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 0.15 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?

The molar concentration of the original Cu²⁺ solution: M₁ = 0.196 M

Explanation:

Given: Reaction 1: Dilution-

Original Cu²⁺ solution: Volume: V₁ = 10.0 mL, Molarity: M₁=?

Diluted Cu²⁺ solution: Volume: V₂ = 75 mL, Molarity: M₂ = ?

Reaction 2: Titration of Diluted Cu²⁺ solution with-

Na₂S₂O₃ solution: Volume: V₃ = 13.05 mL, Molarity: M₃ = 0.15 M

The overall reaction involved is:

2Cu²⁺ + 2Na₂S₂O₃+ 4KI → 2CuI + 4K⁺ + 2NaI + Na₂S₄O₆

In this titration reaction, equal moles of Cu²⁺ and Na₂S₂O₃ reacts.

So to find the concentration of diluted Cu²⁺ solution (M₂), we use the equation:

Now, to find the concentration of the original Cu²⁺ solution (M₁), we use the equation:

Therefore, the molar concentration of the original Cu²⁺ solution: M₁ = 0.196 M

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