Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 16 - Applications of Aqueous Equilibria - Section Problems - Page 711: 106

Answer

(a) $1.1 \times 10^{-5}M$ (b) $1.1 \times 10^{-4}M$ (c) $1.6 \times 10^{-5}M$

Work Step by Step

(a) 1. Write the $K_{sp}$ expression: $ Ba(CrO_4)(s) \lt -- \gt 1Ba^{2+}(aq) + 1Cr{O_4}^{2-}(aq)$ $1.2 \times 10^{-10} = [Ba^{2+}]^ 1[Cr{O_4}^{2-}]^ 1$ 2. Considering a pure solution: $[Ba^{2+}] = 1x$ and $[Cr{O_4}^{2-}] = 1x$ $1.2 \times 10^{-10}= ( 1x)^ 1 \times ( 1x)^ 1$ $1.2 \times 10^{-10} = 1x^ 2$ $1.2 \times 10^{-10} = x^ 2$ $ \sqrt [ 2] {1.2 \times 10^{-10}} = x$ $1.1 \times 10^{-5} = x$ - This is the molar solubility value for this salt. (b) 1. Write the $K_{sp}$ expression: $ Mg(OH)_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2{OH}^{-}(aq)$ $5.6 \times 10^{-12} = [Mg^{2+}]^ 1[{OH}^{-}]^ 2$ 2. Considering a pure solution: $[Mg^{2+}] = 1x$ and $[{OH}^{-}] = 2x$ $5.6 \times 10^{-12}= ( 1x)^ 1 \times ( 2x)^ 2$ $5.6 \times 10^{-12} = 4x^ 3$ $1.4 \times 10^{-12} = x^ 3$ $ \sqrt [ 3] {1.4 \times 10^{-12}} = x$ $1.1 \times 10^{-4} = x$ (c) 1. Write the $K_{sp}$ expression: $ Ag_2SO_3(s) \lt -- \gt 2Ag^{+}(aq) + 1{SO_3}^{2-}(aq)$ $1.5 \times 10^{-13} = [Ag^{+}]^ 2[{SO_3}^{2-}]^ 1$ 2. Considering a pure solution: $[Ag^{+}] = 2x$ and $[{SO_3}^{2-}] = 1x$ $1.5 \times 10^{-13}= ( 2x)^ 2 \times ( 1x)^ 1$ $1.5 \times 10^{-13} = 4x^ 3$ $3.8 \times 10^{-25} = x^ 3$ $ \sqrt [ 3] {3.8 \times 10^{-25}} = x$ $1.6 \times 10^{-5} = x $
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