Problem 1: Basicity Determination

Question: Determine the basicity of phosphorous acid ($H_{3}PO_{3}$).

Solution: Step 1 — Draw the structure of $H_{3}PO_{3}$: The IUPAC structure of $H_{3}PO_{3}$ is HP(=O)(OH)_{2}.

• 1 P-H bond (H directly bonded to P)
• 2 P-OH bonds (H bonded to P through O)
• 1 P=O bond (no H here)

Step 2 — Apply the rule: Basicity = number of P-OH bonds = 2

Step 3 — Verification: $H_{3}PO_{3}$ can be ionized in two steps: $H_{3}PO_{3}$ ⇌ $H^{+}$ + $H_{2}PO_{3}^{-}$ (Ka1 ≈ $5 \times 10^{-2}$) $H_{2}PO_{3}^{-}$ ⇌ $H^{+}$ + $HPO_{3}^{2-}$ (Ka2 ≈ $2 \times 10^{-7}$)

Answer: $H_{3}PO_{3}$ is dibasic (basicity = 2), NOT tribasic despite having 3 hydrogen atoms.