Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 16 - Solubility and Complex-Ion Equilibria - Exercises - Page 781: 27

Answer

a. $1.6 \times 10^{-5} M$ b. $9.3 \times 10^{-5}M$ c. $6.5 \times 10^{-7}M$

Work Step by Step

(a) 1. Write the $K_{sp}$ expression: $ Ag_3PO_4(s) \lt -- \gt 3Ag^{+}(aq) + 1P{O_4}^{3-}(aq)$ $1.8 \times 10^{-18} = [Ag^{+}]^ 3[P{O_4}^{3-}]^ 1$ 2. Considering a pure solution: $[Ag^{+}] = 3x$ and $[P{O_4}^{3-}] = 1x$ $1.8 \times 10^{-18}= ( 3x)^ 3 \times ( 1x)^ 1$ $1.8 \times 10^{-18} = 27x^ 4$ $6.667 \times 10^{-20} = x^ 4$ $ \sqrt [ 4] {6.667 \times 10^{-20}} = x$ $1.607 \times 10^{-5} = x$ - This is the molar solubility value for this salt. (b) 1. Write the $K_{sp}$ expression: $ CaCO_3(s) \lt -- \gt 1Ca^{2+}(aq) + 1C{O_3}^{2-}(aq)$ $8.7 \times 10^{-9} = [Ca^{2+}]^ 1[C{O_3}^{2-}]^ 1$ 2. Considering a pure solution: $[Ca^{2+}] = 1x$ and $[C{O_3}^{2-}] = 1x$ $8.7 \times 10^{-9}= ( 1x)^ 1 \times ( 1x)^ 1$ $8.7 \times 10^{-9} = 1x^ 2$ $8.7 \times 10^{-9} = x^ 2$ $ \sqrt [ 2] {8.7 \times 10^{-9}} = x$ $9.327 \times 10^{-5} = x$ - This is the molar solubility value for this salt. (c) 1. Write the $K_{sp}$ expression: $ Hg_2Cl_2(s) \lt -- \gt 1Hg^{2+}(aq) + 2Cl^-(aq)$ $1.1 \times 10^{-18} = [Hg^{2+}]^ 1[Cl^-]^ 2$ 2. Considering a pure solution: $[Hg^{2+}] = 1x$ and $[Cl^-] = 2x$ $1.1 \times 10^{-18}= ( 1x)^ 1 \times ( 2x)^ 2$ $1.1 \times 10^{-18} = 4x^ 3$ $2.75 \times 10^{-19} = x^ 3$ $ \sqrt [ 3] {2.75 \times 10^{-19}} = x$ $6.503 \times 10^{-7} = x$ - This is the molar solubility value for this salt.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.