A student determines the % NaOCl in a bleach by reacting 5.00 mL of bleach (density 1.05 g/mL) with excess KI in acidic solution. The liberated I2 is titrated to a starch endpoint (blue->colorless) with 0.1000 M Na2S2O3, requiring 26.45 mL. Reactions: OCl- + 2 I- + 2 H+ -> Cl- + I2 + H2O; I2 + 2 S2O3^2- -> 2 I- + S4O6^2-.
What is the mass percent of NaOCl (M = 74.44 g/mol) in the bleach?
- Acheck_circle
1.87%
- B
3.74%
- C
0.94%
- D
9.84%
Explanation
mol S2O3^2- = 0.02645 * 0.1000 = 2.645e-3. mol I2 = 1.323e-3. mol OCl- = 1.323e-3. mass NaOCl = 1.323e-3 * 74.44 = 0.0985 g. Mass bleach = 5.00*1.05 = 5.25 g. % = 0.0985/5.25 = 1.87%.