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Chemistry Calculator Undergraduate

Henderson-Hasselbalch Calculator

Find buffer pH, pKa or the acid-to-base ratio using the Henderson-Hasselbalch equation, with a warning when the buffer runs out of capacity.

Calculator

7.40062

Phosphate 7.21, Tris 8.07, acetate 4.76, HEPES 7.48, bicarbonate 6.35, carbonate 10.33.

Working, with your numbers

  1. pH = pKa + log10([A-] / [HA])
  2. = 7.21 + log10(60.8 / 39.2)
  3. = 7.21 + log10(1.551)
  4. = 7.21 + 0.1906
  5. = 7.401

Values are converted into the units the equation is worked in before the arithmetic.

Base : acid ratio
1.551 : 1
Total buffer
100 mM
Fraction deprotonated
At pH = pKa this is 50%, which is where buffering is strongest.
60.8 %

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The equation

pH=pKa+log⁡10[A−][HA]\mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac{[A^-]}{[HA]}

Henderson (1908) and Hasselbalch (1917)

What the equation says about buffers

The Henderson-Hasselbalch equation, pH = pKa + log₁₀([A⁻]/[HA]), is the acid dissociation constant rearranged into logarithmic form. Its practical value is that pH depends on the ratio of conjugate base to weak acid, not on their absolute amounts. Diluting a buffer twofold halves both terms, the ratio is unchanged, and the pH stays where it was.

When the two species are present in equal amounts the logarithm is zero and pH = pKa. That is the point of maximum buffering, where the acid is exactly 50% deprotonated and a small addition of strong acid or base shifts the ratio least. Moving away from it costs capacity quickly: at pKa ± 1 the ratio is already 10:1 and the solution is about 91% one species, so there is little of the minor component left to absorb further additions.

Worked example

A phosphate buffer at pH 7.40 using the pKa 7.21 pair, made to 100 mM total:

  • Rearrange: log₁₀([A⁻]/[HA]) = 7.40 − 7.21 = 0.19
  • Take the antilog: [A⁻]/[HA] = 10^0.19 = 1.549
  • Split the total: [A⁻] = 100 × 1.549 / 2.549 = 60.8 mM
  • Remainder: [HA] = 100 − 60.8 = 39.2 mM

So 60.8 mM hydrogen phosphate against 39.2 mM dihydrogen phosphate. The ratio 1.549 sets the pH; the choice of 100 mM sets how much added acid or base the buffer can absorb before the pH moves appreciably.

Choosing the pair and the concentration

Select a buffer whose pKa lies within about 0.5 units of the target pH, and then treat concentration as a separate decision. Typical laboratory buffers run from 10 to 100 mM: higher concentrations resist pH change better but contribute more ionic strength, which matters for enzyme kinetics and electrophoresis.

The equation also assumes the concentrations of the two species equal the amounts weighed out. That assumption weakens when the buffer is very dilute or the target pH is far from the pKa, since dissociation of the acid and the water equilibrium then contribute measurably. Verify the final pH with a calibrated meter rather than trusting the calculation alone.

Common mistakes

  • Expecting total concentration to set the pH. A 10 mM and a 100 mM buffer at the same ratio have the same pH. Concentration governs capacity only.
  • Working outside pKa ± 1. Beyond that range the ratio exceeds 10:1 and buffering capacity has largely collapsed, so the solution behaves like a dilute acid or base.
  • Picking the wrong pKa for a polyprotic acid. Phosphoric acid has pKa values of 2.15, 7.21 and 12.32. Use the one nearest the target pH; substituting 2.15 for a pH 7.4 buffer gives a meaningless ratio.
  • Treating pKa as a fixed constant. It shifts with temperature and ionic strength, so a buffer titrated at room temperature will not hold the same pH at 4 °C or 37 °C.
Henderson-Hasselbalch Calculator: the equation pH = pK a + log₁₀([A⁻]/[HA]), solved for any of pH, pKa, [A⁻] and [HA].
The equation the calculator is built on, with its source. Image © ScienceQuest, CC BY 4.0. Free to reuse with credit and a link to this page; how to reuse it. Download PNG

Worked examples

Each one runs through the calculator above, so the arithmetic here is the arithmetic it does.

What is the pH of a buffer of 0.2 M acetic acid and 0.1 M acetate, pKa 4.76?

  1. pH = pKa + log10([A-] / [HA])
  2. = 4.76 + log10(100 / 200)
  3. = 4.76 + log10(0.5)
  4. = 4.76 - 0.301
  5. = 4.459

4.46, or 0.30 below the pKa, because the acid outnumbers the base two to one and log₁₀ 2 is 0.301. That step is worth knowing by heart: every doubling of the ratio moves the pH by 0.3 and every factor of ten by a whole unit, so most buffer answers can be checked without a calculator.

How much sodium acetate does a pH 5 buffer need with 0.1 M acetic acid, pKa 4.76?

  1. [A-] = [HA] x 10^(pH - pKa)
  2. = 100 x 10^(5 - 4.76)
  3. = 100 x 10^0.24
  4. = 173.78 mM

0.174 M, about 1.74 times the acid, since a pH above the pKa needs more base than acid. The exponent is pH minus pKa, and reversing it is the usual slip: 10 to the power of −0.24 gives 0.0575 M, a buffer that would sit at pH 4.52 rather than 5.

Common questions

Why is a buffer strongest when pH equals pKa?

At pH = pKa the acid and its conjugate base are present in equal amounts, so there is a maximum reserve of both to absorb added acid or base. Move one pH unit away and the ratio is already 10:1, leaving little of the minority species. Useful buffering is roughly pKa ± 1.

Does the total buffer concentration change the pH?

No, only the ratio sets the pH. Total concentration sets the capacity: a 100 mM buffer resists roughly ten times more added acid than a 10 mM buffer at the same pH. That is why the equation contains a ratio rather than absolute amounts.