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 0.15 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:

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:

$M_{2}\times V_{2} = M_{3}\times V_{3}$

$\Rightarrow M_{2} = \frac{M_{3}\times V_{3}}{V_{2}} = \frac{0.15\, M\times 13.05\, mL}{75\, mL}$

$\Rightarrow M_{2} = 0.0261\, M$

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

$M_{1}\times V_{1} = M_{2}\times V_{2}$

$\Rightarrow M_{1} = \frac{M_{2}\times V_{2}}{V_{1}} = \frac{0.0261\, M\times 75\, mL}{10\, mL}$

$\Rightarrow M_{1} = 0.196\, M$

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