pH Calculator

pH places the acidity of an aqueous solution on a base-10 logarithmic scale:

$\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}]$

where $[\mathrm{H_3O^+}]$ is the hydronium-ion concentration in mol/L (M). The matching quantity for bases is $\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}]$.

What the scale assumes. These definitions are used for dilute aqueous solutions, where molar concentration is a good stand-in for chemical activity. At 25 °C the ion product of water is $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$, which gives the familiar companion relation:

$\mathrm{pH} + \mathrm{pOH} = 14.00$

That 14.00 belongs to 25 °C only. $K_w$ grows with temperature, so neutral water sits near pH 6.6 at 50 °C — still neutral, just not 7.

Reading the scale. One whole pH unit is a factor of 10 in $[\mathrm{H_3O^+}]$, so pH 3 is ten times more acidic than pH 4 and a hundred times more acidic than pH 5. At 25 °C, below 7 is acidic, 7 is neutral, above 7 is basic.

From concentration to pH

1. Get $[\mathrm{H_3O^+}]$ in mol/L. A strong acid (HCl, $\mathrm{HNO_3}$, $\mathrm{HClO_4}$) ionises completely, so $[\mathrm{H_3O^+}]$ equals its molarity. A strong base such as NaOH gives $[\mathrm{OH^-}]$ instead.
2. Take the negative base-10 log.
3. Convert if needed with $\mathrm{pH} = 14.00 - \mathrm{pOH}$. For 0.025 M NaOH: $\mathrm{pOH} = -\log_{10}(0.025) = 1.60$, so $\mathrm{pH} = 14.00 - 1.60 = 12.40$.

From pH back to concentration

$[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}} \qquad\qquad [\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H_3O^+}]}$

Weak acids

A weak acid ionises only partly, so its formal concentration $C_a$ is not $[\mathrm{H_3O^+}]$. Use its acid dissociation constant $K_a$ with an ICE table. When ionisation stays under about 5%, the approximation

$[\mathrm{H_3O^+}] \approx \sqrt{K_a\,C_a}$

is accurate enough; otherwise solve the full quadratic $x^2 + K_a x - K_a C_a = 0$.

Significant figures

In a pH, only the digits after the decimal point count as significant — the part before it just records the power of ten. A concentration known to 2 significant figures therefore gives a pH quoted to 2 decimal places.

• Treating a weak acid as strong — 0.10 M acetic acid is not pH 1.00. Without $K_a$ the pH of a weak acid cannot be found from molarity alone.
• Ignoring the base's stoichiometry — 0.010 M $\mathrm{Ca(OH)_2}$ releases two hydroxides per formula unit, so $[\mathrm{OH^-}] = 0.020$ M, $\mathrm{pOH} = 1.70$ and $\mathrm{pH} = 12.30$.
• Reporting too many digits — a 2-significant-figure concentration supports pH 2.60, not 2.6021.
• Assuming $\mathrm{pH} + \mathrm{pOH} = 14$ at any temperature — that sum equals $\mathrm{p}K_w$, which is 14.00 only at 25 °C.
• Using $\ln$ instead of $\log_{10}$ — the two differ by a factor of 2.303.
• Forgetting water in very dilute solutions — below about $10^{-6}$ M, water's own autoionisation dominates, so a $10^{-8}$ M strong acid is slightly acidic, not pH 8.