Chemistry

Problem

The active ingredient in household bleach is

\text{NaOCl}

. To find the concentration of

\text{NaOCl}

in a sample of bleach, a

25.00\ \text{cm}^3

sample of bleach was mixed with an excess of iodide ions. The equation for the reaction is as follows:

\text{OCl}^{-}(\text{aq}) + 2\text{I}^{-}(\text{aq}) + 2\text{H}^{+}(\text{aq}) \rightarrow \text{I}_{2}(\text{aq}) + \text{Cl}^{-}(\text{aq}) + \text{H}_{2}\text{O}(\text{l})

The liberated iodine was titrated with

\text{Na}_{2}\text{S}_{2}\text{O}_{3}

solution. The titration required

18.60\ \text{cm}^{3}

0.040\ \text{mol dm}^{-3}

\text{Na}_{2}\text{S}_{2}\text{O}_{3}

solution. The equation for the reaction of iodine with thiosulfate ions is as follows:

2\text{S}_{2}\text{O}_{3}^{2-}(\text{aq}) + \text{I}_{2}(\text{aq}) \rightarrow \text{S}_{4}\text{O}_{6}^{2-}(\text{aq}) + 2\text{I}^{-}(\text{aq})

Calculate the concentration of

\text{NaOCl}

in the bleach.