Question

4. 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 0.20 M Na2S2O3. The equivalence point of the titration was reached when 14.45 mL of Na2S2O3 had been added. What is the molar concentration of Cu2+ in the original 10.0 mL solution?

Answer 1

Answer:

0.289 M

Explanation:

In the rxn1, the ions Cu²⁺ will react with the ions I⁻ that comes from KI, and the ions I⁻ will react with themselves:

Cu²⁺ + 4I⁻ → CuI₂ + I₂

Then, the I₂ will react with the ions Na⁺ that comes from Na₂S₂O₃ at the rxn 2:

Na₂S₂O₃ + I₂ → 2NaI + S₂O₃²⁻

The number of moles of Na₂S₂O₃ is the volume (0.01445 L) multiplied by the concentration:

n = 0.01445*0.20

n = 2.89x10⁻³

By the stoichiometry, 1 mol of Na₂S₂O₃ reacts with 1 mol of I₂, so nI₂ = 2.89x10⁻³ mol. And, by the stoichiometry of the other reaction, 1 mol of I₂ is produced with 1 mol of Cu²⁺, thus, nCu²⁺ = 2.89x10⁻³ mol.

The molar concentration is the number of moles divided by the volume is L. Thus, for the original solution with a volume of 10 mL = 0.01 L:

Concentration of Cu²⁺ = 2.89x10⁻³/0.01 = 0.289 M