Question

Part A:

Part 1: Give balanced equations that are in accord with the following reactions. You will use these reactions to prepare solutions of halogens.
a. The reaction of OCl^- (hypochlorite ion) and Cl^- in acidic solution gives Cl₂ as the only product containing the halogen.

b. The reaction of BrO₃^- (bromate ion) and Br^- in acidic solution gives Br₂ as the only product containing the halogen.

c. The reaction of Cu²⁺ with I^- in neutral solution gives I₂ and insoluble CuI. A neutral solution does not contain either an acid or a base.

Part B: Describe the preparation of 12 mL of a 0.050 M solution of each of the halogens, using the reactions in the preceding question. Available to you are solid Ca(OCl)₂; 0.20 M solutions of NaCl, NaBr, NaBrO₃, and CuSO₄; a 0.40 M solution of NaI; a 0.50 M solution of H₂SO₄; distilled water; and a 10-mL graduated cylinder (where volumes can be read to the nearest 0.1 mL). Calculate the required quantity of each reagent and the volume of water that must be added for dilution.

a. Chlorine water

b. Bromine water

c. Iodine water

pls I need a step by step solution as an answer Not just an answer and will also like to get an answer for for all three not just a or b or c. They are all connected. thanks

Answer 1

Part A

a, OCl^- + Cl^- + 2H⁺ -------> Cl₂ + H₂O

b, BrO₃^- + 5Br^- + 6H⁺ ------> 3Br₂ + 3H₂O

c, 2Cu²⁺ + 4I^- ---------> 2CuI + I₂

Part B

12ml of 0.50 M each halogen solution should be formed

thus, 12 * 0.050 = 0.6 mmol of halogen should be formed

for Cl2 water, OCl^- + Cl^- +  2H⁺ -------> Cl₂ + H₂O

1mol of both OCl^- , Cl^- and 2 mol of  H⁺ is needed

thus, 0.6 mmol of  OCl^- , Cl^- and 1.2 mmol of  H⁺ are required for 0.6 mmol of Cl2

Ca(OCl)2 gives 2 OCl^- hence, for 0.6 mmol of OCl^- , 0.3 mmol of Ca(OCl)2 should be taken i.e., 0.043g

1 mol of NaCl give 1 mol Cl^- , for 0.6 mmol Cl^- ------> 0.6 mmol/0.2M= 3ml

solution of 12 ml is needed to form thus, volume of water = 12-3 = 9 ml

for Br2 water, BrO₃^- + 5Br^- + 6H⁺ ------> 3Br₂ + 3H₂O

similiarly as for Cl2, 0.6 mmol Br2 need to form

thus, 0.2 mmol BrO₃^- ,1 mmol Br^- and  1.2 mmol H⁺ is needed

0.2 mmol BrO₃^- = 0.2 mmol/0.2 M = 1 ml

1 mmol Br^- = 1 mmol/0.2 M = 5 ml

as 1 mol H2SO4 gives 2mol H+

0.6 mmol H2SO4 = 0.6 mmol/0.5 M = 1.2 ml

vol of water = 12 -1 -5 -1.2 = 4.8 ml

For I2 solution, 2Cu²⁺ + 4I^- ---------> 2CuI + I₂

_{0.6 mmol of I2 require 1.2 mmol Cu2+ and 2.4 mmol I-}

_{thus, 1.2 mmol Cu2+ = 1.2/0.2 = 6 ml}

_{2.4 mmol I2 = 2.4/ 0.4 = 6 ml}

_{vol of water = 12-6-6 = 0ml}