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?

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

To find the molar concentration of Cu2+ in the original 10.0 mL solution, we can use the equation for the reaction between Cu2+ and KI:

Cu2+ + 2KI → CuI2 + 2K+

Here's how you can solve the problem step by step:

Step 1: Calculate the moles of Na2S2O3 used in the titration.
Since 10.95 mL of 0.10 M Na2S2O3 was added before reaching the equivalence point, you can calculate the moles of Na2S2O3 used:
moles Na2S2O3 = volume (L) x concentration (mol/L)
moles Na2S2O3 = 0.01095 L x 0.10 mol/L

Step 2: Calculate the moles of Cu2+ reacted.
Since the reaction between Cu2+ and Na2S2O3 is 1:2 stoichiometrically, the moles of Cu2+ reacted is half the moles of Na2S2O3 used:
moles Cu2+ = moles Na2S2O3 / 2

Step 3: Calculate the moles of Cu2+ in the original 10.0 mL of solution.
Since the volume of the original Cu2+ solution is 10.0 mL and you have already diluted it to a total volume of 75 mL, you can calculate the moles of Cu2+ in the original solution:
moles Cu2+ = moles reacted / (10.0 mL / 75 mL)

Step 4: Calculate the molar concentration of Cu2+ in the original solution.
Finally, calculate the molar concentration of Cu2+ by dividing the moles of Cu2+ by the volume of the original solution:
concentration Cu2+ = moles Cu2+ / volume (L)

Now, plug in the values and calculate the concentration of Cu2+ in the original 10.0 mL solution.

Answer 2

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 = ?