Question

A 50.00-mL sample containing La31 was treated with sodium oxalate to precipitate La2(C2O4)3, which was washed, dissolved in acid, and titrated with 18.04 mL of 0.006 363 M KMnO4. Write the titration reaction and find [La31] in the unknown.

Answer 1

Answer:

2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → 2MnSO₄ + 10CO₂ + 8H₂O + K₂SO₄

0,02658g of La in the unknown

Explanation:

The reaction of  La₂(C₂O₄)₃ with acid is:

La₂(C₂O₄)₃ + 6H⁺ → 3H₂C₂O₄ + 2La³⁺

The titration of H₂C₂O₄ with KMnO₄ is:

2KMnO₄ + 5H₂C₂O₄ + 3H₂SO₄ → 2MnSO₄ + 10CO₂ + 8H₂O + K₂SO₄

The moles of KMnO₄ that react are:

0,006363M KMnO₄×0,01804L = 1,148x10⁻⁴moles of KMnO₄

By the titration reaction, 2 moles of KMnO₄ react with 5 moles of H₂C₂O₄, that means:

1,148x10⁻⁴moles of KMnO₄×$\frac{5molesH_{2}C_{2}O_{4}}{2molesKMnO_{4}}$ = 2,870x10⁻⁴ moles of H₂C₂O₄.

In the reaction of La₂(C₂O₄)₃ with acid, 3 moles of H₂C₂O₄ were produced while 2 moles of La³⁺ were produced, that means:

2,870x10⁻⁴ moles H₂C₂O₄× $\frac{2molesLa^{3+}}{3molesH_{2}C_{2}O_{4}}$ = 1,913x10⁻⁴ moles of La³⁺, in grams -Using molar mass of lanthanum-:

1,913x10⁻⁴ moles of La³⁺×$\frac{138,9g}{1mol}$ = 0,02658g of La

There are 0,02658g of La in the unknown

I hope it helps!