The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine; \(\mathrm{ClO}^{-}\) is reduced to \(\mathrm{Cl}^{-}\). The amount of iodine produced by the redox reaction is determined by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ; \mathrm{I}_{2}\) is reduced to \(\mathrm{I}^{-}\). The sodium thiosulfate is oxidized to sodium tetrathionate, \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\). In this analysis, potassium iodide was added in excess to \(5.00 \mathrm{~mL}\) of bleach \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\). If \(25.00 \mathrm{~mL}\) of \(0.0700 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) was required to reduce all the iodine produced by the bleach back to iodide, what is the mass percent of \(\mathrm{NaClO}\) in the bleach?

Using the volume and concentration of sodium thiosulfate, we can calculate the moles of sodium thiosulfate: Moles of \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} = \mathrm{Volume} \times \mathrm{Concentration}\) Moles of \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} = 25.00\times 10^{-3} \mathrm{~L} \times 0.0700 \mathrm{~M} = 1.75\times 10^{-3} \mathrm{~mol}\)

The balanced redox reaction between sodium hypochlorite and iodide ions is given by: 2 \(\mathrm{ClO}^{-}\) + 6 I\(^-\) + 6 \(\mathrm{OH}^{-}\) -> 2 \(\mathrm{Cl}^-\) + 3 I\(_2\) + 6 H\(_2\)O For 3 moles of I\(_2\) produced, we have 2 moles of \(\mathrm{ClO}^-\) consumed.

Now, to calculate the moles of sodium hypochlorite, we first find the moles of I\(_2\) produced: Moles of I\(_2\) = Moles of \(\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} = 1.75 \times 10^{-3} \mathrm{~mol}\). From the balanced redox reaction, we can find the moles of \(\mathrm{ClO}^-\): Moles of \(\mathrm{ClO}^-\) \(= \frac{2}{3}\) \(\times\) Moles of I\(_2\) \(= \frac{2}{3} \times 1.75 \times 10^{-3} = 1.17 \times 10^{-3} \mathrm{~mol}\)

To calculate the mass percent, we first find the mass of sodium hypochlorite in 5.00 mL of bleach: Mass of \(\mathrm{NaClO}\) = Moles of \(\mathrm{ClO}^-\) \(\times\) Molar mass of \(\mathrm{NaClO}\) Mass of \(\mathrm{NaClO}\) = \(1.17 \times 10^{-3} \mathrm{~mol} \times 74.44 \mathrm{~g/mol} = 0.0872 \mathrm{~g}\) Then, calculate the mass of 5.00 mL of bleach: Mass of bleach = Volume of bleach \(\times\) density of bleach Mass of bleach = \(5.00 \times 10^{-3} \mathrm{~L} \times 1.00 \mathrm{~g/cm^3}\times 1000 \mathrm{cm^{3}/L} = 5.00 \mathrm{~g}\) Finally, calculate the mass percent of sodium hypochlorite in the bleach: Mass percent of \(\mathrm{NaClO}\) = \(\frac{\mathrm{Mass \thinspace of \thinspace NaClO}}{\mathrm{Mass \thinspace of \thinspace bleach}} \times 100\) Mass percent of \(\mathrm{NaClO}\) = \(\frac{0.0872 \mathrm{~g}}{5.00 \mathrm{~g}} \times 100 = 1.74 \%\) Therefore, the mass percent of sodium hypochlorite in the bleach is 1.74%.