Answer:
0.03 mol H₂
Explanation:
In a diprotic acid titration, the first equivalence point relates to the equilibrium:
- H₂A + OH⁻ ↔ HA⁻ + H₂O
And the second equivalence point to:
- HA⁻ + OH⁻ ↔ A⁻² + H₂O
We can add those two equations and we're left with:
- H₂A + 2OH ⁻ ↔ A⁻² + 2H₂O
So to calculate the moles of H₂ that were in the original acid solution <u>we use the total volume used</u> (in this case 600 mL).
600 mL ⇒ 600/1000 = 0.6 L
We <u>calculate the moles of LiOH</u>, using its molar concentration:
- 0.1 M * 0.6 L = 0.06 mol LiOH
And now we <u>convert moles of LiOH (or OH⁻) to moles of H₂ </u>using the stoichiometric ratio:
- 0.06 mol LiOH *
= 0.03 mol H₂