When 0.300 g of a diprotic acid was titrated with 0.100 m lioh, 40.0 ml of the lioh solution was needed to reach the second equivalence point. identify the formula of the diprotic acid. h2seh2teh2c2o4h2c4h4o6h2s

H₂C₄H₄O₆

Explanation:

The clue of this question is to find the molar mass of the diprotic acid and compare witht the molars masses of the choices' acid to identify the formula of the diprotic acid.

The procedure is:

Find the number of moles of the base: LiOHUse stoichionetry to infere the number of moles of the acid.Use the formula molar mass = mass in grams / number of moles, to find the molar mass of the diprotic acid.Compare and conclude.

Solution:

1. Number of moles of the base, LiOH:

M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.

2. Stoichiometry:

Since this a neutralization reaction of a diprotic acid with a mono hydroxide base (LiOH), the mole ratio at the second equivalence point is: 2 mol of base / 1 mole of acid; because each mole of LiOH releases 1 mol of OH⁻, while each mole of diprotic acid releases 2 mol of H⁺.

Hence, 0.004 mol LiOH × 1 mol acid / 2 mol LiOH = 0.002 mol acid.

3. Molar mass of the acid:

molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol

4. Molar mass of the possible diprotic acids:

a. H₂Se: 2×1.008 g/mol + 78.96  g/mol = 80.976 g/mol

b. H₂Te: 2×1.008 g/mol + 127.6  g/mol = 129.616 g/mol

c. H₂C₂O₄ ≈ 2×1.008 g/mol + 2×12.011 g/mol + 4×15.999 g/mol ≈ 90.034 g/mol

d. H₂C₄H₄O₆ = 6×1.008 g/mol +  4×12.011 g/mol + 6×15.999 g/mol = 150.086 g/mol ≈ 150 g/mol.

Conclusion: since the molar mass of H₂C₄H₄O₆ acid is 150 g/mol, you conclude that is the diprotic acid whose 0.300 g were titrated with 40.0 ml of 0.100 M LiOH solution.

The mass has escaped into a gas which you can not see so the law stays true.

Iimpact me by the one the yet but him.