Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 16 - Acid-Base Equilibria and Solubility Equilibria - Questions & Problems - Page 767: 16.54

Answer

(a) $ K_{sp} (SrF_2) = (7.8 \times 10^{-10})$ (b) $ K_{sp} (Ag_3PO_4)= (1.8 \times 10^{-18})$

Work Step by Step

(a) 1. Calculate the molar mass $(SrF2)$: 87.62* 1 + 19* 2 = 125.62g/mol 2. Calculate the number of moles $(SrF2)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{0.073}{ 125.62}$ $n(moles) = 5.8\times 10^{- 4}$ 3. Find the concentration in mol/L $(SrF2)$: $5.8 \times 10^{-4}$ mol in 1L: $5.8 \times 10^{-4} M (SrF2)$ 4. Write the $K_{sp}$ expression: $ SrF_2(s) \lt -- \gt 1Sr^{2+}(aq) + 2F^-(aq)$ $ K_{sp} = [Sr^{2+}]^ 1[F^-]^ 2$ 5. Determine the ion concentrations: $[Sr^{2+}] = [SrF_2] * 1 = [5.8 \times 10^{-4}] * 1 = 5.8 \times 10^{-4}$ $[F^-] = [SrF_2] * 2 = 1.2 \times 10^{-3}$ 6. Calculate the $K_{sp}$: $ K_{sp} = (5.8 \times 10^{-4})^ 1 \times (1.2 \times 10^{-3})^ 2$ $ K_{sp} = (5.8 \times 10^{-4}) \times (1.4 \times 10^{-6})$ $ K_{sp} = (7.8 \times 10^{-10})$ ------ (b) 1. Calculate the molar mass $(Ag3PO4)$: 107.87* 3 + 30.97* 1 + 16* 4 ) = 418.58g/mol 2. Calculate the number of moles $(Ag3PO4)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{6.7 \times 10^{-3}}{ 418.58}$ $n(moles) = 1.6\times 10^{- 5}$ 3. Find the concentration in mol/L $(Ag3PO4)$: $1.6 \times 10^{-5}$ mol in 1L: $1.6 \times 10^{-5} M (Ag3PO4)$ 4. Write the $K_{sp}$ expression: $ Ag_3PO_4(s) \lt -- \gt 3Ag^{+}(aq) + 1P{O_4}^{3-}(aq)$ $ K_{sp} = [Ag^{+}]^ 3[P{O_4}^{3-}]^ 1$ 5. Determine the ion concentrations: $[Ag^{+}] = [Ag_3PO_4] * 3 = [1.6 \times 10^{-5}] * 3 = 4.8 \times 10^{-5}$ $[P{O_4}^{3-}] = [Ag_3PO_4] * 1 = 1.6 \times 10^{-5}$ 6. Calculate the $K_{sp}$: $ K_{sp} = (4.8 \times 10^{-5})^ 3 \times (1.6 \times 10^{-5})^ 1$ $ K_{sp} = (1.1 \times 10^{-13}) \times (1.6 \times 10^{-5})$ $ K_{sp} = (1.8 \times 10^{-18})$
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