When ammonium compounds are heated with sodium hydroxide or calcium hydroxide, ammonia is liberated. The ionic equation for the reaction is shown. $$\[ \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \]$$ FA 1 is prepared by heating $$\(10.0 \mathrm{~cm}^{3}\)$$ of aqueous ammonium chloride, $$\(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})\)$$, with $$\(25.0 \mathrm{~cm}^{3}\)$$ of $$\(2.00 \mathrm{moldm}^{-3}\)$$ sodium hydroxide, NaOH . The sodium hydroxide is in excess. The reaction mixture is cooled and diluted to $$\(250 \mathrm{~cm}^{3}\)$$ with distilled water. You will determine the concentration of the aqueous solution of ammonium chloride by titrating the remaining sodium hydroxide from the preparation of FA 1 with a known concentration of sulfuric acid. $$\[ 2 \mathrm{NaOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \]$$ FA 2 is 0.0520 mol dm $$\(^{-3}\)$$ sulfuric acid, $$\(\mathrm{H}_{2} \mathrm{SO}_{4}\)$$. FA 3 is bromophenol blue indicator. From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtain the mean value. $$\[ 25.0 \mathrm{~cm}^{3} \text { of FA } 1 \text { required } \]$$ .......................... $$\(\mathrm{cm}^{3}\)$$ of FA 2.

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