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A solution of sodium oxalate (Na 2C 20 4) in acidic solution is titrated with a solution of potassium permanganate (KMnO 4) according to the following
balanced chemical equation:
2KMnO4(aq) + 8H2SO4(aq) + 5Na2C204(aq) 2MnSO4(aq) - 8H20() +10C02(9) + SNa2SO4(aq) + K2SO4(aq)
What volume of 0.0206 M KMnO4 is required to titrate 0.176 g of Na2C204 dissolved in 50.0 mL of solution?
O a. 8. 54 ml
O b. 500 ml
OC 25 5 ml
O d.3.42 m
e. 63 8 ml
X

Respuesta :

Eduard22sly

The volume of 0.0206 M KMnO₄ required to titrate 0.176 g of Na₂C₂O₄ dissolved in 50.0 mL of solution is 25.5 mL (Option C)

 

How to determine the mole of Na₂C₂O₄

  • Mass of Na₂C₂O₄ = 0.176 g
  • Molar mass of Na₂C₂O₄ = 134 g/mol
  • Mole of Na₂C₂O₄ =?

Mole = mass / molar mass

Mole of Na₂C₂O₄ = 0.176 / 134

Mole of Na₂C₂O₄ = 1.313×10¯³ mole

How to determine the mole of KMnO₄

2KMnO₄ + 8H₂SO₄ + 5Na₂C₂O₄ —> 2MnSO₄ + 8H₂O +10CO₂ + 5Na₂SO₄ + K₂SO₄

From the balanced equation above,

5 moles of Na₂C₂O₄ reacted with 2 moles of KMnO₄

Therefore,

1.313×10¯³ mole of Na₂C₂O₄ will react with = (1.313×10¯³ × 2) / 5 = 5.252×10¯⁴ mole of KMnO₄

How to determine the volume of KMnO₄

  • Mole of KMnO₄ = 5.252×10¯⁴ mole
  • Molarity of KMnO₄ = 0.0206 M
  • Volume of KMnO₄ =?

Volume = mole / molarity

Volume of KMnO₄ = 5.252×10¯⁴ / 0.0206

Volume of KMnO₄ = 0.0255 L

Multiply by 1000 to express in mL

Volume of KMnO₄ = 0.0255 × 1000

Volume of KMnO₄ = 25.5 mL

Learn more about titration:

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