Chemistry Titration

Question

What is the molarity of NaOCl in a solution of bleach if 10.00mL requires 22.35mL of 0.02191M Na2S2O3 to reach the endpoint? Show work. Hint: Work through the I2 first.

 

Reaction Equations that take place:

1.) 2HCl (aq) + 2KI (aq) + NaOCl (aq) ----> I2 (aq) + NaCl (aq) +H2O + 2KCl (aq)

2.) I2 (aq) + 2Na2S2O3 (aq) ----> 2NaI (aq) + Na2S4O6 (aq)

Answer 1

Nice questions

 

Work step-by-step:

 

1. use of Na₂S₂O₃ = 22.35 ml*0.02191 mol/L = 0.02235 L*0.02191 mol/L = 4.897*10^-4 mol

2: mole ratio: Na₂S₂O₃ : I₂ = 2:1

2: mole ratio: I₂: NaOCl = 1:1

so Na₂S₂O₃ : NaOCl = 2:1

 

3: assess mole of NaOCl:

4.897*10^-4 mol of Na₂S₂O₃ has reacted with 4.897*10^-4 / 2 mol NaOCl = 2.448*10^-4 mole NaOCl

 

4: this amount of NaOCl was in a 10.00 ml solution:

Molarity of NaOCl = 2.448*10^-4 mol/10.00 ml = 2.448*10^-4/0.01000 L = 0.02448 mol/L

 

Be aware of use of mL and L!

good luck