Acid Dissociation Constant Calculator

The acid dissociation constant is the equilibrium constant for an acid giving up a proton to water in dilute aqueous solution:

$\mathrm{HA} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-}, \qquad K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$

All three concentrations are equilibrium values in mol/L. The solvent is omitted from the expression because it is present in enormous excess and its concentration is effectively constant.

$K_a$ answers one question: of the acid molecules dissolved, what fraction have handed over a proton at equilibrium? A $K_a$ near $10^{-2}$ describes an acid that is substantially ionised; $10^{-10}$ describes one that is barely ionised at all.

The conjugate base. Every acid's $K_a$ has a partner constant for its conjugate base, and at 25 °C the two are locked together by the ion product of water:

$K_a K_b = K_w = 1.0 \times 10^{-14} \qquad\Longrightarrow\qquad \mathrm{p}K_a + \mathrm{p}K_b = 14.00$

So a stronger acid necessarily has a weaker conjugate base. Both relations assume 25 °C, dilute solution, and that concentration is a good stand-in for activity.

From a measured pH

For a monoprotic weak acid at formal concentration $C_a$, the ionisation produces equal amounts of $\mathrm{H_3O^+}$ and $\mathrm{A^-}$:

1. $x = [\mathrm{H_3O^+}] = 10^{-\mathrm{pH}}$, and $[\mathrm{A^-}] = x$ as well.
2. $[\mathrm{HA}] = C_a - x$ — subtract what ionised.
3. $K_a = \dfrac{x^2}{C_a - x}$.

From percent ionisation

Percent ionisation is $\alpha = x/C_a \times 100\%$, so $x = \alpha C_a / 100$. Substituting gives

$K_a = \frac{C_a \alpha^2}{1 - \alpha}$

with $\alpha$ as a decimal fraction. Note that percent ionisation rises on dilution even though $K_a$ does not move.

From pKa

$K_a = 10^{-\mathrm{p}K_a}$

Converting to Kb

$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{K_a}$

Significant figures

A pH measured to 2 decimal places gives $[\mathrm{H_3O^+}]$ to 2 significant figures, and therefore $K_a$ to 2. Because $K_w$ is quoted to 2 figures, any $K_b$ derived from it inherits that limit.

• Using the formal concentration as $[\mathrm{HA}]$. The denominator is $C_a - x$; skipping the subtraction inflates $K_a$, badly so when ionisation exceeds a few percent.
• Thinking $K_a$ changes with concentration. It does not — only temperature moves it. Percent ionisation and pH do change with dilution.
• Reporting $K_b$ as $14 - K_a$. The additive relation is between the p-values; the constants themselves multiply to $K_w$.
• Quoting a $K_a$ for a strong acid. HCl and $\mathrm{HNO_3}$ ionise essentially completely, so their equilibrium constants are not measurable by these methods in water.
• Ignoring the second ionisation of a polyprotic acid. $\mathrm{H_2SO_4}$ and $\mathrm{H_3PO_4}$ have separate $K_{a1}, K_{a2}, \ldots$, each far smaller than the last.
• Assuming $\mathrm{p}K_a + \mathrm{p}K_b = 14$ at every temperature — that sum is $\mathrm{p}K_w$, which equals 14.00 only at 25 °C.