Easy Stoichiometry

Chemistry Level 4

\(20\) ml of \(\dfrac{1}{60}\text{ M}\) solution of \(\text{KBrO}_{3}\) was added to a definite volume of \(\text{SeO}_{3}^{2-}\) solution . The Bromine gas evolved was removed by boiling and excess of \(\text{KBrO}_{3}\) was back titrated with \(5\) mL of \(\dfrac{1}{25}\text{ M}\) \(\text{NaAsO}_{2}\) .

The Chemical Reactions taking place are as follows :

  • \(\text{SeO}_{3}^{2-} + \text{BrO}_{3}^{-} + \text{H}^{+} \longrightarrow \text{SeO}_{4}^{2-} + \text{Br}_{2} + \text{H}_{2}\text{O}\)

  • \(\text{BrO}_{3}^{-} + \text{AsO}_{2}^{-} + \text{H}_{2}\text{O} \longrightarrow \text{Br}^{-} + \text{AsO}_{4}^{3-} + \text{H}^{+} \)

Calculate the amount (in milligrams) of \(\text{SeO}_{3}^{2-}\) in the solution

Some Important Data:

Atomic masses: K=39, Br=80, As=75, Na=23, O=16 and Se=79


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