Chemistry (7th Edition)

(a) $K_{sp} = 4.37 \times 10^{-9}$ (b) $K_{sp} = 1.10 \times 10^{-12}$ (c) $K_{sp} = 7.0 \times 10^{-7}$ (d) $K_{sp} = 3.0 \times 10^{-16}$
(a) 1. Write the $K_{sp}$ expression: $SrF_2(s) \lt -- \gt 1Sr^{2+}(aq) + 2{F}^{-}(aq)$ $K_{sp} = [Sr^{2+}]^ 1[{F}^{-}]^ 2$ 2. Determine the ion concentrations: $[Sr^{2+}] = [SrF_2] * 1 = [1.03 \times 10^{-3}] * 1 = 1.03 \times 10^{-3}$ $[{F}^{-}] = [SrF_2] * 2 = 2.06 \times 10^{-3}$ 3. Calculate the $K_{sp}$: $K_{sp} = (1.03 \times 10^{-3})^ 1 \times (2.06 \times 10^{-3})^ 2$ $K_{sp} = (1.03 \times 10^{-3}) \times (4.24 \times 10^{-6})$ $K_{sp} = (4.37 \times 10^{-9})$ (b) 1. Write the $K_{sp}$ expression: $CuI(s) \lt -- \gt 1Cu^{2+}(aq) + 1{I}^{-}(aq)$ $K_{sp} = [Cu^{2+}]^ 1[{I}^{-}]^ 1$ 2. Determine the ion concentrations: $[Cu^{2+}] = [CuI] * 1 = [1.05 \times 10^{-6}] * 1 = 1.05 \times 10^{-6}$ $[{I}^{-}] = [CuI] * 1 = 1.05 \times 10^{-6}$ 3. Calculate the $K_{sp}$: $K_{sp} = (1.05 \times 10^{-6})^ 1 \times (1.05 \times 10^{-6})^ 1$ $K_{sp} = (1.05 \times 10^{-6}) \times (1.05 \times 10^{-6})$ $K_{sp} = (1.1 \times 10^{-12})$ (c) 1. Calculate the molar mass: 24.31* 1 + 12.01* 2 + 16* 4 ) = 112.33g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.094}{ 112.33}$ $n(moles) = 8\times 10^{- 4}$ 3. Find the concentration in mol/L: $8 \times 10^{-4}$ mol in 1L: $8 \times 10^{-4} M$ 4. Write the $K_{sp}$ expression: $MgC_2O_4(s) \lt -- \gt 1Mg^{2+}(aq) + 1{C_2O_4}^{2-}(aq)$ $K_{sp} = [Mg^{2+}]^ 1[{C_2O_4}^{2-}]^ 1$ 5. Determine the ion concentrations: $[Mg^{2+}] = [MgC_2O_4] * 1 = [8.36 \times 10^{-4}] * 1 = 8.36 \times 10^{-4}$ $[{C_2O_4}^{2-}] = [MgC_2O_4] * 1 = 8.36 \times 10^{-4}$ 6. Calculate the $K_{sp}$: $K_{sp} = (8.36 \times 10^{-4})^ 1 \times (8.36 \times 10^{-4})^ 1$ $K_{sp} = (8.36 \times 10^{-4}) \times (8.36 \times 10^{-4})$ $K_{sp} = (7 \times 10^{-7})$ (d) 1. Calculate the molar mass: 65.38* 1 + 2 * ( 12.01* 1 + 14.01* 1 ) = 117.42g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 4.95 \times 10^{- 4}}{ 117.42}$ $n(moles) = 4\times 10^{- 6}$ 3. Find the concentration in mol/L: $4 \times 10^{-6}$ mol in 1L: $4 \times 10^{-6} M$ 4. Write the $K_{sp}$ expression: $Zn(CN)_2(s) \lt -- \gt 1Zn^{2+}(aq) + 2{CN}^{-}(aq)$ $K_{sp} = [Zn^{2+}]^ 1[{CN}^{-}]^ 2$ 5. Determine the ion concentrations: $[Zn^{2+}] = [Zn(CN)_2] * 1 = [4.22 \times 10^{-6}] * 1 = 4.22 \times 10^{-6}$ $[{CN}^{-}] = [Zn(CN)_2] * 2 = 8.43 \times 10^{-6}$ 6. Calculate the $K_{sp}$: $K_{sp} = (4.22 \times 10^{-6})^ 1 \times (8.43 \times 10^{-6})^ 2$ $K_{sp} = (4.22 \times 10^{-6}) \times (7.11 \times 10^{-11})$ $K_{sp} = (3 \times 10^{-16})$