Solving Polyprotic Acid pH via the Dominant First Ionization
This video presents the same text shown beside it, spoken and on screen. It adds nothing the text does not say.
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.
Adding H3O+ contributions from every ionization step as if each were comparably sized overstates [H3O+] and so understates the pH.
Builds on
Unlocks
- Nothing yet depends on this.