Question

A sample of bleach was analyzed as in this procedure. The only procedural difference is that the student weighed out the bleach solution used instead of pipetting a certain volume of bleach. The student weighed out 0.634 g of commercial bleach solution. It was found that it required 13.24 mL of 0.0732 M sodium thiosulfate solution to react with the iodine produced. What is the percentage of sodium hypochlorite in this bleach sample

Answer 1

% = 5.69%

Step-by-step explanation:

To do this, we need to write the equations taking place here. First, this is a REDOX reaction where the hypoclorite and thiosulfate solution reacts. The balanced equations are:

ClO⁻ + 2I⁻ + 2H⁺ -------> Cl⁻ + I₂ + H₂O

I₂ + 2S₂O₃²⁻ -----------> 2I⁻ + S₄O₆²⁻

We already have the required volume and concentration of the thiosulfate solution, so we can calculate the moles of thiosulfate. With this moles, we can calculate the moles of hypochlorite, then the mass and finally the %.

The moles of thiosulfate would be:

moles S₂O₃²⁻ = V * M

moles S₂O₃²⁻ = 0.01324 * 0.0732 = 9.69x10⁻⁴ moles

Now according to the above reactions, we can see that

moles I₂ = moles ClO⁻

and

moles I₂ / moles S₂O₃²⁻ = 1/2

Therefore, let's calculate the moles of ClO⁻:

moles ClO⁻ = 9.69x10⁻⁴ / 2 = 4.845x10⁻⁴ moles

Now, we can calculate the mass of these moles, using the molar mass of sodium hypochlorite which is 74.44 g/mol:

m = 74.44 * 4.845x10⁻⁴

m = 0.036 g

Finally the % of this, in the bleach sample would be:

% = 0.036 / 0.634 * 100

% = 5.69%