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Henderson-Hasselbalch Practice: pH, pKa, Ratio, Concentrations

Practice buffer chemistry problems using the Henderson–Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]). Create multiple scenarios, choose what to solve for, and optionally check your own answers.

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Formulas verified by Abbas Kalim Khan, Associate Scientist

Build Your Practice Set

pH = pKa + log₁₀([A⁻]/[HA])

[A⁻] = conjugate base concentration, [HA] = weak acid concentration

Scenarios (1)

Scenario 1

Provide pKa and ratio [A⁻]/[HA] (or both concentrations)

We'll compare your answer to the computed value

For learning and homework practice only; not for clinical, pharmaceutical, or industrial use.

Practice the Henderson–Hasselbalch Equation

Build one or more buffer scenarios by specifying pKa, pH, ratios, or concentrations. Choose what you want to solve for and optionally enter your own answer to check it.

pH = pKa + log₁₀([A⁻]/[HA])

We'll apply the Henderson–Hasselbalch equation for each scenario.

For learning and homework practice only; not for clinical, pharmaceutical, or industrial use.

What This Page Is For

Textbook Henderson-Hasselbalch problems give you pKa and the conjugate-base ratio, then ask for pH. Real exams flip that and ask you to back-solve for pKa from a measured pH and a known ratio. Same equation, but the algebra runs in the other direction and the answer's sensitivity to small errors changes. A 0.1-unit error in pH translates to roughly 25% error in the ratio when solving forward, but a 0.1-unit error in pKa when solving backward. This page generates both flavors and checks your work. The derivation and worked acetate recipe live on the Buffer Maker page.

The variants matter. Real exams don't always hand you a clean ratio and ask for pH. They give you grams of weak acid and grams of conjugate-base salt, or moles after a strong-acid spike, or pH and pKa and want a ratio. Each phrasing tests the same equation but breaks students at a different step. The point of this generator is to expose you to all of them, not to re-teach HH from scratch.

How to Attack HH Problems on an Exam

A few habits will save you marks on every HH problem. Write the equation down first, before plugging numbers in: pH = pKa + log([A⁻]/[HA]). Always that direction. Conjugate base over weak acid. Flip the ratio and you'll be wrong by 2 × log(ratio), which on a pKa of 4.76 with a 3:1 mix is almost a full pH unit.

Sign of the log term is the next checkpoint. pH greater than pKa means more A⁻ than HA. pH lower than pKa means more HA than A⁻. If your computed ratio contradicts that intuition, you have a sign error.

Last habit: moles and concentrations are interchangeable inside the log. Since [A⁻]/[HA] = n(A⁻)/n(HA) when both species share the same solution volume, you don't have to convert to molarity. That saves a step and an error opportunity.

Variants the Generator Throws at You

The generator rotates through four standard problem shapes. The basic one gives you both concentrations plus pKa and asks for pH. A close variant gives you pH and pKa and wants the ratio (or one concentration if the other is known). Then there's the harder version where you start with moles, a strong-acid or strong-base spike gets added, and you have to recompute the new pH after the stoichiometric shift converts some HA to A⁻ (or vice versa).

That last one catches most students. Strong acid converts A⁻ to HA mole for mole. Strong base does the opposite. You can't just plug fresh numbers into HH after an addition; you have to update the moles first, then apply the equation. If your work disagrees with the answer-check by more than 0.02 pH, look at the strong-acid/base bookkeeping before you blame the calculator.

Where Theory and Operational Tools Live

This page does practice. It doesn't re-derive Henderson-Hasselbalch from first principles. The Buffer Maker page does that with worked acetate examples and an explanation of why the ratio cancels volume. If you're rusty on the derivation, read that first and come back.

For operational buffer questions (what's the new pH after I add a specific number of moles of HCl, what's β at this composition), the Buffer Capacity & pH Drift Estimator is the right tool. It's the difference between practicing the equation on synthetic numbers and actually computing a real buffer's response to a perturbation. Both are useful. They're separate pages because they answer separate questions.

Sources

Frequently Asked Questions

What is the Henderson–Hasselbalch equation used for?
The Henderson–Hasselbalch equation is used to calculate the pH of buffer solutions, determine the ratio of conjugate base to weak acid needed for a specific pH, and understand the behavior of weak acid/base systems in chemistry and biochemistry. The equation is: pH = pKa + log₁₀([A⁻]/[HA]), where [A⁻] is the conjugate base concentration and [HA] is the weak acid concentration. It's essential for: (1) Preparing buffer solutions at specific pH values, (2) finding fractional dissociation from pH and pKa, (3) predicting pH after adding strong acid or base, and (4) planning pH-controlled experiments. Above 0.1 M, activity coefficients drift from 1 and the simple ratio form loses accuracy.
When is the Henderson–Hasselbalch equation most accurate?
The equation is most accurate for dilute buffer solutions (typically 0.001–0.1 M) where ionic strength effects are minimal, when the buffer pH is within ±1 unit of the pKa (the effective buffer range), and when the weak acid is not too strong (pKa > 2) or too weak (pKa < 12). Outside these conditions, the equation becomes less accurate due to: (1) Activity coefficient effects at high ionic strengths, (2) One buffer component being very small (low buffering capacity), (3) Water's contribution to pH at extreme values, (4) Non-ideal solution behavior. The equation assumes ideal behavior with activity coefficients ≈ 1, which is valid for dilute solutions. Above 0.5 M total buffer concentration, the Davies or Debye-Hückel correction shifts effective pKa by 0.1 to 0.3 units, large enough to throw off pH targeting in working buffers.
What happens when pH equals pKa?
When pH = pKa, the Henderson–Hasselbalch equation gives: pKa = pKa + log([A⁻]/[HA]), so log([A⁻]/[HA]) = 0, which means [A⁻]/[HA] = 1, or [A⁻] = [HA]. This means the acid is 50% dissociated—half is in the acid form (HA) and half is in the conjugate base form (A⁻). This is the midpoint of a titration curve and the point of maximum buffer capacity. At this point, the buffer can neutralize equal amounts of added acid or base most effectively.
How do I solve for pKa using this equation?
Rearrange the equation: pKa = pH − log([A⁻]/[HA]). You need to know the pH and either the ratio or both concentrations. If you have concentrations, first calculate the ratio: [A⁻]/[HA] = [A⁻] / [HA]. Then calculate pKa: pKa = pH - log(ratio). For example, if pH = 5.0 and [A⁻] = 0.2 M, [HA] = 0.1 M, then ratio = 2.0, and pKa = 5.0 - log(2.0) = 5.0 - 0.301 = 4.70. Input these values in the calculator and select 'Solve for pKa' to get the answer.
Why doesn't this calculator account for temperature?
This is an educational tool that assumes standard conditions (25°C). In reality, pKa values change with temperature according to relationships like: pKa(T) = pKa(25°C) + (ΔH/2.303R) × (1/T - 1/298.15), where ΔH is the enthalpy change and R is the gas constant. For precise work, you would need temperature-corrected pKa values, which is beyond the scope of basic Henderson–Hasselbalch problems. Most textbook problems assume 25°C, so this calculator follows that convention.
Can I use this for polyprotic acids?
The Henderson–Hasselbalch equation applies to each dissociation step of a polyprotic acid separately. For example, phosphoric acid (H₃PO₄) has three pKa values: pKa₁ = 2.15 (H₃PO₄ ⇌ H⁺ + H₂PO₄⁻), pKa₂ = 7.20 (H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻), pKa₃ = 12.35 (HPO₄²⁻ ⇌ H⁺ + PO₄³⁻). You would use the appropriate pKa for the pH range you're working in. For pH around 7, use pKa₂ with H₂PO₄⁻ as HA and HPO₄²⁻ as A⁻.
What tolerance is used for checking my answers?
For pH and pKa values, answers within ±0.02 units are considered correct. For concentrations and ratios, answers within ±3% (relative) or ±0.001 (absolute) are accepted. These tolerances account for rounding differences and minor calculation variations. For example, if the correct pH is 5.06, answers between 5.04 and 5.08 are accepted. If the correct ratio is 2.0, answers between 1.94 and 2.06 (within 3%) are accepted. The tolerance window matters because two honest rounding paths can land a few hundredths apart.
Why are my calculated concentrations different from expected?
Ensure you're using consistent units (all in M, mM, or µM). The calculator treats all concentration inputs as the same unit. Also check that you've provided enough information—you need at least two knowns to solve for the third. Common issues: (1) Mixing units (e.g., M and mM), (2) Using wrong ratio direction ([HA]/[A⁻] instead of [A⁻]/[HA]), (3) Not providing enough information to solve, (4) Calculation errors in manual work. The calculator shows all calculated values—compare them to your expectations to identify discrepancies.
How do I calculate percent ionization from this equation?
Percent ionization (% A⁻) = [A⁻]/([HA] + [A⁻]) × 100 = ratio/(1 + ratio) × 100. The calculator automatically shows percent acid and percent base in the results. At pH = pKa, the percent ionization is 50% (ratio = 1, so 1/(1+1) = 0.5 = 50%). When pH = pKa + 1 (ratio = 10), percent ionization = 10/(1+10) = 91%. When pH = pKa - 1 (ratio = 0.1), percent ionization = 0.1/(1+0.1) = 9%.

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