## Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education

# Chapter 19 - Problems - Page 871: 19.76

#### Answer

(a) $Solubility = 1.7 \times 10^{-3}M$ (b) $Solubility = 2.0 \times 10^{-4}M$

#### Work Step by Step

(a) 1. Write the $K_{sp}$ expression: $Ca(IO_3)_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2IO{_3}^-(aq)$ $7.1 \times 10^{-7} = [Ca^{2+}]^ 1[IO{_3}^-]^ 2$ $7.1 \times 10^{-7} = (0.06 + S)^ 1( 2S)^ 2$ 2. Find the molar solubility. Since 'S' has a very small value, we can approximate: $[Ca^{2+}] = 0.06$ $7.1 \times 10^{-7}= (0.06)^ 1 \times ( 2S)^ 2$ $7.1 \times 10^{-7}= (0.06)^ 1 \times ( 2S)^ 2$ $7.1 \times 10^{-7}= 0.06 \times ( 2S)^ 2$ $\frac{7.1 \times 10^{-7}}{0.06} = ( 2S)^ 2$ $1.2 \times 10^{-5} = ( 2S)^ 2$ $\sqrt [ 2] {1.2 \times 10^{-5}} = 2S$ $3.4 \times 10^{-3} = 2S$ $1.7 \times 10^{-3} = S$ ---- (b) 3. Write the $K_{sp}$ expression: $Ca(IO_3)_2(s) \lt -- \gt 2{IO_3}^{-}(aq) + 1Ca^{2+}(aq)$ $7.1 \times 10^{-7} = [{IO_3}^{-}]^ 2[Ca^{2+}]^ 1$ $7.1 \times 10^{-7} = (0.06 + S)^ 2( 1S)^ 1$ 4. Find the molar solubility. Since 'S' has a very small value, we can approximate: $[{IO_3}^{-}] = 0.06$ $7.1 \times 10^{-7}= (0.06)^ 2 \times ( 1S)^ 1$ $7.1 \times 10^{-7}= (0.06)^ 2 \times ( 1S)^ 1$ $7.1 \times 10^{-7}= 3.6 \times 10^{-3} \times ( 1S)^ 1$ $\frac{7.1 \times 10^{-7}}{3.6 \times 10^{-3}} = ( 1S)^ 1$ $2.0 \times 10^{-4} = ( 1S)^ 1$ $2.0 \times 10^{-4} = 1S$

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