## Chemistry: The Central Science (13th Edition)

Published by Prentice Hall

# Chapter 17 - Additional Aspects of Aqueous Equilibria - Exercises - Page 770: 17.53c

#### Answer

$Ba(IO_3)_2$ $Solubility: 5.313 \times 10^{-4}M$

#### Work Step by Step

1. Write the Kps formula: $K_{ps} = [Ba^{+2}]*[{IO_3}^-]^2$ 2. Calculate x: $[Ba(IO_3)_2] = x$ $[Ba^{2+}] = x$ $[{IO_3}^-] = 2x$ $K_{ps} = x * (2x)^2$ $K_{ps} = x * 4x^2$ $6 \times 10^{-10} = 4x^3$ $x^3 = 1.5 \times 10^{-10}$ $x = 5.313 \times 10^{-4} M$

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