Chemistry

pH Calculator

Calculate pH and pOH for strong/weak acids and bases and simple buffers (Henderson–Hasselbalch).

pH / acid–base calculator

Monoprotic strong/weak acid–base · buffer (25 °C, Kw = 1.0×10⁻¹⁴)

pH formula strip

pH = −log₁₀[H⁺] · pOH = −log₁₀[OH⁻] · pH + pOH = 14 (25 °C, Kw = 10⁻¹⁴)

Henderson–Hasselbalch (buffer approx.): pH = pKa + log₁₀([A⁻]/[HA])

pH

2.875

pOH

11.12

[H⁺] (M)

0.001333

[OH⁻] (M)

7.504e-12

Ka = x² / (C − x)

  • Solved with acid mass balance, charge balance, and water autoionization.

What is the pH Calculator?

pH measures the acidity of an aqueous solution on a logarithmic scale using the pH formula pH = −log₁₀[H⁺] (more precisely, activity of H₃O⁺, approximated by concentration in dilute solutions). Strong acids and bases dissociate essentially completely; weak acids and bases only partially, so Ka or Kb is required. Buffers contain a weak acid and its conjugate base and resist pH change; their pH is estimated with the Henderson–Hasselbalch equation pH = pKa + log₁₀([A⁻]/[HA]).

The pH Calculator covers five classroom cases at 25 °C (Kw = 1.0×10⁻¹⁴): strong acid, strong base, weak acid, weak base, and a simple HA/A⁻ buffer, with presets for acetate, phosphate-like, and ammonium buffers. Strong electrolytes include water’s autoionization so extremely dilute solutions do not report nonsense pH values far past 7. The tool compares the Henderson–Hasselbalch estimate to a charge-balance solver for buffers.

Always match the mode to the chemistry. Acetic acid is weak (use Ka); HCl is strong. A mixture of acetic acid and sodium acetate is a buffer, not a single weak-acid problem.

  • Strong acid/base: start from complete dissociation; include water when C is tiny
  • Weak acid/base: solve Ka or Kb = x²/(C−x) with the quadratic
  • Buffer: pH = pKa + log₁₀([A⁻]/[HA])

Mathematical / chemical formulas

Core relations used by the calculator (25 °C):

pH = −log₁₀[H⁺]
pOH = −log₁₀[OH⁻]
[H⁺][OH⁻] = Kw = 1.0×10⁻¹⁴

Weak acid: Ka = x² / (C − x)
Weak base: Kb = x² / (C − x)
Buffer: pH = pKa + log₁₀([A⁻]/[HA])
pKa = −log₁₀(Ka)
  • Polyprotic acids and activity corrections are outside this tool’s scope.
  • If Ka or Kb ≥ 1, treat the species as strong instead.

Step-by-step example: pH of 0.10 M acetic acid (Ka = 1.8×10⁻⁵)

Find the pH of 0.10 M CH₃COOH using the weak-acid quadratic.

  1. Ka = x²/(0.10 − x) = 1.8×10⁻⁵.
  2. x = (−Ka + √(Ka² + 4 Ka C))/2 ≈ 1.33×10⁻³ M.
  3. pH = −log₁₀(1.33×10⁻³) ≈ 2.88.
  4. Check: x/C ≈ 1.3% < 5%, so the x ≪ C shortcut would also be roughly OK here.

Choose Weak acid, concentration 0.10, Ka 1.8e-5 in the pH Calculator. Expect pH near 2.88.

Frequently asked questions

What is the Henderson–Hasselbalch equation used for?

It estimates buffer pH from pKa and the ratio of conjugate base to weak acid: pH = pKa + log₁₀([A⁻]/[HA]). Use it for classroom buffer problems and phosphate/acetate-style examples. The calculator also solves a fuller charge-balance model so you can see when the approximation is excellent.

Why is the pH of 1.0×10⁻⁸ M HCl not 8?

Water contributes [H⁺] as well. A charge-balance treatment gives a pH slightly below 7, not an alkaline value. The strong-acid mode includes that correction.

When is Henderson–Hasselbalch valid?

When both HA and A⁻ are present at concentrations much larger than [H⁺] and [OH⁻], typically in the 0.01–1 M range for common buffers. It is an approximation, not an exact charge-balance solution.

How do Ka and Kb relate for a conjugate pair?

Ka × Kb = Kw at the same temperature. If you know Ka for acetic acid, Kb for acetate is Kw/Ka.

References & further reading

Standards bodies, university open courseware, and peer-reviewed references that align with the methods used on this page.

Keep learning with more calculators and study guides on Online Science Tools.

Practice problems & worked examples

Practice alongside the pH calculator above. Each problem includes a full worked solution so you can check your reasoning step by step.

Practice problem 1

Strong acid pH

What is the pH of 0.010 M HCl?

Show solution

Worked solution

  1. Strong acid: [H⁺] ≈ 0.010 M.
  2. pH = −log₁₀(0.010) = 2.00.

Answer: pH = 2.00

Practice problem 2

Weak acid quadratic

Estimate the pH of 0.10 M acetic acid (Ka = 1.8×10⁻⁵).

Show solution

Worked solution

  1. x ≈ √(Ka C) = √(1.8×10⁻⁶) ≈ 1.34×10⁻³ (shortcut).
  2. Quadratic gives x ≈ 1.33×10⁻³; pH ≈ 2.88.

Answer: pH ≈ 2.88

Practice problem 3

Buffer pH

A buffer has [HA] = 0.10 M and [A⁻] = 0.10 M with Ka = 1.8×10⁻⁵. Find pH.

Show solution

Worked solution

  1. pKa = 4.74.
  2. pH = pKa + log([A⁻]/[HA]) = 4.74 + log(1) = 4.74.

Answer: pH ≈ 4.74

Practice problem 4

Phosphate-like buffer

For a buffer with pKa = 7.20, [HA] = 0.050 M, [A⁻] = 0.050 M, estimate pH with Henderson–Hasselbalch.

Show solution

Worked solution

  1. pH = 7.20 + log(0.050/0.050) = 7.20.

Answer: pH ≈ 7.20

Practice problem 5

pH formula check

If [H⁺] = 3.0×10⁻⁴ M, what are pH and pOH at 25 °C?

Show solution

Worked solution

  1. pH = −log(3.0×10⁻⁴) ≈ 3.52.
  2. pOH = 14 − 3.52 ≈ 10.48.

Answer: pH ≈ 3.52; pOH ≈ 10.48

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