A weak acid ionizes only partly, so it reaches an equilibrium with water. Its equilibrium constant measures its strength. This builds on Brønsted–Lowry pairs and on pH.
Key idea
For $\mathrm{HA + H_2O \rightleftharpoons H_3O^+ + A^-}$, $K_a = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$, and $\text{p}K_a = -\log K_a$. A larger $K_a$ (smaller $\text{p}K_a$) means a stronger acid. For a base, $K_b = \dfrac{[\mathrm{HB^+}][\mathrm{OH^-}]}{[\mathrm{B}]}$. For any conjugate pair, $K_a \times K_b = K_w$, so $\text{p}K_a + \text{p}K_b = 14.00$ at $25\ ^\circ\text{C}$.
Why it works
Add the acid's ionization to its conjugate base's reaction with water:
$\mathrm{HA + H_2O \rightleftharpoons H_3O^+ + A^-}$ ($K_a$)
$\mathrm{A^- + H_2O \rightleftharpoons HA + OH^-}$ ($K_b$)
The sum is $\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-}$, whose constant is $K_w$. Adding reactions multiplies their constants, so $K_aK_b = K_w$. The stronger the acid, the weaker its conjugate base, and the product is always fixed.
$\mathrm{HA + H_2O \rightleftharpoons H_3O^+ + A^-}$ ($K_a$)
$\mathrm{A^- + H_2O \rightleftharpoons HA + OH^-}$ ($K_b$)
The sum is $\mathrm{2H_2O \rightleftharpoons H_3O^+ + OH^-}$, whose constant is $K_w$. Adding reactions multiplies their constants, so $K_aK_b = K_w$. The stronger the acid, the weaker its conjugate base, and the product is always fixed.
Worked examples
Worked example: From Ka to pKa
Acetic acid has $K_a = 1.8 \times 10^{-5}$. Find its $\text{p}K_a$.
- $\text{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74$.
- Check the size: $10^{-5}$ alone would give 5; the factor 1.8 makes $K_a$ bigger, so $\text{p}K_a$ is a bit less than 5.
Answer: $\text{p}K_a = 4.74$.
Worked example: Kb of a conjugate base
Find $K_b$ for acetate ion, $\mathrm{CH_3COO^-}$, at $25\ ^\circ\text{C}$.
- $K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}}$.
- $K_b = 5.6 \times 10^{-10}$: acetate is a very weak base.
Answer: $5.6 \times 10^{-10}$.
Worked example: Ranking by pKa
HF has $\text{p}K_a = 3.17$ and HCN has $\text{p}K_a = 9.31$. Which is the stronger acid, and by what factor in $K_a$?
- Smaller $\text{p}K_a$ means larger $K_a$: HF is stronger.
- The difference is $9.31 - 3.17 = 6.14$ units, so HF's $K_a$ is $10^{6.14} \approx 1.4 \times 10^{6}$ times larger.
Answer: HF, by a factor of about a million.
Strong acids have $K_a$ values far above 1, so large that they are hard to measure; in water they all look equally strong, because each one gives its proton to water completely. Weak acids have $K_a$ below 1, and their $\text{p}K_a$ values usually lie between about 2 and 12, a span of ten powers of ten in strength.
Common mistake
Thinking a larger $\text{p}K_a$ means a stronger acid. The "p" means minus the log, so large $\text{p}K_a$ means small $K_a$: a weak acid. Check: strong-ish weak acids such as HF have $\text{p}K_a$ near 3; very weak ones such as HCN are near 9.
Values of $K_a$ and $K_b$ in these lessons are at $25\ ^\circ\text{C}$ from the tables in the back of OpenStax Chemistry 2e.