Problem: A sample of a new cleaning product, "Joe's Famous Bleach Cleaner," with a mass of 44.0g , was diluted with an acetic acid 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 0.200M sodium thiosulfate, Na2S2O2, containing the ion S2O32?. A volume of 45.0mL of sodium thiosulfate, the titrant, was needed to turn the solution colorless.Part A:What is the percentage composition by mass of NaClO in the bleach product?Analysis of bleach involves two sequential redox reactions: First, bleach is reacted in acid solution with excess iodide anion to produce yellow-colored iodine:ClO-+2H++2I-→I2+Cl-+H2OThen, to determine how much of the iodine was formed, the solution is titrated with sodium thiosulfate solution:I2+2S2O32- → 2I-+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.

FREE Expert Solution

To determine percentage composition by mass of NaClO in the bleach product:

$\boxed{\mathbf{mass}\mathbf{ }\mathbf{percent}\mathbf{ }\mathbf{NaClO}\mathbf{=}\frac{\mathbf{mass}\mathbf{ }\mathbf{NaClO}}{\mathbf{mass}\mathbf{ }\mathbf{bleach}}\mathbf{\times }\mathbf{100}}$

Redox reaction 1: ClO^-+ 2H⁺+2I^- → I₂+Cl^-+H₂O

• NaClO → Na⁺ + ClO^-

Titration/ Redox reaction 2: I₂+2S₂O₃^2- → 2I^-+S₄O₆²-

Recall that at the equivalence point of a titration:

$\boxed{\mathbf{moles}\mathbf{ }\mathbf{acid}\mathbf{=}\mathbf{moles}\mathbf{ }\mathbf{base}}$

Also, recall that moles = molarity × volume. 

This means:

$\boxed{{\left(\mathbf{MV}\right)}_{\mathbf{acid}}\mathbf{=}{\left(\mathbf{MV}\right)}_{\mathbf{base}}}$

Step 1: Moles Na₂S₂O₃

$\mathbf{mol}\mathbf{ }{\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{=}\left(\mathbf{45}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{L}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\right)\mathbf{\times }\frac{\mathbf{0}\mathbf{.}\mathbf{200}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{Na}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}}{\cancel{\mathbf{L}}}$

moles Na₂S₂O₃ =0.009 mol Na₂S₂O₃