ChemBench

Henderson-Hasselbalch Calculator

The pH of any acid/base buffer.

pKa

[A⁻] conjugate base (mol/L)

[HA] weak acid (mol/L)

4.76

What you entered

pKa: 4.76[A⁻] conjugate base (mol/L): 0.1[HA] weak acid (mol/L): 0.1

1
Ratio [A⁻] ÷ [HA]
0.1 ÷ 0.1= 1
2
pH = pKa + log₁₀([A⁻]/[HA])
4.76 + log₁₀(1)= 4.76

Result

pH: 4.76

This buffer has a pH of 4.76.

Buffer pH from pKa and the base-to-acid ratio

pH equals pKa plus the base-10 logarithm of conjugate base over weak acid. When the two are equal the logarithm is zero, so pH equals pKa exactly.

Buffer system	pKa	[Base]/[Acid]	pH
Acetic acid / acetate	4.76	1 : 2	4.46
Acetic acid / acetate	4.76	1 : 1	4.76
Acetic acid / acetate	4.76	2 : 1	5.06
Carbonic acid / bicarbonate	6.35	20 : 1	7.65
Dihydrogen phosphate	7.21	1 : 1	7.21
Dihydrogen phosphate	7.21	3 : 1	7.69
Ammonium / ammonia	9.25	1 : 4	8.65

A buffer works best within about one pH unit of its pKa, which is why the carbonate row matters: blood holds pH 7.4 on a pKa 6.35 buffer only because the body continuously vents CO2 to keep the ratio near 20:1.

pKa sets the baseline, the ratio adjusts it

When a buffer's conjugate base and weak acid are in equal concentration, pH exactly equals pKa — any imbalance in the ratio shifts pH up or down from that midpoint via the log term.

Why buffers resist pH change

This equation shows why buffers work: because pH depends on the log (not the raw ratio) of concentrations, even a several-fold change in the acid/base ratio only shifts pH by about 1 unit — this is what makes blood and lab buffers so stable.

Frequently asked questions

I need a buffer at pH 5.0 using acetic acid (pKa 4.76) — what ratio of acetate to acetic acid do I need?

pH = pKa + log([A-]/[HA]) → 5.0 = 4.76 + log(ratio) → log(ratio) = 0.24 → ratio = 1.74. Use 1.74 mol acetate for every 1 mol acetic acid.

How is this different from the pH calculator?

The pH calculator converts a known [H+] to pH. Henderson-Hasselbalch predicts pH from a weak acid's pKa and its conjugate base/acid ratio — it is specifically for buffer solutions, not strong acids or pure water.

Does this equation work for strong acids like HCl?

No. Henderson-Hasselbalch assumes a weak acid/conjugate base equilibrium. Strong acids dissociate completely and do not form buffers. For strong acid pH, just use pH = -log[H+] in the pH calculator.

What happens outside the buffer range?

Buffers work best within about 1 pH unit of the pKa. Beyond that range, one component is nearly exhausted and the solution loses its buffering capacity — adding a small amount of acid or base then causes large pH swings.

Can I use Henderson-Hasselbalch for bases?

Yes. For a base buffer, use pKb and the conjugate acid/base ratio to find pOH, then convert: pH = 14 - pOH. Alternatively, use the pKa of the conjugate acid directly.