Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - 17-4 Solubility of Salts - Review & Check for Section 17-4 - Page 666: 4

Answer

The solubility of $PbSO_4$ in this solution will be equal to : $1 \times 10^{-7}M$ Correct answer: $(a)$

Work Step by Step

1. Write the $K_{sp}$ expression: $ PbS{O_4}(s) \lt -- \gt 1S{O_4}^{2-}(aq) + 1Pb^{2+}(aq)$ $2.5 \times 10^{-8} = [S{O_4}^{2-}]^ 1[Pb^{2+}]^ 1$ $2.5 \times 10^{-8} = (0.25 + S)^ 1( 1S)^ 1$ 2. Find the molar solubility. Since 'S' has a very small value, we can approximate: $[S{O_4}^{2-}] = 0.25$ $2.5 \times 10^{-8}= (0.25)^ 1 \times ( 1S)^ 1$ $2.5 \times 10^{-8}= (0.25)^ 1 \times ( 1S)^ 1$ $2.5 \times 10^{-8}= 0.25 \times ( 1S)^ 1$ $ \frac{2.5 \times 10^{-8}}{0.25} = ( 1S)^ 1$ $1 \times 10^{-7} = S$
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