## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677c: 48

#### Answer

$K_{sp} (Ca(OH)_2) = (5.47 \times 10^{-5})$

#### Work Step by Step

1. Calculate the hydroxide ion concentration: pH + pOH = 14 12.68 + pOH = 14 pOH = 1.32 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 1.32}$ $[OH^-] = 4.79 \times 10^{- 2}$ 2. Write the $K_{sp}$ expression: $Ca(OH)_2(s) \lt -- \gt 1Ca^{2+}(aq) + 2OH^{-}(aq)$ $K_{sp} = [Ca^{2+}]^ 1[OH^{-}]^ 2$ 3. Determine the ion concentrations: $[OH^{-}] = [Ca(OH)_2] * 2 = 0.0479$ $[Ca(OH)_2] = \frac{0.0479}{2} = 0.0239$ $[Ca^{2+}] = [Ca(OH)_2] * 1 = [0.0239] * 1 = 0.0239$ 4. Calculate the $K_{sp}$: $K_{sp} = (0.0239)^ 1 \times (0.0479)^ 2$ $K_{sp} = (0.0239) \times (2.29 \times 10^{-3})$ $K_{sp} = (5.47 \times 10^{-5})$

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.