Sulfite in wine was measured by the following procedure: To 50.0mL of wine were

Question

Sulfite in wine was measured by the following procedure: To 50.0mL of wine were added 5.00mL of a solution containing (0.8043g KlO3 + 5g Kl)/100 mL. Acidification with 1.0mL of 6.0M sulfuric acid quantitatively converts lO3 to l3. The l3 reacts with sulfite ion to generate sulfate ion, leaving excess l3. The excess l3 is titrated with 12.86mL of 0.04818 M NaS2O3 to reach a starch end point. Write a balanced equation for the reaction between I3 and SO3, and find the concentration of sulfite in the wine in mol/L and mg/L.

Explanation / Answer

Balanced equation : SO3^2- + I3^- ---> SO4^2- + 3I^-

moles of Na2S2O3 used to titrate excess I3- = 0.04818 M x 0.01286 ml = 6.2 x 10^-4 mols

moles of I3- = 6.2 x 10^-4/2 = 3.1 x 10^-4 mols

moles of IO3- = 0.8043/214.001 = 3.76 x 10^-3 mols

moles of I- = 5/166.003 = 0.03 mols

1 mole of IO3- reacts with 5 mols of I-, so 3.76 x 10^-4 mols IO3- would require = 0.019 mols of I-

since I- available is excess, IO3- is the limiting reagent

moles of I3- formed = 2 x 3.76 x 10^-4 = 7.52 x 10^-4 mols

moles of I3- reacted with SO3^2- = 7.52 x 10^-4 - 3.1 x 10^-4 = 4.42 x 10^-4 mols

moles of SO3^2- in sample = 4.42 x 10^-4 mols

concentration of SO3^2- in wine = 4.42 x 10^-4 x 0.005/0.105 = 2.10 x 10^-5 mol/L

In mg/L = 2.10 x 10^-5 x 1000 x 80.0632 = 1.68 mg/L

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