pKa Calculation Practice Problems With Worked Answers

Use this pKa calculator to practice estimating acid–base equilibrium pH from pKa and the ratio of conjugate base to acid. Enter the pKa and the concentrations of HA and A⁻; the calculator returns pH using the Henderson–Hasselbalch equation, plus the ratio used in the calculation.

Formula

The Henderson–Hasselbalch equation estimates the pH of a buffer from its acid’s pKa and the ratio of conjugate base to weak acid:

pH = pKa + log₁₀([A⁻] / [HA])

• pH is the calculated acidity on the pH scale (unitless).
• pKa is the negative base-10 logarithm of the acid dissociation constant, Ka (unitless). It depends on the particular acid and conditions.
• [A⁻] is the concentration of the conjugate base, in mol/L.
• [HA] is the concentration of the weak acid, in mol/L.

The calculator uses concentrations in the ratio. Since the two concentrations have the same units, those units cancel. The ratio itself is also unitless. In a more rigorous treatment, activities replace concentrations; the concentration form is a useful approximation for many dilute aqueous buffer exercises.

Worked Example

For the calculator’s defaults, pKa = 4.76, [HA] = 0.10 mol/L, and [A⁻] = 0.10 mol/L.

1. Find the base-to-acid ratio: [A⁻]/[HA] = 0.10 mol/L ÷ 0.10 mol/L = 1.
2. Take its base-10 logarithm: log₁₀(1) = 0.
3. Substitute: pH = 4.76 + 0 = 4.76.

The calculated pH is 4.76, and the ratio is 1.00. This matches the calculator outputs to two decimal places.

Assumptions

• The solution is an aqueous buffer containing a weak acid and its conjugate base.
• The concentrations are used as approximations to activities; the solution is dilute enough for that approximation to be useful.
• The pKa entered applies to the acid and the solution conditions. No temperature adjustment is calculated.
• The acid–base pair is treated as a single relevant equilibrium, without additional reactions or other buffering species.

How to Solve pKa Practice Problems

First identify the conjugate pair. HA is the protonated acid; A⁻ is the form remaining after it loses a proton. Then locate the pKa for that specific acid under the conditions stated in the problem. Finally, make sure the information gives the ratio [A⁻]/[HA], either directly or through concentrations or mole amounts.

If a problem gives moles rather than concentrations, you can use the mole ratio when both forms occupy the same final solution volume: each concentration is its amount divided by that volume, so the common volume cancels. If the volumes differ or the amounts are measured before a reaction, determine the final amounts and conditions before using the equation.

• Equal acid and base amounts give a ratio of 1, so log₁₀(1) = 0 and pH = pKa.
• If [A⁻] is ten times [HA], log₁₀(10) = 1, so pH = pKa + 1.
• If [A⁻] is one-tenth [HA], log₁₀(0.1) = −1, so pH = pKa − 1.

These patterns are useful mental checks. Increasing the base-to-acid ratio raises the calculated pH; decreasing it lowers the pH. The pKa is not itself the pH in every solution—it equals the calculated pH specifically when the conjugate forms are present in equal amounts.

Practice Problems With Answers

Try each calculation before reading its answer. These examples use the Henderson–Hasselbalch approximation and assume that the supplied pKa is appropriate.

1. Equal amounts: An acid has pKa 6.35, and [A⁻]/[HA] = 1. Since log₁₀(1) = 0, pH = 6.35.
2. Base exceeds acid: pKa = 4.20, [A⁻] = 0.30 mol/L, and [HA] = 0.10 mol/L. The ratio is 3; log₁₀(3) ≈ 0.4771, so pH ≈ 4.6771, or 4.68.
3. Acid exceeds base: pKa = 9.25, [A⁻] = 0.020 mol/L, and [HA] = 0.20 mol/L. The ratio is 0.1; log₁₀(0.1) = −1, so pH = 8.25.
4. Find the ratio: A buffer has pH 5.10 and pKa 4.50. Rearranging gives log₁₀([A⁻]/[HA]) = 0.60. Therefore, [A⁻]/[HA] = 10^0.60 ≈ 3.98. There is about four times as much conjugate base as acid.

When the Equation Works—and When It Does Not

The equation is intended for a weak acid and its conjugate base in the same solution, especially in buffer calculations. It is most useful when both forms are present in appreciable amounts and the solution is not so concentrated that activity effects dominate. It is not a universal shortcut for finding the pH of any acid or base solution.

For a very dilute weak acid with little conjugate base initially present, the acid’s dissociation and water’s contribution may need to be handled with an equilibrium calculation instead. Strong acids and strong bases are generally treated using their own dissociation assumptions, not by inserting an arbitrary ratio into this buffer equation. Polyprotic acids have multiple pKa values and multiple equilibria; choose the relevant conjugate pair for the pH range and problem setup.

Real pH can also differ from this estimate because pKa and activity depend on temperature and solution composition. In laboratory work, the pH meter reading is the real check; temperature and ionic strength shift pH. Follow your lab method, the safety data sheet (SDS) and your supervisor.

Common Calculation Mistakes

• Reversing the ratio: The equation uses base over acid, [A⁻]/[HA]. Reversing it changes the logarithm’s sign.
• Using the wrong pKa: A molecule with multiple ionizable groups can have several pKa values. Use the one for the equilibrium represented by HA ⇌ H⁺ + A⁻.
• Mixing units inconsistently: Convert both concentrations to the same units before dividing. A ratio of concentrations in different units is not ready for the logarithm.
• Rounding too early: Keep extra digits through the ratio and logarithm, then round the final pH to a sensible precision.
• Assuming pKa determines pH alone: The ratio matters. At equal concentrations pH equals pKa, but unequal proportions shift the result.

Frequently Asked Questions

What is the pKa equation for calculating pH?
The Henderson–Hasselbalch equation is pH = pKa + log₁₀([A⁻]/[HA]). It estimates buffer pH from the acid’s pKa and the conjugate-base-to-acid ratio.

When does pH equal pKa?
When the conjugate base and weak acid have equal concentrations, their ratio is 1 and its logarithm is zero. The equation then gives pH = pKa.

How do I find pKa from pH and concentrations?
Rearrange the equation: pKa = pH − log₁₀([A⁻]/[HA]). Use the base-to-acid ratio and consistent concentration units.

Can I use moles instead of molar concentrations?
Yes, if both forms are in the same final volume, because the common volume cancels in their concentration ratio. Otherwise, account for the actual final concentrations.

Is pKa the same as pH?
No. pKa describes an acid’s dissociation equilibrium; pH describes a solution. They are equal in this equation only when [A⁻] and [HA] are equal.

Related

• pH Calculator From Concentration: pH = −log[H+]
• Monosodium Phosphate Buffer Calculator: NaH2PO4 and Na2HPO4 for a Target pH
• Tris Buffer pH Calculation: Henderson–Hasselbalch for Tris and Tris-HCl

in pka | pKa | - | 4.76 | - | -
in ha | Acid concentration, HA | mol/L | 0.10 | 0.000000000001 | -
in a | Conjugate-base concentration, A- | mol/L | 0.10 | 0.000000000001 | -
out ph | pH | - | pka + log10(a / ha) | 2
out ratio | A- to HA ratio | - | a / ha | 2