Question

Imagine it took 300 mL of 0.1 M LiOH to reach the first equivalence point and an additional 300 mL of 0.1 M LiOH (600 mL total) to reach the second equivalence point. How many moles of H2 were in the original acid solution

Answer 1

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 we use the total volume used (in this case 600 mL).

600 mL ⇒ 600/1000 = 0.6 L

We calculate the moles of LiOH, using its molar concentration:

• 0.1 M * 0.6 L = 0.06 mol LiOH

And now we convert moles of LiOH (or OH⁻) to moles of H₂ using the  stoichiometric ratio:

• 0.06 mol LiOH * $\frac{1molH_{2}A}{2molLiOH}$ = 0.03 mol H₂