Question

Bleach contains the active ingredient NaClO. Analysis of bleach involves two sequential redox reactions: First, bleach is reacted in acidic solution with excess iodide anion to produce yellow-colored iodine: ClO− + 2 H+ + 2 I− → I2 + Cl− + H2O Then, to determine how much of the iodine was formed, the solution is titrated with sodium thiosulfate solution: I2 + 2 S2O32− → 2 I− + S4O62− A bit of starch is added to the titration reaction. Starch is intensely blue in the presence of I2. The solution thus turns from deep blue to colorless at the reaction equivalence point. A sample of a new cleaning product, "Joe's Famous Bleach Cleaner," with a mass of 60 g , was diluted with an acidic solution containing excess I−. A small amount of starch indicator solution was then added, turning the solution a deep bluish-purple. The solution was then titrated with 0.117 M sodium thiosulfate, Na2S2O3, containing the ion S2O32−. A volume of 43 mL of sodium thiosulfate, the titrant, was needed to turn the solution colorless. What is the percentage composition by mass of NaClO in the bleach product?

Answer 1

Answer:

0,31%

Explanation:

For the reaction:

I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻

0,043 L × 0,117 M of sodium tiosulfate = 5,031x10⁻³ moles of S₂O₃²⁻

5,031x10⁻³ moles of S₂O₃²⁻ × $\frac{1 I_2 mol}{2 S_{2}O_3 mol^{2-}}$ = 2,5156x10⁻³ moles of I₂

These moles of I₂ were produced from:

ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O

2,5156x10⁻³ moles of I₂ ≡ moles of NaClO

2,5156x10⁻³ moles of NaClO ×$\frac{74,44 g}{1mol}$ = 0,187 g of NaClO

Thus, percentage composition by mass is:

$\frac{0,187 g of NaClO}{60 g Of Bleach} x 100$ = 0,31%

I hope it helps!