Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 15 - Additional Aqueous Equilibria - Questions for Review and Thought - Topical Questions - Page 693c: 53

Answer

$ K_{sp} (PbBr_2) = (8.616 \times 10^{-10})$

Work Step by Step

1. Calculate the molar mass: 207.2* 1 + 79.9* 2 = 367g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.022}{ 367}$ $n(moles) = 5.995\times 10^{- 5}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $ C(mol/L) = \frac{ 5.995\times 10^{- 5}}{ 0.1} $ $C(mol/L) = 5.995\times 10^{- 4}M$ 4. Write the $K_{sp}$ expression: $ PbBr_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Br^-(aq)$ $ K_{sp} = [Pb^{2+}]^ 1[Br^-]^ 2$ 5. Determine the ions concentrations: $[Pb^{2+}] = [PbBr_2] * 1 = [5.995 \times 10^{-4}] * 1 = 5.995 \times 10^{-4}$ $[Br^-] = [PbBr_2] * 2 = 1.199 \times 10^{-3}$ 6. Calculate the $K_{sp}$: $ K_{sp} = (5.995 \times 10^{-4})^ 1 \times (1.199 \times 10^{-3})^ 2$ $ K_{sp} = (5.995 \times 10^{-4}) \times (1.437 \times 10^{-6})$ $ K_{sp} = (8.616 \times 10^{-10})$
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