Question

What is the molarity of a solution of s2o32- if 58 ml were required to titrate 42 ml of a 0.2M solution of IO3-

Answer 1

Number of moles of IO₃^- solution , n = Molarity x volume in L

                                                     = 0.2M x 0.042L

                                                     = 0.0084 moles

The balanced reaction between IO₃^- & S₂O₃^2- is

IO₃^- + 6 S₂O₃^2- + 6 H⁺ 3 S₄O₆2- + I^- + 3 H₂O

From the balanced equation ,

1 mole of IO₃^- requires 6 moles of S₂O₃^2-

0.0084 moles of IO₃^- requires M moles of S₂O₃^2-

M = ( 0.0084x6) / 1

    = 0.0504 moles

Given the volume of S₂O₃^2- reacted is 58 mL = 0.058 L

So Molarity of S₂O₃^2- is , M = Number of moles / volume of S₂O₃^2- in L

                                         = 0.0504 mol / 0.058 L

                                         = 0.87 M

Therefore the Molarity of S₂O₃^2- is 0.87 M