Question

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 0.15 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?

Answer 1

Answer:

Molar concentration: 0,0489M

Explanation:

In this titration of Cu²⁺ you add an excess of I⁻ that reacts with Cu²⁺ producing I₂, this I₂ reacts with Na₂S₂O₃. If you know the I₂ that reacts with Na₂S₂O₃ you can know the I⁻ that reacts with Cu²⁺ and, thus, the quantity of Cu²⁺. The reactions are:

2Cu²⁺ + 4I⁻ → 2CuI + I₂

I₂ + 2S₂O₃⁻ → S₄O₆ + 2I⁻

Moles of S₂O₃⁻ are:

0,01305L×0,15M = 1,96x10⁻³ moles of S₂O₃⁻.

Moles of I₂ are:

1,96x10⁻³ moles of S₂O₃⁻× ( 1 mole of I₂ /  2 moles of S₂O₃⁻) = 9,79x10⁻⁴ moles of I₂

Moles of Cu²⁺ are:

9,79x10⁻⁴ moles of I₂×( 2 moles of Cu²⁺ / 4 moles of I₂) = 4,89x10⁻⁴ moles of Cu²⁺

As volume of the solution was 10,0mL = 0,0100L, the molar concentration of the original solution is:

4,89x10⁻⁴ moles of Cu²⁺ / 0,0100L = 0,0489M