka of acetic acid, 1.8 × 10⁻⁵); IUPAC pKa of acetic acid via PubChem. Related: pH calculator, ICE table calculator, molarity calculator and dilution calculator.