How many milliliters of 0.350 M Na2S2O3 solution are needed to titrate 2.423 g of I2 to the equivalence point?

Question

How many milliliters of 0.350 M Na2S2O3 solution are needed to titrate 2.423 g of I2 to the equivalence point?
I2(aq) + 2 S2O32-(aq) S4O62-(aq) + 2 I -(aq)

Answer 1

Hello Annie

 

I₂ (aq) + 2S₂O₃^2- (aq) → S₄O₆^2- (aq) + 2I^- (aq)

 

Start with 2.423 g of I₂ and use stoichiometry to convert it to moles of S₂O₃^2-.  Then using the given molarity of Na₂S₂O₃ (0.350 M) and knowing that Molarity = moles / Liter, we can rearrange this equation to solve for the volume in Liters.  Finally, convert Liters to mL using the conversion factor 1 L = 1000 mL.

 

For the stoichiometry calculation, molar mass of I₂ = 253.81 g/mol, and the mole ratio of I₂ to S₂O₃^2- is 1:2.

 

2.423 g I₂ x (1 mol I₂ / 253.81 g I₂) x (2 mol S₂O₃^2- / 1 mol I₂) = 0.01916 mol S₂O₃^2-

 

Molarity = mol / L

 

L = mol / M

 

   = 0.01916 / 0.350 = 0.05474 L

 

0.05474 L x (1000 mL / 1 L) = 54.74 mL