Question

ClO(aq) + 3I(aq) + H₂O(I) ------> I₃(aq) + CI(aq) + 2OH(aq) (Equation 1)

I₃(aq) + 2S₂O₃^2-(aq) -----> 3I(aq) + S₄O₆^2- (aq) (Equation 2)

IO₃^- (aq) + 8I^-(aq) + 6H⁺(aq) ----> 3I₃^-(aq) + 3H₂O(I) (Equation 3)  

Part A: To standardize a thiosulfate (S₂O₃^2-) solution, 10.00 mL of a  1.199×10⁻² MM   KIO₃ solutionare treated with excess KI to generate I₃^- according to Equation 3. How many moles of I₃^- are generated?

Part B: If  5.99 mL of the S₂O₃^2- solution are required to reach the endpoint of the titration, what is the molarity of S₂O₃^2- in the solution? (Use Equation 2.)

Part C:  2.11 g of bleach solution are treated with excess KI to generate I₃^- according to Equation 1 on page 161 in the lab manual. 27.02 mL of the standardized S₂O₃^2- solution in Part B are required to reach the endpoint of the titration. How many moles of S₂O₃^2- is this? (Use Equation 2.)

Part D: How many moles of I₃^- are involved in the reaction? (Use Equation 2.)  

Part E: How many moles of ClO^- are involved in the reaction? (Use Equation 1.)

Part F: How many moles of NaClO are involved in the reaction?

Part G: What is the mass percent of NaClO in the bleach solution? Recall, the mass of bleach used in the reaction is 2.11 g .

Answer 1

Part A

The equation (3) is IO₃^- + 8 I^- + 6H⁺ 3 I₃^- + 3 H₂O

Given that 10 mL of 1.199 x 10^-2 M KIO₃ solution used.

KIO₃ K⁺ + IO₃^- thus 1 mole of KIO₃ = 1 mole of IO₃^-

Therefore

10 mL of 1.199 x 10^-2 M KIO₃ = 10 mL x 1.199 x 10^-2 mole / 1000 mL = 1.199 x 10^-4 mole of IO₃^-

as per equation (3) one mole of IO₃^- will generate 3 moles of I₃^- .

therefore moles of I₃^- = 3 x 1.199 x 10^-4 moles

                             = 3,597 x 10^-4 moles.

Thus the mole of I₃^- generated = 3.597 x 10^-4 moles.

Part B

The given equation (2) is   I₃^- + 2 S₂O₃^2- 3 I^- + S₄O₆

mole of I₃^- = 3.597 x 10^-4

mole of S₂O₃^2- required for complete reaction as per the equation (2) is 2 x mole of I₃^-

                                                                                        = 2 x 3.597 x 10^-4

                                                                                        = 7.194 x 10^-4 moles

Let the molarity of S₂O₃^2- = y, then 5.99 mL of this solution provides 7.194 x 10^-4 moles

that is 5.99 mL * y mole/1000 mL = 7.194 x 10^-4

                          y = 0.7194 / 5.99 = 0.120 M

Thus the molarity of S₂O₃^2- in the solution = 0.120 M

Part C

from part B, the molarity of S₂O₃^2- = 0.120 M

27.02 mL of 0.12 M S₂O₃^2- = 27.02 mL x 0.120 moles / 1000 mL

                                       = 3.242 x 10^-3 moles of S₂O₃^2-

Thus the number of moles of S₂O₃^2- reacted with the I₃^- generated from 2.11 g of bleach soluion is 3.242 x 10^-3 moles.

Part D

The given equation (2) is   I₃^- + 2 S₂O₃^2- 3 I^- + S₄O₆

as per the equation (2) two moles of S₂O₃^2- is equivalent to one mole of I₃^-

therefore 3.242 x 10^-3 moles of S₂O₃^2- is equivalent to 3.242 x 10^-3 / 2 moles of I₃^-

                                                                            = 1.621 x 10^-3 moles,

Thus 1.621 x 10^-3 moles of I₃^- involved in this reaction.

Part E

The given equation (1) is

         ClO^- + 3 I^- + H₂O I₃^- + Cl^- + 2 OH^-

As per equation (1), one mole of I₃^- generated from 1 mole of ClO^-

Therefore 1.621 x 10^-3 moles of I₃^- generated from 1.621 x 10^-3 moles of ClO^-

Thus 1.621 x 10^-3 moles of ClO^- involved in this reaction.

Part F

NaClO Na⁺ + ClO^-

one mole of ClO^- generated from one mole of NaClO

therefore 1.621 x 10^-3 moles of ClO^- is equivalent to 1.621 x 10^-3 moles of NaClO

Thus the number of moles of NaClO involved in this reaction is 1.621 x 10^-3 moles

Paft G

molecular mass of NaClO = 23 + 35.5 + 16 = 74.5 g / mol

Therefore mass of 1.621 x 10^-3 moles of NaClO = 74.5 g mol^-1 x 1.621 x 10^-3 mol

                                                                     = 0.1207 g

This much NaClO present in 2.11 g of the bleach solution

Therefore mass % of NaClO = [0.1207 / 2.11 ] x 100

                                         = 5.72 %

Thus the mass % of NaClO in the bleach solution is 5.72 %