Question

Shana, a student in the Analytical Chemistry class, performed a Redox Titration to determine the content of hypochlorite ions (% w/w) in an unknown solution. First, Shana prepared a primary standard solution of KIO3 by dissolving 0.4991 g dry KIO3 in 250.00 mL DI water. Next, using the procedure in your lab manual, Shana used 50.00 mL of this primary standard solution to standardize the Na2S2O3 titrant, and 26.25 mL of titrant was consumed. Next, Shana prepared the unknown solution by diluting 12.8624 g of the solution in 250.00 mL of DI water. A 50.00 mL portion of this dilute unknown solution was titrated using the procedure in your lab manual, and 20.42 mL of Na2S2O3 solution was consumed. Answer the following questions about this titration. Recall the stoichiometry of the NET chemical equation for this reaction. Note: use appropriate significant figures in your answer! Calculate the concentration (M) of Na2S2O3 titrant.

Answer 1

Concentration of Na₂SO₃ is 0.0533M

Step-by-step explanation:

To obtain concentration of Na₂SO₃ it is necessary to obtain moles of primary standard, KIO₃ in the standarization:

0.4991g × (1mol / 214g) = 2.332x10⁻³ mol KIO₃ / 0.2500L = 9.329x10⁻³M

As you take 50.00mL of this solution, moles are:

0.05000L × (9.329x10⁻³mol / L) = 4.664x10⁻⁴ moles KIO₃

Now, the reaction of Na₂SO₃ with KIO₃ is:

KIO₃(aq) + 3 Na₂SO₃(aq) → KI(aq) + 3 Na₂SO₄(aq)

That means 1 mole of KIO₃ reacts with 3 moles of Na₂SO₃. Thus, moles of Na₂SO₃ that reacts are:

4.664x10⁻⁴ moles KIO₃ × (3 mol Na₂SO₃ / 1 mol KIO₃) = 1.399x10⁻³ mol Na₂SO₃

As 26.25mL of titrant were consumed, concentration of Na₂SO₃ is:

1.399x10⁻³ mol Na₂SO₃ / 0.02625L = 0.0533M