Solving Polyprotic Acid pH via the Dominant First Ionization

Solving for a polyprotic acid solution's pH means solving the first ionization step alone with an ICE table, since later steps contribute so little additional H3O+, given how much smaller each successive Ka is, that they can be checked afterward and then safely neglected from the final answer.

For 0.10 M H2CO3, Ka1 = 4.3 × 10⁻⁷, Ka2 = 4.8 × 10⁻¹¹: solving the first step alone gives [H3O+] ≈ 2.1 × 10⁻⁴ M, pH ≈ 3.68. The second step, starting from that [H3O+], contributes an amount of additional H3O+ on the order of Ka2 itself, negligible next to 2.1 × 10⁻⁴ M already present.