The amount of i3-(aq) in a solution can be determined by titration with a solution containing a known concentration of s2o3^2-(aq) (thiosulfate ion).

the determination is based on the net ionic equation

2s2o3^2-(aq)+i3^-(aq) > s4o6^2-(aq)+31^-(aq)

given that it requires 35.8 ml of 0.350 m na2s2o3(aq) to titrate a 30.0-ml sample of i3^-(aq) calculate the molarity of i3^-(aq) in the solution

molarity of I3^-(aq) in the solution = 0.21 M

Explanation:

We are given the balanced chemical reaction and the volume and molarity of the  Na₂S₂O₃ so we can calculate the moles of the thiosulfate that were required, and then can calculate the molarity the  I₃⁻ by the definition of molarity.

First lets  convert the volume of Na₂S₂O₃ to liters:

35.8 mL x 1 L/1000 mL = 0.0358 L

# moles Na₂S₂O₃ = 0.350 mol/L x 0.0358 L = 0.0125 mol

From the stoichiometry of the reaction we know 2 mol Na₂S₂O₃ s react with 1 mol I₃⁻ , therefore mol of I₃⁻  will be given by

1 mol I₃⁻ / 2 mol Na₂S₂O₃  x 0.0125 mol Na₂S₂O₃ = 0.0063 mol  I₃⁻

and its molarity is:

0.0063 mol  I₃⁻ / 0.030 L = 0.21 M

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.