Question

Use these 2 reactions to answer Question 4:

2Cu²⁺ + 4KI → 2CuI + I₂ + 4K⁺ (rxn 1)

I₂ + 2Na₂S₂O₃ → 2NaI + Na₂S₄O₆ (rxn 2)

For rxn 1, 10.0 mL of a Cu²⁺ solution of unknown concentration was placed in a 250 mL Erlenmeyer flask. An excess of KI solution was added. Indicator was added and the solution was diluted with H₂O to a total volume of 75 mL. For rxn 2, the solution from rxn 1 was titrated with 0.35 M Na₂S₂O₃. The equivalence point of the titration was reached when 13.05 mL of Na₂S₂O₃ had been added. What is the molar concentration of Cu²⁺ in the original 10.0 mL solution?

Cu²⁺ molarity =

Answer 1

The balanced equations are Number of moles of Na2S203 M*V 0.35 M * 13.05 mL 4.5675 mmol Accordingobalacdqon, So, 4.5675 mmol of Na2S203 would react with 4.5675 mmol2.28375 mmol of 2 Number of moles of 12 produced in reaction 1 2.28375 mmol According to balanced equation (1), So, 2.28375 mmol of 12 would be produced from 2.28375 mmol * 2/1 4.5675 mmol 2 moles of Na2S203 reacts with 1 mole of 12 1 mol of l, is produced fro. mols ofCu of Cu+2 Number of moles of Cu*2 4.5675 mmo Volume of the solution 10 mL Molarity of Cu+2-moles/volume = 4.5675 mmol/10 mL = 0.45675 M Molarity of Cu*2 solution 0.4568 M