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Answer:
0.00500M of Na₂C₂O₄
Explanation:
<em>When are dissolved in 150 mL of 1.0 M H2SO4.</em>
<em />
We can solve this problem finding molarity of sodium oxalate: That is, moles of Na2C2O4 per liter of solution. Thus, we need to convert the 0.1005g to moles using molar mass of sodium oxalate (134g/mol) and dividing in the 0.150L of the solution:
0.1005g * (1mol / 134g) = 7.5x10⁻⁴ moles of Na₂C₂O₄
In 0.150L:
7.5x10⁻⁴ moles of Na₂C₂O₄ / 0.150L =
<h3>0.00500M of Na₂C₂O₄</h3>
Answer:
Explanation:
The clue of this question is to find the molar mass of the <em>diprotic acid</em> 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: LiOH
- Use 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.
<u>Solution:</u>
<u>1. Number of moles of the base, LiOH:</u>
- M = n / V in liter ⇒ n = M × V = 0.100 M × 40.0 ml × 1 liter / 1,000 ml = 0.004 mol LiOH.
<u>2. Stoichiometry:</u>
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.
<u>3. Molar mass of the acid:</u>
- molar mass = mass in grams / number of moles = 0.300 g / 0.002 mol = 150. g/mol
<u>4. Molar mass of the possible diprotic acids:</u>
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.
<u>Conclusion:</u> 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.
