Buffer Capacity Calculator
Compute a monoprotic buffer's capacity β at a chosen pH using the Van Slyke equation, or run it in reverse to find the concentration a buffer actually needs to hit a target capacity — with a distinct visual for each mode and full step-by-step work.
Background
Buffer capacity β = d(nacid/base)/d(pH), per liter, measures how strongly a buffer resists a pH change. For a monoprotic HA/A⁻ pair at 25 °C, βbuffer = 2.303·C·(Ka[H⁺])/(Ka+[H⁺])², which peaks right at pH = pKa and falls off the further you move away. Because this formula is exactly linear in C, it can be solved in reverse: given a target β at a chosen pH, dividing by the capacity a 1 mol/L solution of the same buffer would produce there gives the concentration you actually need — a genuinely practical formulation question most buffer capacity tools skip entirely.
How to use this calculator
- Choose Compute Capacity to find β from a known concentration, or Required Concentration to work backward from a target β.
- Provide either pKa or Ka directly — whichever you have on hand.
- Fill in the mode-specific fields (concentration and pH, or target β and pH).
- Decide whether to include the water contribution βwater — it matters most at very low or very high pH.
- Click Calculate to see the result, the matching visual, and the full step-by-step work.
How this calculator works
The Henderson–Hasselbalch equation splits the total concentration C into [HA] and [A⁻] at the working pH, using the ratio R = 10^(pH−pKa).
The Van Slyke equation then gives the buffer's own contribution to capacity, βbuffer, which is exactly proportional to C at any fixed pH — doubling the concentration exactly doubles the capacity.
Because βbuffer is proportional to C, computing the capacity a 1 mol/L solution would produce at that pH gives a conversion factor: dividing any target β by that factor gives the concentration needed to reach it.
The optional water term βwater depends only on pH, not on the buffer at all — it's largest at very acidic or very basic pH, and negligible near neutral pH.
If a target β already includes water, that fixed water contribution is subtracted off first before solving for the required buffer concentration — and if water alone already meets or exceeds the target, no added buffer is needed at all.
Formulas & Equations Used
Henderson–Hasselbalch: pH = pKa + log₁₀([A⁻]/[HA])
Split at pH: R = 10^(pH−pKa), [A⁻] = C·R/(1+R), [HA] = C/(1+R)
Van Slyke (buffer pair): βbuffer = 2.303·C·(Ka[H⁺])/(Ka+[H⁺])²
Water term (optional): βwater = 2.303·([H⁺]+[OH⁻]), giving βtotal = βbuffer + βwater
Required concentration (reverse): C = (βtarget − βwater) / k, where k = 2.303·Ka[H⁺]/(Ka+[H⁺])² is the capacity a 1 mol/L buffer would give
Universal peak fact: at pH = pKa, k always equals 2.303/4 ≈ 0.576, the largest capacity-per-concentration any monoprotic buffer can reach
Example Problems & Step-by-Step Solutions
Example 1 — Acetate at its pKa
pKa=4.76, C=0.100 M, pH=4.76.
Step: R=1, so [HA]=[A⁻]=0.0500 M. k=2.303/4=0.5758.
Result: βbuffer = 0.100×0.5758 = 0.0576 mol·L⁻¹·pH⁻¹.
Example 2 — Moving away from pKa
Same acetate buffer, but pH=6.76 (2 units above pKa).
Step: k drops sharply since [H⁺] ≪ Ka now.
Result: βbuffer falls to roughly 1/40th of its value at pH=pKa — capacity collapses fast away from pKa.
Example 3 — Required concentration, on target
pKa=4.76, target βbuffer=0.05, pH=4.76.
Step: k=0.5758 at this pH. C = 0.05/0.5758.
Result: C ≈ 0.0868 M is the minimum concentration needed.
Example 4 — Required concentration, off target
Same target β=0.05, but pH=6.76 (2 units off pKa).
Step: k is now about 25× smaller, so C must be about 25× larger.
Result: C ≈ 2.21 M — a concentration that's rarely even achievable, showing why buffers are chosen near their pKa.
Example 5 — When water dominates
pKa=3.75, C=0.001 M (1 mM), pH=3.75.
Step: βbuffer≈0.000576, but βwater≈0.000410 at this acidic pH.
Result: water supplies over 40% of the total capacity — far too large to ignore for such a dilute buffer.
Example 6 — An impossible target
pKa=7.00, target βtotal=1×10⁻⁶, pH=2.00.
Step: at pH=2, βwater alone is about 0.023 — already far above the target.
Result: no buffer needed — water alone already exceeds this target at that pH.
Frequently Asked Questions
When should I include the water term?
Include it whenever you're working at a very low or very high pH, or with a fairly dilute buffer — both make βwater a meaningful fraction of the total. Near pH≈pKa with a reasonably concentrated buffer, it's usually small enough to ignore.
Which concentration do I enter for C?
Use the total buffer pair concentration, C = [HA] + [A⁻], at the pH you care about — not just one side of the pair.
What does "required concentration" actually solve for?
It answers a practical formulation question: given a target buffering strength at a specific pH, how much of this buffer pair do you actually need to dissolve? Most buffer calculators only run the forward direction, even when they use "target" in their name.
Why would a target be "impossible" to reach with added buffer?
If your target already includes water's own contribution, and that water contribution alone meets or exceeds the target at that pH, then no amount of added buffer is needed — asking for less capacity than water already provides isn't a real constraint.
Why does capacity peak exactly at pH = pKa?
The Van Slyke formula's denominator, (Ka+[H⁺])², is smallest relative to the numerator exactly when [H⁺]=Ka, which is precisely pH=pKa. At that point, the capacity-per-concentration factor always equals 2.303/4, regardless of which buffer you're using.
Does this handle polyprotic buffers?
This tool uses a monoprotic model. For a polyprotic system, sum the Van Slyke term for each relevant dissociation step separately.
All calculators