CalcOpenly
English

pH calculator for acids, bases and buffers

Calculate pH, pOH, [H⁺] and [OH⁻] from a concentration, a strong acid or base, a weak acid's Ka or pKa, or a Henderson–Hasselbalch buffer.

Drafted with AI writing toolsUpdated Checked examples: 8

= 0.000018
Try
pH
pH: 2.88
Decimal places: 2; Nearest, ties to even
pOH
11.12
[H⁺]
0.00133267mol/L
[OH⁻]
7.50373 × 10⁻¹²mol/L
Solution is
Acidic
pH with the √(Ka·C) shortcut
2.872
Shortcut error in [H⁺]
0.67%
Fraction ionised
1.33%
Uses Kw = 1.0 × 10⁻¹⁴ (25 °C) and concentrations in place of activities.

pH 2.88 (acidic): [H⁺] = 0.001333 mol/L and [OH⁻] = 7.504 × 10⁻¹² mol/L; the acid is 1.33% ionised, and the √(Ka·C) shortcut would be off by 0.67%.

pH scale

Strongly acidicAcidicNear neutralBasicStrongly basic03681114neutral, pure waterpH 2.88
How it's calculated S
  1. Equilibrium HA ⇌ H⁺ + A⁻

    Ka=x2C−x⇒x2+Kax−KaC=0,Ka=0.000018, C=0.1K_a = \frac{x^2}{C - x} \Rightarrow x^2 + K_a x - K_aC = 0,\quad K_a = 0.000018,\ C = 0.1
  2. Exact root

    x=−Ka+Ka2+4KaC2=0.00133267 mol/Lx = \frac{-K_a + \sqrt{K_a^2 + 4K_aC}}{2} = 0.00133267\ \mathrm{mol/L}

    Water's own ionisation is neglected: fine while [H⁺] ≫ 10⁻⁷ mol/L.

  3. Compare the shortcut

    KaC=0.00134164⇒pH≈2.8724 (0.673% high in [H+])\sqrt{K_aC} = 0.00134164 \Rightarrow \text{pH} \approx 2.8724\ (0.673\%\ \text{high in }[\mathrm{H^+}])
  4. pOH and [OH⁻] at 25 °C

    pH=2.87528,pOH=14−pH=11.1247,[OH−]=7.50373×10−12 mol/L\text{pH} = 2.87528,\quad \text{pOH} = 14 - \text{pH} = 11.1247,\quad [\mathrm{OH^-}] = 7.50373 \times 10^{-12}\ \mathrm{mol/L}

About the pH calculator

pH is the negative base-10 logarithm of the hydrogen-ion concentration, pH = −log₁₀[H⁺]. At 25 °C the ion product of water, Kw, is 1.0 × 10⁻¹⁴, so pH + pOH = 14. Strong acids and bases are treated as fully dissociated, with water's own ions included; weak acids are solved exactly from Ka with the quadratic x² + Ka·x − Ka·C = 0; buffers use the Henderson–Hasselbalch equation.

The default, 0.1 mol/L acetic acid with Ka = 1.8 × 10⁻⁵, gives pH 2.88 with 1.33 % of the acid ionised. The √(Ka·C) shortcut taught in class gives pH 2.872, because it overstates [H⁺] by 0.67 %.

Concentrations stand in for activities, which works best for dilute solutions. Results below pH 0 or above 14 are possible for concentrated acids and bases, but there the simple formula is only an estimate.

How to calculate pH from concentration

pH is the negative base-10 logarithm of the hydrogen-ion concentration: pH = −log₁₀[H⁺]. A solution with [H⁺] = 0.001 mol/L has pH 3.00, and at 25 °C its pOH is 14 − 3 = 11.00. Strong acids give [H⁺] directly; weak acids also need their Ka.

Each pH unit is a factor of ten

Because pH is a logarithm, pH 3 has ten times the hydrogen ions of pH 4 and a thousand times those of pH 6. To go back from pH to concentration, use [H⁺] = 10^(−pH): pH 7 is 10⁻⁷ mol/L.

Hydroxide follows the same pattern, pOH = −log₁₀[OH⁻]. At 25 °C the two ions are tied by [H⁺][OH⁻] = Kw = 1.0 × 10⁻¹⁴, which is why pH + pOH = 14.

pH of a strong acid or base

A strong acid dissociates completely, so [H⁺] is its concentration times the protons each unit releases. 0.01 M HCl gives [H⁺] = 0.01 mol/L and pH 2.00.

For a strong base, find pOH first. 0.05 M NaOH gives [OH⁻] = 0.05 mol/L, so pOH = 1.30 and pH = 14 − 1.30 = 12.70.

Below about 10⁻⁶ mol/L, water's own ions start to matter. 1 × 10⁻⁸ M HCl has pH 6.98, not pH 8: solving [H⁺]² − C[H⁺] − Kw = 0 gives [H⁺] = 1.05 × 10⁻⁷ mol/L.

Only the first proton of sulfuric acid is fully strong. Counting two protons for 0.01 M H₂SO₄ gives pH 1.70, which is an upper bound on its acidity.

pH of a weak acid from Ka

A weak acid HA only partly ionizes (OpenStax). If C is the starting concentration and x is [H⁺] at equilibrium, Ka = x²/(C − x), which rearranges to x² + Ka·x − Ka·C = 0.

For 0.1 M acetic acid with Ka = 1.8 × 10⁻⁵:

  1. Take the positive root: x = (−Ka + √(Ka² + 4·Ka·C)) ÷ 2 = 1.333 × 10⁻³ mol/L.
  2. pH = −log₁₀(1.333 × 10⁻³) = 2.88.
  3. The fraction ionized is x ÷ C = 1.33%.

The classroom shortcut x ≈ √(Ka·C) gives pH 2.872 here, overstating [H⁺] by only 0.67%, because just 1.33% of the acid ionizes.

The shortcut fails when much of the acid ionizes. For 0.001 M hydrofluoric acid (Ka = 6.8 × 10⁻⁴), the quadratic gives pH 3.26 with 55% ionized, while the shortcut gives 3.08.

Since pKa = −log₁₀Ka, Ka = 10^(−pKa). Tables round differently: Ka = 1.8 × 10⁻⁵ is pKa 4.74, while the often-quoted pKa 4.76 is Ka = 1.74 × 10⁻⁵. Use the constant your problem states.

Buffer pH with the Henderson–Hasselbalch equation

A buffer mixes a weak acid with its conjugate base: pH = pKa + log₁₀([A⁻]/[HA]). Equal concentrations give pH = pKa, so a 1:1 acetate buffer with pKa 4.76 sits at pH 4.76.

Doubling the base, [A⁻] = 2[HA], raises the pH to 5.06, and a tenfold excess gives 5.76. Diluting both parts equally keeps the pH but lowers how much added acid or base the buffer can absorb.

The equation treats the stated concentrations as equilibrium values. It is reliable when both are far above Ka and the base-to-acid ratio stays between 0.1 and 10 (OpenStax buffers).

Assumptions behind the numbers

Kw = 1.0 × 10⁻¹⁴ is the 25 °C value from the IUPAC Gold Book, so neutral water is pH 7 only at that temperature. At 50 °C, pKw is 13.26 and neutral water has pH 6.63.

Concentrations stand in for ion activities, which suits dilute solutions. A 2 M strong acid gives pH −0.30 by the formula, but a pH electrode in such a solution reads a different value. See OpenStax on pH and pOH for the definitions.

pH of four typical solutions

Default weak acid is 0.1 M acetic acid with Ka = 1.8 × 10⁻⁵. Hydrogen ions 0.001 M is a known [H⁺]. Dilute strong acid is 1 × 10⁻⁸ M HCl, and equal buffer is 0.1 M of each component with pKa 4.76.

ScenariopH
Default weak acid2.88
Hydrogen ions 0.001 M3.00
Dilute strong acid6.98
Equal buffer4.76

Common mistakes

  • Using a weak acid's concentration as [H⁺]. 0.1 M acetic acid is pH 2.88, not pH 1.
  • Ignoring water in a very dilute acid. Adding acid never makes water basic: 10⁻⁸ M HCl is pH 6.98.
  • Mixing Ka and pKa from different tables. 1.8 × 10⁻⁵ and pKa 4.76 are slightly different constants.
  • Treating a pH change as small. A fall from pH 5 to pH 3 is a hundredfold rise in [H⁺].
  • Counting every proton of a polyprotic acid as strong. Sulfuric acid's second proton is only partly dissociated.
  • Using pH + pOH = 14 at another temperature. The sum is 14 at 25 °C.

Units and constants

  • Concentrations can be in mol/L, mmol/L or µmol/L; [H⁺] and [OH⁻] are reported in mol/L.
  • For strong acids and bases, set the ions released per formula unit from 1 to 3, such as 2 for Ba(OH)₂.
  • Weak-acid strength can be entered as Ka or pKa; buffers use pKa.
  • A pH from −2 to 16 converts back to [H⁺] and [OH⁻].

To prepare a solution at a target concentration first, use the dilution and concentration calculator. The molar mass calculator turns grams of solute into moles.

Worked examples

[H⁺] = 0.001 mol/L

Start from
[H⁺] concentration
Concentration
0.001
Concentration unit
mol/L
pH
3.00
pOH
11.00
Solution is
Acidic

Checked against: −log₁₀(10⁻³) = 3; pOH = 14 − 3

pH 7 is neutral

Start from
pH value
pH
7
[H⁺]
1 × 10⁻⁷ mol/L
[OH⁻]
1 × 10⁻⁷ mol/L
Solution is
Neutral

Checked against: 10⁻⁷ mol/L at 25 °C (Kw = 10⁻¹⁴)

0.01 M HCl

Start from
Strong acid concentration
Concentration
0.01
Concentration unit
mol/L
H⁺ or OH⁻ released per formula unit
1
pH
2.00

Checked against: [H⁺] = (C + √(C² + 4Kw))/2 = 0.010000000001 ⇒ pH 2.0000

Very dilute HCl (1e-8 M) is still acidic

Start from
Strong acid concentration
Concentration
1e-8
Concentration unit
mol/L
H⁺ or OH⁻ released per formula unit
1
pH
6.98
[H⁺]
1.05125 × 10⁻⁷ mol/L

Checked against: [H⁺] = (10⁻⁸ + √(10⁻¹⁶ + 4×10⁻¹⁴))/2 = 1.0512492×10⁻⁷ ⇒ pH 6.97829 (not 8)

Questions

How do you calculate pH from concentration?

For a strong acid, [H⁺] equals the acid concentration times the protons each unit releases, and pH = −log₁₀[H⁺]: 0.01 M HCl has pH 2.00. For a strong base, take pOH = −log₁₀[OH⁻] and subtract it from 14: 0.005 M Ba(OH)₂ releases 0.01 M OH⁻, so pOH is 2 and pH is 12. Weak acids only partly ionise, so they also need Ka.

Why is the pH of pure water 7?

Water ionises slightly, H₂O ⇌ H⁺ + OH⁻, and at 25 °C the product [H⁺][OH⁻] equals Kw = 1.0 × 10⁻¹⁴. In pure water the two concentrations are equal, so each is 10⁻⁷ mol/L and the pH is 7. Kw grows with temperature: at 50 °C pKw is 13.26, so neutral water has pH 6.63. This calculator uses the 25 °C value.

What is the difference between Ka and pKa?

Ka is the acid dissociation constant, [H⁺][A⁻]/[HA] at equilibrium, and pKa = −log₁₀Ka. A lower pKa means a stronger acid, and each unit of pKa is a factor of 10 in Ka. Acetic acid has pKa 4.76, so Ka = 1.74 × 10⁻⁵ (often rounded to 1.8 × 10⁻⁵); hydrofluoric acid, Ka = 6.8 × 10⁻⁴, has pKa 3.17.

When is the Henderson–Hasselbalch equation accurate?

pH = pKa + log₁₀([A⁻]/[HA]) holds when the acid and its conjugate base are both far more concentrated than Ka and [H⁺], and their ratio lies between 0.1 and 10, which keeps the pH within one unit of pKa. Outside that ratio the buffer has little capacity left, and the calculator warns you. Equal concentrations give pH = pKa, so a 1:1 acetate buffer sits at 4.76.

Can pH be negative?

Yes. A 2 mol/L solution of a strong acid such as HCl has [H⁺] of about 2 mol/L, and −log₁₀2 gives pH −0.30. At concentrations like this, ion activities differ markedly from concentrations, so a pH electrode reads a different value. Treat any calculated pH below 0 or above 14 as an estimate rather than a measurement.

About this calculator

pH=−log⁡10[H+], pH+pOH=14;x2+Kax−KaC=0;pH=pKa+log⁡10[A−][HA]\text{pH} = -\log_{10}[\mathrm{H^+}],\ \text{pH} + \text{pOH} = 14;\quad x^2 + K_a x - K_aC = 0;\quad \text{pH} = \text{p}K_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}

Sources

  1. OpenStax Chemistry 2e, §14.2 pH and pOH; §14.3 Relative strengths of acids and bases; §14.6 Buffers
  2. IUPAC Gold Book — pH; ionic product of water (Kw = 1.0 × 10⁻¹⁴ at 25 °C)

Checked against references

This calculator includes 8 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.

Put this calculator on your site

Paste this where you want the calculator to appear. It is free to use, it stays in step with corrections made here, and the credit line below it links back so your readers can find the full version.

Height
Preview the embed

Keep the credit line. It is the only condition, and it is what keeps these calculators free.