# pH calculator

> Convert pH, pOH, [H+] and [OH−]; find the pH of strong and weak acids and bases exactly; and design buffers with Henderson–Hasselbalch. 25 °C. Free.

URL: https://www.labcanvas.com/tools/ph-calculator

pH = −log₁₀[H⁺] and, at 25 °C, pH + pOH = 14. This calculator converts between pH, pOH, [H⁺] and [OH⁻], gives the pH of strong and weak acids and bases (solving the full equilibrium, including water), and calculates buffer pH or the base-to-acid ratio for a target pH with the Henderson–Hasselbalch equation.

## Quick answer

- **What it does:** Converts between pH, pOH, [H⁺] and [OH⁻], gives the exact pH of strong and weak acids and bases, and designs buffers with Henderson–Hasselbalch, at 25 °C.
- **Formula:** pH = −log₁₀[H⁺]; Kw = 1.0 × 10⁻¹⁴; weak acid: [H⁺]³ + Ka[H⁺]² − (KaC + Kw)[H⁺] − KaKw = 0; buffer pH = pKa + log₁₀([A⁻]/[HA]).
- **Example:** 0.1 M acetic acid (pKa 4.76) gives [H⁺] = 1.31 × 10⁻³ M, so pH = 2.88 with 1.3 % ionised.

Free, in the browser, no account needed. Updated 2026-11-08.

## Features

- pH ⇄ pOH ⇄ [H⁺] ⇄ [OH⁻]
- Strong acids and bases, including very dilute ones
- Weak acids and bases from \(\mathrm{p}K_\mathrm{a}\) (exact solution)
- Buffer pH and ratio for a target pH
- \(\mathrm{p}K_\mathrm{a}\) presets: acetate, phosphate, Tris, HEPES…

## How to use it

1. **Choose a calculator**: Convert a value, or calculate for a strong acid/base, a weak acid/base or a buffer.
2. **Enter the values**: Type the concentration and, for weak acids, bases and buffers, the \(\mathrm{p}K_\mathrm{a}\) (or pick a common one).
3. **Read the result**: Get the pH with pOH, [H⁺] and [OH⁻], plus the fraction ionised or the buffer ratio.

## At a glance

| | |
| --- | --- |
| Converts | pH, pOH, [H⁺] and [OH⁻] (\(K_\mathrm{w} = 1.0\times10^{-14}\) at 25 °C) |
| Strong acids and bases | Fully dissociated, with water’s own ions, so very dilute solutions come out right |
| Weak acids and bases | Solved exactly from the charge balance, from \(\mathrm{p}K_\mathrm{a}\) |
| Buffers | \(\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10}([\mathrm{A^-}]/[\mathrm{HA}])\), and the ratio for a target pH |
| Limit | Concentrations stand in for activities, which is accurate for dilute solutions |

## How it works

$$\begin{aligned} \mathrm{pH} &= -\log_{10}[\mathrm{H^+}] \\ \mathrm{pOH} &= -\log_{10}[\mathrm{OH^-}] \\ [\mathrm{H^+}][\mathrm{OH^-}] &= K_\mathrm{w} = 1.0\times10^{-14}\ \ (25\ ^\circ\mathrm{C}) \end{aligned}$$

$$\begin{aligned} &\text{Weak acid:} \\ &[\mathrm{H^+}]^3 + K_\mathrm{a}[\mathrm{H^+}]^2 \\ &\quad {}- (K_\mathrm{a}C + K_\mathrm{w})[\mathrm{H^+}] - K_\mathrm{a}K_\mathrm{w} = 0 \end{aligned}$$

$$\text{Buffer: } \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$

Strong acids are treated as fully dissociated with water’s own ions included, so a \(10^{-8}\) M HCl solution correctly gives a pH just below 7. Weak acids and bases are solved exactly from the charge balance rather than with the \(\sqrt{K_\mathrm{a}C}\) shortcut. Concentrations stand in for activities, which is accurate for dilute solutions.

## Worked example

0.1 M acetic acid (\(\mathrm{p}K_\mathrm{a}\) 4.76): \([\mathrm{H^+}] = 1.31\times10^{-3}\ \mathrm{M}\), so \(\mathbf{pH = 2.88}\), with 1.3 % of the acid ionised. A phosphate buffer at pH 7.4 (\(\mathrm{p}K_\mathrm{a}\) 7.20) needs \([\ce{HPO4^2-}]/[\ce{H2PO4-}] = 10^{0.2} = 1.58\).

## Questions

### How do you calculate pH from concentration?

For a strong acid, \(\mathrm{pH} = -\log_{10}\) of the acid concentration (times the number of protons released). For a weak acid you also need \(K_\mathrm{a}\): \([\mathrm{H^+}] \approx \sqrt{K_\mathrm{a} \times C}\) when the acid is only slightly ionised, or solve the equilibrium exactly, as this calculator does.

### What is the Henderson–Hasselbalch equation?

\(\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log_{10}([\mathrm{A^-}]/[\mathrm{HA}])\). It gives the pH of a buffer from the ratio of base to acid forms, and buffers work best within one pH unit of the \(\mathrm{p}K_\mathrm{a}\).

### Why is pH + pOH = 14?

Because water’s ion product \(K_\mathrm{w} = [\mathrm{H^+}][\mathrm{OH^-}] = 1.0\times10^{-14}\) at 25 °C. At other temperatures \(K_\mathrm{w}\) changes, so the sum is not exactly 14.

### Can pH be negative?

Yes, for very concentrated strong acids (above 1 M), although activities then differ from concentrations and measured values deviate from the simple formula.

### Does Tris buffer pH change with temperature?

Yes, strongly: Tris’s \(\mathrm{p}K_\mathrm{a}\) drops by about 0.03 per °C, so set the pH at the temperature you will use it.