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pH Calculator

ChemistrySolutions and Concentration

For a strong acid or base, the molarity of H+ or OH- it releases (e.g. 0.02 M for 0.01 M Ba(OH)2).

pH —
Solution is
—
pOH
—
[H+]
—
[OH-]
—
Position on the 0–14 pH scale
The chart appears with your result

pH is the negative base-10 logarithm of the hydrogen ion concentration: pH = −log10[H+]. For 0.01 M hydrochloric acid, which dissociates completely, [H+] is 0.01 mol/L and the pH is 2.00. At 25 °C pH and pOH add to 14, so the same solution has a pOH of 12.00.

About this tool

Chemistry students checking homework, lab technicians preparing buffers and teachers writing answer keys all need the same conversion: from a concentration to pH. This calculator covers six starting points. Give it the molarity of a strong acid or strong base, a weak acid with its Ka, a weak base with its Kb, or a measured [H+] or [OH-], and it returns pH, pOH, both ion concentrations in scientific notation, and whether the solution is acidic, neutral or basic. Weak acids and bases are solved with the full quadratic rather than the square-root shortcut, so dilute solutions stay correct, and strong acids include water's own ions, so 1 × 10^-8 M HCl reads 6.98 rather than an impossible 8. The limit: everything assumes 25 °C and ideal behaviour, so above roughly 1 mol/L the figure departs from what a pH meter shows.

How to use it

  1. Pick what you know

    Choose strong acid, strong base, weak acid, weak base, or a known [H+] or [OH-] from the list.

  2. Enter the concentration

    Type the molarity in mol/L. For Ba(OH)2 or H2SO4-style species, enter the hydroxide or proton molarity released.

  3. Add Ka or Kb for weak species

    The dissociation constant box appears only for weak acids and bases; scientific notation such as 1.8e-5 is accepted.

  4. Read pH and the scale

    The result shows pH, pOH, both ion concentrations, the working, and a marker on the 0–14 scale.

Examples

0.01 M hydrochloric acid

What you know
Strong acid molarity
Concentration
0.01

Result pH: 2.00
Solution is: Acidic
pOH: 12.00
[H+]: 1.00 × 10^-2 mol/L
[OH-]: 1.00 × 10^-12 mol/L

  1. [H+] = (C + √(C² + 4Kw)) ÷ 2 = 1.00 × 10^-2 mol/L
  2. pH = −log10(1.00 × 10^-2 mol/L) = 2
  3. pOH = 14 − pH = 12

HCl gives up every proton, so [H+] equals the molarity and pH is −log10(0.01) = 2.00.

0.1 M acetic acid, Ka 1.8 × 10^-5

What you know
Weak acid + Ka
Concentration
0.1
Dissociation constant (Ka or Kb)
1.8E-5

Result pH: 2.88
Solution is: Acidic
pOH: 11.12
[H+]: 1.33 × 10^-3 mol/L
[OH-]: 7.50 × 10^-12 mol/L

  1. Solve x² ÷ (C − x) = Ka with C = 1.00 × 10^-1 mol/L, Ka = 0.000018
  2. x = (−Ka + √(Ka² + 4·Ka·C)) ÷ 2 = 1.33 × 10^-3 mol/L = [H+]
  3. Percent ionised = x ÷ C × 100 = 1.33%
  4. pH = −log10(1.33 × 10^-3 mol/L) = 2.88
  5. pOH = 14 − pH = 11.12

Solving the quadratic gives [H+] = 1.33 × 10^-3 M, so only about 1.3% of the acid ionises and the pH is 2.88.

[H+] of 1 × 10^-7 M

What you know
[H+] concentration
Concentration
1.0E-7

Result pH: 7.00
Solution is: Neutral
pOH: 7.00
[H+]: 1.00 × 10^-7 mol/L
[OH-]: 1.00 × 10^-7 mol/L

  1. [H+] = 1.00 × 10^-7 mol/L (given)
  2. pH = −log10(1.00 × 10^-7 mol/L) = 7
  3. pOH = 14 − pH = 7

This is pure water at 25 °C: [H+] equals [OH-] and both pH and pOH are 7.00.

How it is calculated

pH = −log10[H+]; pOH = 14 − pH; weak acid: [H+] = (−Ka + √(Ka² + 4·Ka·C)) ÷ 2; strong acid: [H+] = (C + √(C² + 4Kw)) ÷ 2

[H+]
hydrogen (hydronium) ion concentration, mol/L
C
analytical concentration of the acid or base, mol/L
Ka, Kb
acid or base dissociation constant
Kw
ion product of water, 1.0 × 10^-14 at 25 °C

For a weak acid HA the equilibrium x² ÷ (C − x) = Ka is solved exactly for x = [H+]; weak bases use the same equation with Kb, giving [OH-]. Strong acids and bases are treated as fully dissociated, with water's autoionisation added so very dilute solutions approach pH 7 instead of crossing it. When [OH-] is known, [H+] = Kw ÷ [OH-]. All values assume 25 °C and treat concentration as activity.

When not to use it

  • Buffers containing both an acid and its conjugate base need the Henderson–Hasselbalch equation instead.
  • Polyprotic acids such as phosphoric acid have several Ka steps that this single-equilibrium model ignores.
  • Concentrated solutions above about 1 mol/L, where activity coefficients matter, are not modelled.

Common mistakes

  • Entering the formula concentration of Ca(OH)2 instead of twice it for the hydroxide released.
  • Typing pKa where Ka is expected; convert with Ka = 10^(−pKa) first.
  • Assuming pH + pOH = 14 at body or boiling temperatures, where Kw is different.

Frequently asked questions

Why does 1 × 10^-8 M HCl give a pH just below 7 rather than 8?

Water itself supplies about 1 × 10^-7 M of H+, which is ten times more than the acid adds. Ignoring it gives pH 8, which would mean adding acid made water basic. Including Kw in the quadratic gives [H+] of about 1.05 × 10^-7 M and a pH of 6.98.

How is a weak acid's pH different from a strong acid's at the same molarity?

A strong acid releases all its protons, so 0.1 M HCl has pH 1.00. Acetic acid at 0.1 M with Ka 1.8 × 10^-5 ionises only about 1.3%, giving [H+] near 1.33 × 10^-3 M and pH 2.88, almost two units higher.

Can pH be negative or above 14?

Yes. A 2 M strong acid gives a calculated pH of −0.30. In such concentrated solutions the ions do not behave ideally, so measured values differ, and the calculator flags any result outside 0–14.

What do I enter for a weak base like ammonia?

Pick the weak base option, enter the ammonia molarity and its Kb, 1.8 × 10^-5 at 25 °C. The calculator solves for [OH-], converts it to [H+] with Kw, and reports a pH of about 11.13 for 0.1 M ammonia.

Why not use the square-root approximation x = √(Ka·C)?

That shortcut assumes x is tiny compared with C. It fails for fairly strong weak acids or very dilute ones, sometimes by tenths of a pH unit. The exact quadratic costs nothing extra here, so it is always used.

Does temperature change the result?

Kw rises with temperature, to about 5.5 × 10^-14 at 50 °C, so neutral pH falls to roughly 6.6. This calculator fixes Kw at 1.0 × 10^-14, the 25 °C value used in most textbooks and exam questions.

Cite this tool

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APA

Tools Directory Editorial Team. (2026, October 17). pH Calculator. Tools Directory. Retrieved October 27, 2026, from https://tools.milunir.com/chemistry/solutions-and-concentration/ph-calculator/

MLA

Tools Directory Editorial Team. “pH Calculator.” Tools Directory, 17 Oct. 2026, https://tools.milunir.com/chemistry/solutions-and-concentration/ph-calculator/. Accessed 27 Oct. 2026.

Chicago

Tools Directory Editorial Team. “pH Calculator.” Tools Directory. Last modified October 17, 2026. Accessed October 27, 2026. https://tools.milunir.com/chemistry/solutions-and-concentration/ph-calculator/.

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