Some acids can give up more than one proton: carbonic acid ($\mathrm{H_2CO_3}$), sulfurous acid, phosphoric acid ($\mathrm{H_3PO_4}$), and vitamin C. Each proton comes off in its own step, each a weak-acid equilibrium with its own $K_a$.
Key idea
The constants fall steeply: $K_{a1} \gg K_{a2} \gg K_{a3}$, usually by a factor of $10^4$ or more. So the pH comes almost entirely from the first step, treated as a single weak acid. In a solution of the acid alone, the concentration of the ion made in the second step, $[\mathrm{A^{2-}}]$, is about equal to $K_{a2}$.
Why it works
Taking a proton from a neutral molecule is easier than taking one from an anion that already has a negative charge, and harder still from a 2− ion: the leftover charge holds the next proton back. Also, the $\mathrm{H_3O^+}$ from step 1 pushes step 2 backward (a common-ion effect). In $K_{a2} = \dfrac{[\mathrm{H_3O^+}][\mathrm{A^{2-}}]}{[\mathrm{HA^-}]}$, step 1 makes $[\mathrm{H_3O^+}] \approx [\mathrm{HA^-}]$, so they cancel and $[\mathrm{A^{2-}}] \approx K_{a2}$.
Worked examples
Worked example: Vitamin C
Ascorbic acid, $\mathrm{H_2C_6H_6O_6}$, has $K_{a1} = 8.0 \times 10^{-5}$ and $K_{a2} = 1.6 \times 10^{-12}$. For a $0.10\ \text{M}$ solution, find the pH and the concentration of $\mathrm{C_6H_6O_6^{2-}}$.
- Step 1 only: $x = \sqrt{(8.0 \times 10^{-5})(0.10)} = 2.8 \times 10^{-3}\ \text{M}$ (2.8%, fine). $\text{pH} = 2.55$.
- Step 2 adds about $10^{-12}\ \text{M}$ of $\mathrm{H_3O^+}$, nothing on this scale.
- $[\mathrm{C_6H_6O_6^{2-}}] \approx K_{a2} = 1.6 \times 10^{-12}\ \text{M}$.
Answer: $\text{pH} = 2.55$; $[\mathrm{C_6H_6O_6^{2-}}] \approx 1.6 \times 10^{-12}\ \text{M}$.
Worked example: Phosphoric acid
$\mathrm{H_3PO_4}$: $K_{a1} = 7.5 \times 10^{-3}$, $K_{a2} = 6.2 \times 10^{-8}$, $K_{a3} = 4.2 \times 10^{-13}$. Find the pH of $0.10\ \text{M}$ $\mathrm{H_3PO_4}$ and $[\mathrm{HPO_4^{2-}}]$.
- $K_{a1}$ is large, so solve exactly: $x^2 + (7.5 \times 10^{-3})x - 7.5 \times 10^{-4} = 0$ gives $x = 0.0239\ \text{M}$.
- $\text{pH} = -\log(0.0239) = 1.62$.
- $[\mathrm{HPO_4^{2-}}] \approx K_{a2} = 6.2 \times 10^{-8}\ \text{M}$; $\mathrm{PO_4^{3-}}$ is far smaller still.
Answer: $\text{pH} = 1.62$; $[\mathrm{HPO_4^{2-}}] \approx 6.2 \times 10^{-8}\ \text{M}$.
Common mistake
Adding up the $\mathrm{H_3O^+}$ from every step as if each step ionized as much as the first, or as if a diprotic acid gave $2C$ of $\mathrm{H_3O^+}$. Later steps start from a far smaller amount and have far smaller constants. Check: compare $K_{a2}$ with your $[\mathrm{H_3O^+}]$ from step 1; it will be many powers of ten smaller.
Sulfuric acid is the exception you meet most: its first proton is strong. Its second, $K_{a2} = 1.2 \times 10^{-2}$, is weak but large enough that it must be solved as its own equilibrium in dilute solutions.