Ka and pKa Calculator

For a weak acid ionising in dilute aqueous solution,

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

• $K_a$ — the acid dissociation constant, a fixed number at a fixed temperature. Water does not appear because it is the solvent, present in vast excess.
• $[\;]$ — equilibrium molar concentrations, in mol/L.

Because $K_a$ values span many orders of magnitude, they are usually reported on a log scale:

$\mathrm{p}K_a = -\log_{10} K_a \qquad\Longleftrightarrow\qquad K_a = 10^{-\mathrm{p}K_a}$

A larger $K_a$ means a stronger acid, and because of the minus sign a smaller $\mathrm{p}K_a$ means a stronger acid. Acetic acid, $K_a = 1.8 \times 10^{-5}$, has $\mathrm{p}K_a = 4.74$.

Assumed conditions. These relations are used for dilute aqueous solutions at 25 °C, where molarity stands in for activity and $K_w = 1.0 \times 10^{-14}$. $K_a$ itself is temperature-dependent, so a value quoted without a temperature is a 25 °C value. The equilibrium expression applies to weak acids; a strong acid ionises essentially completely and has no useful $K_a$.

1. Ka to pKa and back

Take the negative base-10 log, or the power of ten. On the log scale only the digits after the decimal point are significant, so a $K_a$ with 2 significant figures gives a $\mathrm{p}K_a$ with 2 decimal places.

2. pH from Ka

Set up an ICE table for a formal concentration $C_a$ with $x = [\mathrm{H_3O^+}]$:

$K_a = \frac{x^2}{C_a - x}$

If ionisation is under about 5%, the shortcut $x \approx \sqrt{K_a C_a}$ is accurate; otherwise solve the quadratic $x^2 + K_a x - K_a C_a = 0$. Then $\mathrm{pH} = -\log_{10} x$.

3. Ka from a measured pH

Reverse the same equation. From the pH get $x = 10^{-\mathrm{pH}}$, then

$K_a = \frac{x^2}{C_a - x}$

using the formal concentration for $C_a$, not the equilibrium one.

4. Buffers: Henderson-Hasselbalch

When an acid and its conjugate base are both present in appreciable amounts,

$\mathrm{pH} = \mathrm{p}K_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$

Only the ratio matters, so diluting a buffer barely changes its pH. The equation assumes both concentrations are much larger than $[\mathrm{H_3O^+}]$, which fails when the ratio is extreme or the buffer is very dilute.

• Mixing up the direction of strength. High $K_a$, low $\mathrm{p}K_a$, strong acid. The minus sign flips the ordering.
• Putting equilibrium concentrations into $C_a$. In $K_a = x^2/(C_a - x)$, $C_a$ is the amount of acid dissolved; the subtraction is what converts it to the equilibrium value.
• Using $\sqrt{K_a C_a}$ when ionisation is large. Check that $x/C_a < 5\%$; for a fairly strong weak acid in a dilute solution it will not be, and the quadratic is required.
• Reporting too many digits. A pH read to 2 decimal places supports a $K_a$ with 2 significant figures.
• Applying Henderson-Hasselbalch to a solution of the acid alone. Without the conjugate base present, the ratio is undefined; use the ICE table instead.
• Forgetting $\mathrm{p}K_a + \mathrm{p}K_b = 14.00$ holds only at 25 °C, since it comes from $K_w$.