pH Calculator — Strong & Weak Acid, Strong & Weak Base

The pH scale measures the free hydrogen-ion activity in aqueous solution. This calculator handles strong acids, strong bases, weak acids (via Ka or pKa), and weak bases (via Kb or pKb) — with the full mass-balance quadratic that keeps very dilute solutions in the physical range.

pH Calculator

Pick a mode and enter concentration (plus pKa or pKb for weak acids/bases). Solved with the full quadratic — no dilute-limit surprises.

Mode

All four modes are shown in Show the work above once a compute succeeds. Kw = 1.0 × 10⁻¹⁴ at 25 °C (CRC Handbook 97e).

Definitions

pH = −log₁₀[H⁺] and pOH = −log₁₀[OH⁻]. At 25 °C, K_w = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴, so pH + pOH = 14.00.

Strong acids and bases

For a strong acid at C ≳ 10⁻⁶ M, essentially every molecule ionises, so [H⁺] ≈ C and pH ≈ −log₁₀(C). Below that concentration, water’s own autoionisation matters and the full mass balance is needed: [H⁺]² − C·[H⁺] − K_w = 0. The positive root [H⁺] = (C + √(C² + 4K_w))/2 is what this calculator uses uniformly — no branch on concentration, no slip out of the physical range for 10⁻⁸ M HCl.

Weak acids — the ICE quadratic

For a weak acid HA with dissociation constant KaK_a, let x = [H⁺] = [A⁻]. Mass balance says [HA] = C − x, and Ka=x2C−xK_a = \dfrac{x^2}{C - x}. Rearranging gives x2+Kax−KaC=0x^2 + K_a x - K_a C = 0, whose positive root is x=−Ka+Ka2+4KaC2x = \dfrac{-K_a + \sqrt{K_a^2 + 4 K_a C}}{2}. Then pH = −log₁₀(x). Weak bases follow the same form in [OH⁻].

When the ICE approximation breaks

The exact quadratic above always solves correctly, but three regimes warrant care:

  • Dissociation exceeds ~5 % — the acid is behaving like a strong acid at that concentration. The calculator warns; the answer is still right.
  • Ka⋅C≲10KwK_a \cdot C \lesssim 10 K_w — water autoionization contributes to [H+][\mathrm{H^+}]. Uncommon for typical bench concentrations.
  • Multiple ionizable protons (H₂SO₄, H₃PO₄, H₂CO₃) — treat each pKa separately, or handle as a coupled system. This calculator does the single-ionization case; use it iteratively for polyprotic acids at your discretion.

Reference pKa values (25 °C)

AcidpKaKa
Trichloroacetic0.662.2 × 10⁻¹
Chloroacetic2.871.35 × 10⁻³
Hydrofluoric (HF)3.147.2 × 10⁻⁴
Formic (HCOOH)3.751.78 × 10⁻⁴
Acetic (CH₃COOH)4.761.74 × 10⁻⁵
Carbonic (pKa₁)6.354.45 × 10⁻⁷
Phosphoric (pKa₂)7.206.3 × 10⁻⁸
Ammonium (NH₄⁺)9.245.75 × 10⁻¹⁰
Carbonic (pKa₂)10.334.7 × 10⁻¹¹
Phosphate (pKa₃)12.354.5 × 10⁻¹³

Values from Harris QCA 10e Table 9-2 and the CRC Handbook 97th ed. dissociation-constant tables.

Common mistakes

  1. Using −log⁡(C)-\log(C) for very dilute strong acids. Below 10−610^{-6} M, water’s autoionization contributes. The full quadratic keeps you in the physical range.
  2. Confusing KaK_a and pKa\mathrm{p}K_a. KaK_a is dimensionless (activity ratio); pKa=−log⁡10Ka\mathrm{p}K_a = -\log_{10} K_a. This calculator accepts pKa/pKb — enter the value, not its negative log.
  3. Applying pKa to a base. A weak base uses KbK_b (or pKb). If you only have pKa of the conjugate acid, convert: pKa+pKb=pKw=14.00\mathrm{p}K_a + \mathrm{p}K_b = \mathrm{p}K_w = 14.00 at 25 °C.
  4. Ignoring temperature. KwK_w roughly doubles every 15 °C. All values in this calculator are 25 °C — meaningful for cold-room work.

Practice problems

Attempt each on paper, then expand the worked solution to check your arithmetic and sig-figs.

Problem 1 Easy What is the pH of 0.010 M HCl?

Answer: pH = 2.00

Strong acid — direct log
C (HCl)
0.010 M
Type
Strong acid
  1. Formula
  2. Substitute
  3. Note

    At 0.010 M the water contribution to [H⁺] is negligible (Kw ≪ C²) so the simple pH = -log(C) form is accurate to well under 0.01 pH units.

  4. Result
Problem 2 Easy What is the pH of 0.10 M NaOH?

Answer: pH = 13.00

Strong base — via pOH
C (NaOH)
0.10 M
Type
Strong base
  1. Formula
  2. Substitute
  3. Result
Problem 3 Intermediate What is the pH of 0.10 M acetic acid (pKa = 4.76)?

Answer: pH ≈ 2.87

Weak acid — ICE approximation
C
0.10 M
Ka
1.74 × 10⁻⁵
  1. Formula
  2. Substitute
  3. Note

    Dissociation is about 1.3 %, well inside the 5 % rule so the simple x = √(Ka·C) form matches the exact quadratic to 0.01 pH.

  4. Result
Problem 4 Intermediate What is the pH of 0.050 M ammonia (pKb = 4.75)?

Answer: pH ≈ 10.97

Weak base — ICE in [OH⁻]
C
0.050 M
Kb
1.78 × 10⁻⁵
  1. Formula
  2. Substitute
  3. Result

Frequently asked questions

How do I calculate pH from a molar concentration?
For a strong acid like HCl, pH = -log₁₀[H⁺], and [H⁺] equals the acid's molar concentration when the solution is not extremely dilute. For weak acids, use the ICE-table quadratic Ka = x²/(C - x) to solve for [H⁺]. This calculator does that for you and warns when the ICE approximation breaks down.
What's the difference between pH and pOH?
pH measures the concentration of H⁺; pOH measures the concentration of OH⁻. At 25 °C in water, pH + pOH = 14.00 because Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴. Calculating either gives you the other.
Why does the calculator show a warning at very low concentrations?
Below about 10⁻⁶ M, water's own autoionization contributes meaningfully to [H⁺]. Naively taking pH = -log(C) for 10⁻⁸ M HCl would give pH = 8, which is basic — physically wrong for an acid. This calculator solves the full mass-balance quadratic so the answer stays in the physical range even at dilute limits.
What Ka or Kb should I use for a weak acid or base?
Textbook Ka and pKa values are 25 °C values at infinite dilution. Real solutions at bench ionic strength drift by 0.1–0.2 pKa units — fine for cell culture and general lab work, matters for careful analytical work. The presets in this tool ship the common values from Harris QCA 10e Table 9-2 and the CRC Handbook 97e.

Sources

  • Harris, Quantitative Chemical Analysis, 10th ed. (2020), §8-1 through §8-4
  • Zumdahl & Zumdahl, Chemistry, 10th ed. (2018), §14 — pH definitions and pKa/pKb
  • CRC Handbook of Chemistry and Physics, 97th ed. (2016) — Ionization Constant of Water
  • IUPAC Gold Book — pH

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