referring to the reaction in Question #2, 38.4 ml of a 0.150 M KMnO4 solution is...

referring to the reaction in Question #2, 38.4 ml of a 0.150 M KMnO4 solution is required to titrate 25.2 mL of a sodium oxalate. What is the concentration ofoxalate ion in the solution ?

Expert Solution

Number of moles of KMnO₄ is , n = Molarity x volume in L

= 0.150 M x (38.4 / 1000) L

= 5.76x10^-3 mol

The balanced equation is :

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

According to the balanced equation ,

2 moles of KMnO₄ reacts with 5 moles of Na₂C₂O₄

5.76x10^-3 mol of KMnO₄ reacts with M moles of Na₂C₂O₄

M = ( 5x5.76x10^-3 ) / 2

= 0.0144 moles

We know that Molarity of Na₂C₂O₄ , M = number of moles / volume in L

= 0.0144 mol / ( 25.2 /1000)L

= 0.571 M

Therefore the molarity of Na₂C₂O₄ solution is 0.571 M

queen_honey_blossom answered 6 months ago

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