Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 15 - Chemical Equilibrium - Exercises - Problems - Page 569: 81

Answer

The $K_{sp}$ for $MgF_2$ is equal to $7.0 \times 10^{-11}$

Work Step by Step

1. Write the $K_{sp}$ expression, and find its value: $ MgF_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2F^{-}(aq)$ $K_{sp} = [Mg^{2+}]^ 1[F^{-}]^ 2$ $K_{sp} = 2.6 \times 10^{-4} \times (5.2 \times 10^{-4})^2$ $K_{sp} = 2.6 \times 10^{-4} \times 2.7 \times 10^{-7} = 7.0 \times 10^{-11}$
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