Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 17 - Additional Aspects of Aqueous Equilibria - Exercises - Page 770: 17.57a

Answer

$[AgBr]$ solubility in a water solution: $ 7.071 \times 10^{-7}M$

Work Step by Step

$Kps (AgBr) = 5 \times 10^{-13}$ 1. Write the Kps equation: $Kps (AgBr) = [Ag^+] * [Br^-]$ $[Ag^+] = [Br^-] = x$ $Kps (AgBr) = x * x$ 2. Find x: $5 \times 10^{-13} = x^2$ $x = \sqrt {5 \times 10^{-13}}$ $x = 7.071 \times 10^{-7}$ 3. Since $[Ag^+] = [AgBr] = x$ $[AgBr] = 7.071 \times 10^{-7}$
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