pH Calculator

The calculator applies simplified concentration-based equations to selected aqueous acid, base, and buffer models. At non-dilute concentrations, ion activities may differ from concentrations; temperature and other equilibria can also affect real solutions.

1. Fundamental Logarithmic Scaling

For an ideal dilute solution, the tool approximates pH and pOH using the hydrogen- and hydroxide-ion concentrations:

$$\text{pH} = -\log_{10}[\text{H}^+]$$

$$\text{pOH} = -\log_{10}[\text{OH}^-]$$

The tool assumes $K_w = 1.0 \times 10^{-14}$, a common approximation for water at $25^\circ\text{C}$:

$$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}$$

Taking the negative logarithm yields the operational framework:

$$\text{pH} + \text{pOH} = 14$$

This sum applies to the tool's assumed 25°C model.

2. Strong Electrolytes & Water Compensation

At very low concentrations, water's own ionization can affect the result for a strong acid or base. The calculator accounts for this with a simplified quadratic model. For an acid, it uses:

$$[\text{H}^+]^2 - C_a[\text{H}^+] - K_w = 0$$

Where $C_a$ is the entered concentration multiplied by the selected valence. The quadratic provides an estimate within this simplified model:

$$[\text{H}^+] = \frac{C_a + \sqrt{C_a^2 + 4K_w}}{2}$$

3. Weak Electrolyte Dissociation Framework

Weak acids and bases do not dissociate completely. The model uses an equilibrium expression to estimate how much dissociates:

$$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{[\text{H}^+]^2}{C_a - [\text{H}^+]}$$

The tool uses a quadratic concentration model:

$$[\text{H}^+]^2 + K_a[\text{H}^+] - K_a C_a = 0$$

The result is an estimate under the model assumptions. The calculator also reports a degree-of-ionization estimate ($\alpha$):

$$\alpha = \frac{[\text{H}^+]}{C_a}$$

4. Buffer Architectures

For buffers, the tool uses the Henderson–Hasselbalch equation:

$$\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)$$

$$\text{pOH} = \text{p}K_b + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Base}]}\right)$$