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pH, pOH and Buffer Calculator

Calculate the pH of strong and weak acids and bases, and of buffer solutions. You can also convert between pH, pOH, [H⁺] and [OH⁻], or between Ka, pKa, Kb and pKb. Weak acids and buffers are solved exactly, and the usual shortcuts are shown with a check of whether they are valid.

Chemistry · Acids and basesUpdated Sources cited

Calculation
Acid or base?
e.g. 0.1 or 1e-3
e.g. 1.8e-5 or, as pK, 4.74

Result

pH

2.88

pOH
11.12
[H⁺]
0.00133 mol/L
[OH⁻]
7.5 × 10⁻¹² mol/L
Percentage ionised
1.33%
Assumptions behind this result
  • Temperature 25 °C, so Kw = 1.0 × 10⁻¹⁴ and pH + pOH = 14.00.
  • Concentrations are used in place of activities (accurate for dilute solutions).

Step-by-step solution

  1. Step 1: Write the equilibrium expression

    A weak acid HA only partly dissociates: HA ⇌ H⁺ + A⁻. Let x be the equilibrium [H⁺] produced.

    Ka=x2C−x=1.8×10−5,C=0.1 mol/LK_a = \frac{x^2}{C - x} = 1.8 \times 10^{-5},\quad C = 0.1\ \text{mol/L}
  2. Step 2: The usual approximation

    If x is much smaller than C, then C − x ≈ C and x ≈ √(Ka·C). The check: x/C = 1.3%, which is below 5%, so the approximation is acceptable.

    x≈KaC=0.001342x \approx \sqrt{K_a C} = 0.001342
  3. Step 3: Solve exactly

    The calculator does not rely on the approximation. Solving the quadratic gives x = 0.001333; including water's own ionisation as well gives the final value.

    x3+Kax2−(KaC+Kw)x−KaKw=0⇒x=0.001333 mol/Lx^3 + K_ax^2 - (K_aC + K_w)x - K_aK_w = 0 \Rightarrow x = 0.001333\ \text{mol/L}
  4. Step 4: Percentage ionised

    Only 1.33% of the acid is ionised at equilibrium.

    α=[A−]C=1.333%\alpha = \frac{[\text{A}^-]}{C} = 1.333\%
  5. Step 5: Find the remaining quantities

    Use pH = −log[H⁺], [H⁺][OH⁻] = Kw and pH + pOH = 14.

    pH=2.88,pOH=11.12,[OH−]=7.504×10−12 mol/L\text{pH} = 2.88,\quad \text{pOH} = 11.12,\quad [\text{OH}^-] = 7.504 \times 10^{-12}\ \text{mol/L}

What the calculator does

  • Strong acids and bases: complete dissociation, including bases that release two hydroxide ions, such as Ba(OH)₂.
  • Weak acids and bases: the exact equilibrium pH from Ka or Kb (or pKa, pKb), with the percentage ionised.
  • Buffers: Henderson–Hasselbalch, checked against an exact calculation.
  • Conversions: between pH, pOH, [H⁺] and [OH⁻], and between Ka, pKa, Kb and pKb.

The formulas

pH=−log⁡10[H+],[H+][OH−]=Kw=1.0×10−14,pH+pOH=14.00\text{pH} = -\log_{10}[\text{H}^+], \quad [\text{H}^+][\text{OH}^-] = K_w = 1.0 \times 10^{-14}, \quad \text{pH} + \text{pOH} = 14.00

Weak acids

For HA ⇌ H⁺ + A⁻ with starting concentration C and equilibrium [H⁺] = x:

Ka=x2C−xK_a = \frac{x^2}{C - x}

If x is small compared with C, then x ≈ √(Ka·C). The calculator does not rely on this: it solves the equilibrium exactly, including the contribution of water, and reports whether the approximation would have been acceptable.

Buffers: the Henderson–Hasselbalch equation

pH=pKa+log⁡10[A−][HA]\text{pH} = \text{p}K_a + \log_{10}\frac{[\text{A}^-]}{[\text{HA}]}

This works best when the ratio [A⁻]/[HA] is between 0.1 and 10 and both concentrations are much larger than Ka.

Worked example: 0.10 M acetic acid

Acetic acid has Ka = 1.8 × 10⁻⁵.

  1. Approximation: x ≈ √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ M, which is 1.3% of C. That is below 5%, so the approximation is valid.
  2. Exact solution: [H⁺] = 1.333 × 10⁻³ M.
  3. pH: −log(1.33 × 10⁻³) = 2.88.
  4. Percentage ionised: 1.33%. Most of the acid stays as HA molecules.

Common mistakes

  • Treating a weak acid as strong. For 0.10 M acetic acid, assuming [H⁺] = 0.10 M gives pH 1, not the correct 2.88. (Textbooks often quote 2.87, from the square-root approximation; the exact value is 2.875.)
  • Using the approximation when x is not small. For fairly strong weak acids or dilute solutions, x/C can exceed 5%.
  • Confusing Ka and pKa when entering values. The calculator lets you choose which you are giving.
  • Forgetting that pH + pOH = 14 only at 25 °C. Kw changes with temperature.

Practical applications

  • Biology and medicine. Blood is held at pH 7.35–7.45 by the carbonic acid–bicarbonate buffer.
  • Food and drink. Acidity affects taste, preservation and safety.
  • Environmental science. Acid rain, ocean acidification and soil pH.
  • Laboratory work. Preparing buffers for experiments and titrations.

Assumptions and limitations

All calculations are for water at 25 °C (Kw = 1.0 × 10⁻¹⁴) and use concentrations in place of activities. That is accurate for dilute solutions, roughly below 0.1 mol/L, and less so above. Polyprotic acids, such as phosphoric acid with its successive dissociations, are not yet supported. For sulfuric acid the second dissociation is not complete, so “2 H⁺ per formula” overestimates its acidity.

Frequently asked questions

Why isn't the pH of 10⁻⁸ M HCl equal to 8?

An acid solution cannot be basic. At such low concentration, the H⁺ from water itself (about 10⁻⁷ M) is larger than the H⁺ from the acid. Including both gives pH 6.98, slightly acidic as it must be.

When can I use the "x is small" approximation for weak acids?

When the ionised fraction is below about 5% of the starting concentration. The calculator checks this for you. It always reports the exact result, so the approximation never affects the answer.

Why is the buffer result different from Henderson–Hasselbalch?

Henderson–Hasselbalch assumes the acid and base concentrations barely change on dissociation. That fails for very dilute buffers or fairly strong acids. The calculator solves the full charge balance and warns when the two differ.

What is the relationship between Ka and Kb?

For an acid and its conjugate base, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C, so pKa + pKb = 14.00.

Can pH be negative?

Formally yes, for very concentrated strong acids, but at such concentrations activities differ greatly from concentrations and pH = −log[H⁺] no longer describes the solution well.

References

  1. Buck, R. P. et al. (2002). Measurement of pH. Definition, standards, and procedures (IUPAC Recommendations 2002). Pure and Applied Chemistry, 74(11), 2169–2200.
  2. Harris, D. C. & Lucy, C. A. (2020). Quantitative Chemical Analysis (10th ed.), chapters 8–9. W. H. Freeman.
  3. Atkins, P., Jones, L. & Laverman, L. (2016). Chemical Principles (7th ed.), chapter 6. W. H. Freeman.