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Chapter 16: Problem 32

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid–base properties. a. \(P b I_{2}, K_{s p}=1.4 \times 10^{-8}\) b. \(\operatorname{CdCO}_{3}, K_{s p}=5.2 \times 10^{-12}\) c. $\operatorname{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{s p}=1 \times 10^{-31}$

Short Answer

Expert verified
The solubility of the compounds are as follows: a. \(PbI_{2}\): \(7.1 \times 10^{-4}\) M b. \(CdCO_{3}\): \(2.3 \times 10^{-6}\) M c. \(Sr_3(PO_4)_2\): \(8.9 \times 10^{-11}\) M

Step by step solution

01

Write the dissolution equation for each compound

. For each of the three compounds, we will write a balanced chemical equation representing its dissolution in water. a. \(PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)\) b. \(CdCO_{3}(s) \rightleftharpoons Cd^{2+}(aq) + CO_{3}^{2-}(aq)\) c. \(Sr_3(PO_4)_2(s) \rightleftharpoons 3Sr^{2+}(aq) + 2PO_{4}^{3-}(aq)\)
02

Express the solubility in terms of variables

: For each compound, we will express the concentration of ions in terms of the solubility (S). a. \([Pb^{2+}] = S\) and \([I^-] = 2S\) b. \([Cd^{2+}] = S\) and \([CO_{3}^{2-}] = S\) c. \([Sr^{2+}] = 3S\) and \([PO_{4}^{3-}] = 2S\)
03

Write the \(K_{sp}\) expression for each compound

: Using the solubility variables from Step 2, write the \(K_{sp}\) expressions for each of the compounds: a. \(K_{sp} = [Pb^{2+}][I^-]^2 = S(2S)^2\) b. \(K_{sp} = [Cd^{2+}][CO_{3}^{2-}] = S^2\) c. \(K_{sp} = [Sr^{2+}]^3[PO_{4}^{3-}]^2 = (3S)^3(2S)^2\)
04

Solve for solubility (S) in each case using the given \(K_{sp}\) values

: Using the given \(K_{sp}\) values, solve for S: a. \(1.4 \times 10^{-8} = S(2S)^2 \Rightarrow S = 7.1 \times 10^{-4}\,\text{M}\) b. \(5.2 \times 10^{-12} = S^2 \Rightarrow S = 2.3 \times 10^{-6}\,\text{M}\) c. \(1 \times 10^{-31} = (3S)^3(2S)^2 \Rightarrow S = 8.9 \times 10^{-11}\,\text{M}\) Now we have the solubility of each compound in moles per liter: a. The solubility of \(PbI_{2}\) is \(7.1 \times 10^{-4}\) M. b. The solubility of \(CdCO_{3}\) is \(2.3 \times 10^{-6}\) M. c. The solubility of \(Sr_3(PO_4)_2\) is \(8.9 \times 10^{-11}\) M.

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Most popular questions from this chapter

What is the maximum possible concentration of \(\mathrm{Ni}^{2+}\) ion in water at \(25^{\circ} \mathrm{C}\) that is saturated with \(0.10 M\) $\mathrm{H}_{2} \mathrm{S}\( and maintained at \)\mathrm{pH} 3.0\( with \)\mathrm{HCl} ?$

Calculate the solubility of $\mathrm{AgCN}(s)\left(K_{\mathrm{sp}}=2.2 \times 10^{-12}\right)\( in a solution containing \)1.0 M\( \)\mathrm{H}^{+} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN} \text { is } 6.2 \times 10^{-10} .\right)$

A solution is prepared by mixing 100.0 \(\mathrm{mL}\) of \(1.0 \times 10^{-4} M\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and 100.0 \(\mathrm{mL}\) of $8.0 M\( \)\mathrm{NaF}.$ $$\mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}^{+}(a q) \quad K_{1}=7.9 \times 10^{4}$$ $$\operatorname{Be} \mathrm{F}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{2}(a q) \quad K_{2}=5.8 \times 10^{3}$$ $$\operatorname{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \operatorname{Be} \mathrm{F}_{3}^{-}(a q) \quad K_{3}=6.1 \times 10^{2}$$ $$\operatorname{Be} \mathrm{F}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons \mathrm{BeF}_{4}^{2-}(a q) \qquad K_{4}=2.7 \times 10^{1}$$ Calculate the equilibrium concentrations of $\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+},\( \)\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-},$ and \(\mathrm{BeF}_{4}^{2-}\) in this solution.

Will a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) form if \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(1.0 \mathrm{L}\) of \(5.0 \mathrm{MNH}_{3} ?\) $$\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{7}$$ $$\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=5.9 \times 10^{-15}$$

Calculate the molar solubility of $\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-15}.$
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