The concentration of SO32- in a solution is determined by titration with a 0.1355 M permanganate solution. The balanced net ionic equation for the reaction is: 2MnO4-(aq) + 5SO32-(aq) + 6H3O+(aq) 2Mn2+(aq) + 5SO42-(aq) + 9H2O(l) (a) If 23.89 mL of the 0.1355 M permanganate solution are needed to react completely with 25.00 mL of the SO32- solution, what is the concentration of the SO32- solution? M (b) Which of the two solutions was in the buret during the titration? (c) Suppose at the end of the titration, the solution containing the Mn2+ ion is transferred to a volumetric flask and diluted to 400. mL. What is the concentration of Mn2+ in the diluted solution? M

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-03-26T02:40:05+01:00

Answer:

a) 0.3237M of the SO₃⁻ solution.

b) Permanganate solution

c) 0.008093M

Explanation:

a) Based in the reaction, 2 moles of MnO₄⁻ react with 5 moles of SO₃⁻. Moles of permanganate used in titration are:

0.02389L×0.1355mol/L = 0.003237moles of MnO₄⁻. The moles of SO₃⁻ are:

0.003237moles of MnO₄⁻ × (5 moles of SO₃⁻ / 2 moles of MnO₄⁻) = 0.008093 moles of SO₃⁻

As volume of the solution is 25.00mL, the concentration of the solution is:

0.008093 moles of SO₃⁻ / 0.02500L = 0.3237M of the SO₃⁻ solution.

b) In a titration, the solution that is in the buret (Titrant), is the solution of known concentration, for the problem, permanganate solution

c) The moles of the Mn²⁺ solution are the same of MnO₄⁻. That is:

0.003237moles. As the volume of the new solution is 400mL, the concentration of the solution is:

[ Mn²⁺] = 0.003237moles / 0.400L = 0.008093M