Easy Stoichiometry

Chemistry Level 4

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

The Chemical Reactions taking place are as follows :

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

  • BrO 3 − + AsO 2 − + H 2 O ⟶ Br − + AsO 4 3 − + H + \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 SeO 3 2 − \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


The answer is 84.67.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

First of all we have to balance the Chemical Equations :

  • 5 S e O 3 2 − + 2 B r O 3 − + 2 H + ⟶ 5 S e O 4 2 − + B r 2 + H 2 O 5SeO_{3}^{2-} + 2BrO_{3}^{-} + 2H^{+} \longrightarrow 5SeO_{4}^{2-} + Br_{2} + H_{2}O

  • B r O 3 − + 3 A s O 2 − + 3 H 2 O ⟶ B r − + 3 A s O 4 3 − + 6 H + BrO_{3}^{-} + 3AsO_{2}^{-} + 3H_{2}O \longrightarrow Br^{-} + 3AsO_{4}^{3-} + 6H^{+}

Then next comes some calculation which yields the following results :

  • Total Millimoles of B r O 3 − = 20 ⋅ 1 60 = 1 3 BrO_{3}^{-} = 20\cdot \frac{1}{60} = \frac{1}{3}

  • Millimoles of B r O 3 − BrO_{3}^{-} Backtitrated = 5 ⋅ 1 25 ⋅ 1 3 = 1 15 = 5\cdot \frac{1}{25}\cdot \frac{1}{3} = \frac{1}{15}

  • Millimoles of B r O 3 − BrO_{3}^{-} consumed for S e O 3 2 − = 1 3 − 1 15 = 4 15 SeO_{3}^{2-} = \frac{1}{3}-\frac{1}{15} = \frac{4}{15}

We have the following Molar Relation from the 1 s t 1^{st} reaction 5 millimoles of S e O 3 2 − = 2 millimoles of B r O 3 − ⇒ millimoles of S e O 3 2 − = 4 15 ⋅ 5 2 = 2 3 5 \quad \text{millimoles of} \quad SeO_{3}^{2-} = 2 \quad \text{millimoles of } \quad BrO_{3}^{-} \\ \Rightarrow \quad \text{millimoles of} \quad SeO_{3}^{2-} = \frac{4}{15}\cdot \frac{5}{2} = \frac{2}{3}

∴ \therefore We get the mass of S e O 3 2 − = 2 3 ⋅ 127 = 84.67 SeO_{3}^{2-}= \frac{2}{3}\cdot 127 = 84.67 mg .

4 Helpful 0 Interesting 1 Brilliant 0 Confused

But you have written 5.1 ml reacted in the question but in your solution you are using your value as 5 ml.

Ronak Agarwal - 6 years, 3 months ago

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Sorry , I had decided mid-way that using 5 will facilitate calculations , I've changed the value in the Question :)

A Former Brilliant Member - 6 years, 3 months ago

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You should have written that Titrated completely and not just titrated.

Md Zuhair - 2 years, 7 months ago

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