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

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.