## Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall

# Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic: 45b

#### Answer

Its formula is $IrBr_{3}.4H_{2}O$

#### Work Step by Step

The cation of iridium(III) bromide tetrahydrate has a charge of (III), i.e., it is $Ir^{3+}$. The anion is bromide, i.e $Br^{-}$. To balance the charges, 3 $Br^{-}$ combines with 1 $Ir^{3+}$. Then it also has 4 (tetra) water molecules of hydration. So, its formula is $IrBr_{3}.4H_{2}O$

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