Question

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 0.10 M Na2S2O3. The equivalence point of the titration was reached when 10.95 mL of Na2S2O3 had been added. What is the molar concentration of Cu2+ in the original 10.0 mL solution?

Answer 1

2Cu^2+ + 4KI ==> 2CuI + I2 + 4K^+, then the liberated i2 is titrated with Na2S2O3.
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-

mols S2O3^2- = M x L = 0.1M x 0.01095 = ?
From the thiosulfate reaction you know mols I2 = 1/2 mols thiosulfate.
From the Cu^2+ rxn you know mols Cu^2+ = 2 x mols I2.
Then (Cu^2+) = mols/L = mols Cu^2+/0.0100 = ?