1. In trial 1, technique 1 A, 5 mL of the 0.20 M KI solution is mixed with 10 mL of the 0.005 M Na2S2O3 (in a 0.4% starch) and 15 mL of the 0.20 M KCl and 20 mL of the 0.10 M K2S2O8 resulting in a total volume of the reaction mixture of 50 mL. What is the iodide and persulfate ion concentration in the reaction mixture

Iodide concentration: 0.02M I⁻

Persulfate concentration: 0.02M S₂O₈²⁻

Explanation:

In chemistry, a dilution is the process in which the concentration of solute is reduced by addition of more solvent.

Iodide (KI → I⁻), has a concentration of 0.20M. If you add 5mL resulting in a final volume of 50mL, the final concentration decreases in:

0.20M × (5mL /50mL) = 0.02M I⁻

The persulfate ion (K₂S₂O₈ → S₂O₈²⁻) has a concentration of 0.10M in 10mL that result in a final volume of 50mL. That means final concentration is:

0.10M  × (10mL /50mL) = 0.02M S₂O₈²⁻

I hope it helps!

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