Sodium sulfite is oxidised when it reacts with excess iodine. $$\[ \mathrm{Na}_{2} \mathrm{SO}_{3}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{HI}(\mathrm{aq}) \]$$ The remaining iodine is then titrated using aqueous sodium thiosulfate. $$\[ \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}(\mathrm{aq})+2 \mathrm{NaI}(\mathrm{aq}) \]$$ You will determine the integer value of $$\(x\)$$ in the formula of hydrated sodium sulfite, $$\(\mathrm{Na}_{2} \mathrm{SO}_{3} \cdot \mathrm{HH}_{2} \mathrm{O}\)$$, by titration. FA 4 is aqueous sodium thiosulfate containing $$\(14.24 \mathrm{~g}_{\text {of }} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)$$ in $$\(1.00 \mathrm{dm}^{3}\)$$. FA 5 is aqueous iodine, prepared as shown. - $$\(\quad 5.00 \mathrm{~g}\)$$ of hydrated sodium sulfite is added to $$\(600 \mathrm{~cm}^{3}\)$$ of $$\(0.100 \mathrm{~mol} \mathrm{dm}^{-3}\)$$ aqueous iodine. - The mixture is allowed to stand to ensure that all the sodium sulfite has been oxidised. - The mixture containing the remaining iodine is made up to $$\(1.00 \mathrm{dm}^{3}\)$$ with distilled water. FA 6 is starch indicator. From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtain the mean value. $$\(25.0 \mathrm{~cm}^{3}\)$$ of FA 5 required .......................... $$\(\mathrm{cm}^{3}\)$$ of FA 4.

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