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?

Molar concentration: 0,0489M

Explanation:

In this titration of Cu²⁺ you add an excess of I⁻ that reacts with Cu²⁺ producing I₂, this I₂ reacts with Na₂S₂O₃. If you know the I₂ that reacts with Na₂S₂O₃ you can know the I⁻ that reacts with Cu²⁺ and, thus, the quantity of Cu²⁺. The reactions are:

2Cu²⁺ + 4I⁻ → 2CuI + I₂

I₂ + 2S₂O₃⁻ → S₄O₆ + 2I⁻

Moles of S₂O₃⁻ are:

0,01305L×0,15M = 1,96x10⁻³ moles of S₂O₃⁻.

Moles of I₂ are:

1,96x10⁻³ moles of S₂O₃⁻× ( 1 mole of I₂ /  2 moles of S₂O₃⁻) = 9,79x10⁻⁴ moles of I₂

Moles of Cu²⁺ are:

9,79x10⁻⁴ moles of I₂×( 2 moles of Cu²⁺ / 4 moles of I₂) = 4,89x10⁻⁴ moles of Cu²⁺

As volume of the solution was 10,0mL = 0,0100L, the molar concentration of the original solution is:

4,89x10⁻⁴ moles of Cu²⁺ / 0,0100L = 0,0489M

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."