Chemistry · Grades 10–12
Diprotic-acid stoichiometry
Use the balanced equation, not an assumed one-to-one ratio. Here each formula unit supplies two hydroxide ions.
Rule and assumptions
Use the balanced equation, not an assumed one-to-one ratio. Here each formula unit supplies two hydroxide ions.
H₂A + 2OH⁻ → A²⁻ + 2H₂O; n(OH⁻) = 2n(H₂A)
Worked examples
In the supplied complete neutralization equation H₂A + 2OH⁻ → A²⁻ + 2H₂O, how much OH⁻ reacts with 0.5 mol H₂A? Round to 3 decimal places if needed. Answer in mol.
H₂A + 2OH⁻ → A²⁻ + 2H₂O; n(OH⁻) = 2n(H₂A) 2 × 0.5 = 1
1 mol
In the supplied complete neutralization equation H₂A + 2OH⁻ → A²⁻ + 2H₂O, how much OH⁻ reacts with 0.09 mol H₂A? Round to 3 decimal places if needed. Answer in mol.
H₂A + 2OH⁻ → A²⁻ + 2H₂O; n(OH⁻) = 2n(H₂A) 2 × 0.09 = 0.18
0.18 mol
Try two questions
In the supplied complete neutralization equation H₂A + 2OH⁻ → A²⁻ + 2H₂O, how much OH⁻ reacts with 0.01 mol H₂A? Round to 3 decimal places if needed. Answer in mol.
In the supplied complete neutralization equation H₂A + 2OH⁻ → A²⁻ + 2H₂O, how much OH⁻ reacts with 0.34 mol H₂A? Round to 3 decimal places if needed. Answer in mol.
Answers and reasoning
Answers and marking notes
- 0.02 mol
H₂A + 2OH⁻ → A²⁻ + 2H₂O; n(OH⁻) = 2n(H₂A) 2 × 0.01 = 0.02
- 0.68 mol
H₂A + 2OH⁻ → A²⁻ + 2H₂O; n(OH⁻) = 2n(H₂A) 2 × 0.34 = 0.68
Further reading and review references
These are reference links, not copied exercises, institutional endorsement or proof of independent review.
Learning preview · independent educational and translation review required. Suggested grades are not official curriculum alignment.