Assume that the only reducing agent present in a particular wastewater is SO2−3. 5SO2−3(aq)+2MnO−4(aq)+6H+(aq)→5SO24−(aq)+2Mn+2(aq)+3H2O(l)?
2018-03-18 12:42 pm
If a 29.35-mL sample of this wastewater requires 33.82 mL of 2.244×10−2 M KMnO4 for its titration, what is the molarity of SO2−3 in the wastewater?
回答 (1)
2018-03-19 12:42 am
The balanced equation for the reaction :
5SO₃²⁻(aq) + 2MnO₄⁻(aq) + 6H⁺(aq) → ……
Mole ratio SO₃²⁻ : KMnO₄ = SO₃²⁻ : MnO₄⁻ = 5 : 2
No. of milli-moles of KMnO₄ = (2.244 mmol/mL) × (33.82 mL) = 0.7589 mmol
No. of milli-moles of SO₃²⁻ = (0.7589 mmol) × (5/2) = 1.897 mmol
Molarity of SO₃²⁻ = (1.897 mmol) / (29.35 mL) = 6.463 × 10⁻² M
5SO₃²⁻(aq) + 2MnO₄⁻(aq) + 6H⁺(aq) → ……
Mole ratio SO₃²⁻ : KMnO₄ = SO₃²⁻ : MnO₄⁻ = 5 : 2
No. of milli-moles of KMnO₄ = (2.244 mmol/mL) × (33.82 mL) = 0.7589 mmol
No. of milli-moles of SO₃²⁻ = (0.7589 mmol) × (5/2) = 1.897 mmol
Molarity of SO₃²⁻ = (1.897 mmol) / (29.35 mL) = 6.463 × 10⁻² M
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