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Common Ion Effect

What will be ...

Question

What will be the molarity of $I O_{3}^{⊝}$ ions in the solution after completion of the reaction?

0.152

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0.081

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0.41

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0.03

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Solution

The correct option is C $0.03$
$2 K I O_{3} + P b (N O_{3})_{2} ⟶ P b (I O_{3})_{2} + 2 K N O_{3}$
$0.8 \times 15$ $35 \times 0.15$
$= 12$ $= 5.25$

$[I O_{3}^{⊝}] = \frac{1.5 m m o l}{(15 + 35) m L} = 0.03 M$

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