H+ to pH Converter

Agarapu Ramesh — Editor and content reviewer

Convert between hydrogen ion concentration and pH, then see pOH, [OH-], the acid or base classification (acidic, neutral, or basic), the autoionisation product Kw, and a colour-coded pH scale marker. Useful for chemistry homework, lab calculations, water quality checks, and biochemistry buffer planning where you need to move quickly between molarity and logarithmic pH values.

What can you enter?

ModeInputOutput
[H+] to pHHydrogen ion concentration in mol/LpH, pOH, [OH-], acid/base class
pH to [H+]pH value such as 2, 7, or 11.35[H+] concentration, pOH, [OH-], acid/base class

Formula used

pH = -log10[H+]
[H+] = 10^(-pH)
pH + pOH = 14 at 25 C

Worked examples

If [H+] = 1.0e-3 M, pH = -log10(1.0e-3) = 3. The solution is acidic.

If pH = 7, [H+] = 10^-7 M and pOH = 7. This is neutral at 25 C.

If pH = 11, [H+] = 10^-11 M. The solution is basic because [H+] is low and [OH-] is higher.

How the pH scale behaves

The pH scale is logarithmic. A change of 1 pH unit means a tenfold change in hydrogen ion concentration. pH 3 has ten times more H+ than pH 4 and one hundred times more H+ than pH 5.

pH[H+] MCommon interpretation
11e-1Very acidic
31e-3Acidic
71e-7Neutral at 25 C
111e-11Basic
141e-14Very basic

Where this converter is useful

Common mistakes

How to use the pH result in lab work

Use the pH and concentration result as a consistency check before writing a lab answer. First confirm that hydrogen ion concentration is entered in mol per litre. Then compare the pH class with the expected chemistry: strong acids should normally sit at low pH, pure water is near pH 7 at 25 C, and basic solutions should show low [H+] with higher [OH-]. If your calculated value contradicts the solution type, the most likely causes are a missing exponent, a concentration entered in the wrong unit, or a pH value typed into the concentration mode.

For buffered solutions, remember that this converter shows the direct pH and ion relationship. It does not model buffer capacity, activity coefficients, temperature shifts, or mixed acid-base equilibria. For homework, it is ideal for checking logarithm steps. For lab records, use it alongside measured pH, calibration notes, and the actual solution preparation method.

Related Chemistry Tools

pH and pOH CalculatorHenderson-Hasselbalch CalculatorKa, Kb, and pKa ConverterMolarity CalculatorDilution Calculator

H+ to pH Converter FAQs

How to convert from pH to H+?

pH is defined as the negative logarithm (base 10) of the H+ ion concentration in mol/L. To go from pH to [H+], simply take the antilog: [H+] = 10−pH. Example 1: pH = 3 → [H+] = 10-3 = 1 × 10-3 M. Example 2: pH = 4.5 → [H+] = 10-4·5 ≈ 3.16 × 10-5 M. The lower the pH, the higher the [H+] — note that a pH change of 1 unit means a 10-fold change in H+ concentration. That is why even small pH changes can have huge biological consequences. [H+] = 10− pH mol L-1

How do you convert H+ to pH?

Take the negative logarithm of [H+]: pH = −log10[H+]. Example 1: [H+] = 1 × 10-5 M → pH = −log(10-5) = 5. Example 2: [H+] = 2 × 10-3 M → pH = −log(2 × 10-3) = 3 − log 2 = 3 − 0.301 = 2.7. Always express [H+] in mol/L before taking the log. Pure water has [H+] = 10-7 → pH = 7 (neutral). Acidic solutions have pH < 7, basic solutions have pH > 7. pH = − log10 [H+]

How much H+ to decrease pH from 7 to 6?

At pH 7, [H+] = 10-7 M. At pH 6, [H+] = 10-6 M. The increase needed = 10-6 − 10-7 = 9 × 10-7 M. So in 1 litre, you need to add 9 × 10-7 mol of H+ ions, which is about 0.9 µmol. That is an extremely small amount, illustrating how little acid is required to shift pH by one unit in unbuffered water. In real solutions with buffers present, much more acid would be needed because the buffer absorbs added H+.

How to calculate 2.5 pH to H+?

Apply [H+] = 10−pH. So [H+] = 10−2.5. Compute: 10-2 × 10-0·5 = 0.01 × 0.3162 ≈ 3.16 × 10-3 M, or 0.00316 mol/L. To get this, you can split the exponent: 10−2.5 = 10−3 × 100.5 = 0.001 × 3.162 = 3.162 × 10-3 M. This [H+] corresponds to a moderately acidic solution — typical of black coffee or vinegar diluted twenty-fold. [H+] at pH 2.5 = 10−2.5 ≈ 3.16 × 10-3 mol L-1