Determination of the vanadium:oxalate ratio in the complex?

2018-03-20 6:48 pm
Both the oxalate ligands and the vanadium are oxidised by the KMnO4 with the oxalate decomposing and leaving
the system as CO2(g). Therefore, the first titration step titrates both the oxalate and the vanadium.
1. Dissolve about 0.3 g (weighed accurately) of the complex salt in at least 60 ml of
sulfuric acid (2 M). Dilute this solution to 100 ml with distilled water. This step destroys the complex, liberates C2O4
2– and produces
VO2+, both of which can be titrated with KMnO4.
2. Take a 25 mL aliquot and heat the solution to about 80°C on a hot plate.
3. Titrate this hot solution with potassium permanganate (0.01 M) until a bright pink colour is first seen and doesn’t fade on standing for about 30 seconds.
The vanadium must be reduced back down in order to perform a second KMnO4 titration this time
with just the vanadium being oxidised as there is no oxalate left in the system. The reduction is
achieved by reacting the oxidised vanadium with sodium sulfite. To ensure there is no excess
sodium sulfite to interfere with the second half of the titration the solution is boiled to drive off SO2.
4. Add approximately 0.1 g of sodium sulfite to the oxidised solution and then boil the solution
for 5 minutes until all SO2
is expelled and then titrate at ~ 60°C

IN THIS PROCESS, 1. how do I caculate the moles of oxalate from the difference between first and second titrations. 2, Calculate the no of moles of vanadium from second titration??
更新1:

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回答 (1)

2018-03-20 7:54 pm
✔ 最佳答案
1. The complex is destroyed. The central V(IV) ion is convert to VO²⁺ ion and free C₂O₄²⁻ ion.

3. The titration of VO²⁺ and C₂O₄²⁻ ions against KMnO₄ solution (the first titration):
5VO²⁺(aq) + MnO₄⁻ + H₂O(l) → 5VO₂⁺(aq) + Mn²⁺(aq) + 2H⁺(aq) …… [1]
5C₂O₄²⁻(aq) + 2MnO₄⁻(aq) + 16H⁺(aq) → 10CO₂(g) + 2Mn²⁺(aq) + 8H₂O(l) …… [2]
Suppose that V₁ mL of 0.01 M KMnO₄ solution is required in the first titration.

4. VO₂⁺ ion is quantitatively reduced back to VO²⁺ ion by an excess Na₂SO₃. All the SO₂ formed is boiled off.

5. The titration of VO²⁺ ion formed in step 4 against KMnO₄ solution (the second titration):
5VO²⁺(aq) + MnO₄⁻ + H₂O(l) → 5VO₂⁺(aq) + Mn²⁺(aq) + 2H⁺(aq) …… [1]
Suppose that V₂ mL of 0.01 M KMnO₄ solution is required in the second titration.


Consider the second titration (step 5):
No. of moles of MnO₄⁻ reacted = (0.01 mol/L) × (V₂/1000 L) = 0.00001V₂ mol
Refer to equation [1], mole ratio VO²⁺ : MnO₄⁻ = 5 : 1
No. of moles of VO²⁺ = (0.00001V₂ mol) × 5 = 0.00005V₂ mol
No. of moles of V in the complex = No. of moles of VO²⁺ = 0.00005V₂ mol

Consider the first titration (Step 3):
Total no. of moles of MnO₄⁻ reacted = (0.01 mol/L) × (V₁/1000 L) = 0.00001V₁ mol
No. of moles of MnO₄⁻ reacted with C₂O₄²⁻ = (0.00001V₁ - 0.00001V₂) mol
Refer to equation [2], mole ratio C₂O₄²⁻ : MnO₄⁻ = 5 : 2
No. of moles of C₂O₄²⁻ = [(0.00001V₁ - 0.00001V₂) mol] × (5/2) = 0.000025(V₁ - V₂) mol


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