An unknown concentration of sodium thiosulfate, Na2S2O3, is used to titrate a standardized solution of KIO3 with excess KI present. Suppose 15.55 mL of the Na2S2O3 solution is required to titrate the iodine formed from 21.45 mL of 0.0135 M KIO3. What is the molarity of the Na2S2O3 solution?

0.112M is molarity of the Na₂S₂O₃ solution

Explanation:

First, the KIO₃ reacts with KI as follows:

KIO3 (aq) + 5 KI (aq) + 6 HCl (aq) → 3 I2 (s) + 6 KCl (aq) + 3 H2O (l)

Then, the I₂ produced reacts with the Na₂S₂O₃ as follows:

2 Na2S2O3 + I2 → Na2S4O6 + 2 NaI

Thus, to solve this question, we need to obtain the moles of KIO₃ that react, then, the moles of I₂ produced -Using the chemical reaction- to find the initial moles of Na₂S₂O₃. With these moles and the volume we can find molarity of the solution:

Moles KIO₃:

21.45mL = 0.02145L * (0.0135mol/L) = 2.896x10⁻⁴ moles KIO₃

Moles I₂:

2.896x10⁻⁴ moles KIO₃ * (3mol I₂ / 1mol KIO₃) = 8.687x10⁻⁴ moles I₂

Moles Na₂S₂O₃:

8.687x10⁻⁴ moles I₂ * (2mol Na₂S₂O₃ / 1mol I₂) = 1.737x10⁻³ moles Na₂S₂O₃

As the volume of the titrated solution is 15.55mL, molarity is:

1.737x10⁻³ moles Na₂S₂O₃ / 0.01555L =

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2