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Chapter 16: Problem 33

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{CaF}_{2}(s), K_{\mathrm{sp}}=4.0 \times 10^{-11}\), or \(\mathrm{BaF}_{2}(s), K_{\mathrm{sp}}=2.4 \times 10^{-5}\) b. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s), K_{\infty}=1.3 \times 10^{-32}\), or \(\mathrm{FePO}_{4}(s), K_{\mathrm{sp}}=1.0 \times 10^{-22}\)

Short Answer

Expert verified
a. The molar solubility of CaF2 is \(x = 2.1 \times 10^{-4}\) M and the molar solubility of BaF2 is \(y = 6.9 \times 10^{-3}\) M. Thus, CaF2(s) has the smallest molar solubility. b. The molar solubility of Ca3(PO4)2 is \(p = 1.6 \times 10^{-8}\) M and the molar solubility of FePO4 is \(q = 1.0 \times 10^{-11}\) M. Thus, FePO4(s) has the smallest molar solubility.

Step by step solution

01

Solving for molar solubility of CaF2

Given the balanced chemical equation for the dissolution of CaF2: \[ \mathrm{CaF}_{2}(s) \rightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^-(aq) \] Let the molar solubility of CaF2 be 'x'. Hence, at equilibrium: \[ [\mathrm{Ca}^{2+}] = x \] \[ [\mathrm{F}^-] = 2x \] The Ksp expression for CaF2 is given by: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2 \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 4.0 \times 10^{-11} = (x)(2x)^2 \] Solve for 'x', which gives the molar solubility of CaF2.
02

Solving for molar solubility of BaF2

Given the balanced chemical equation for the dissolution of BaF2: \[ \mathrm{BaF}_{2}(s) \rightarrow \mathrm{Ba}^{2+}(aq) + 2\mathrm{F}^-(aq) \] Let the molar solubility of BaF2 be 'y'. Hence, at equilibrium: \[ [\mathrm{Ba}^{2+}] = y \] \[ [\mathrm{F}^-] = 2y \] The Ksp expression for BaF2 is given by: \[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^-]^2 \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 2.4 \times 10^{-5} = (y)(2y)^2 \] Solve for 'y', which gives the molar solubility of BaF2. Upon calculating the molar solubility of CaF2(s) and BaF2(s), we can determine which one has the smaller molar solubility.
03

Solving for molar solubility of Ca3(PO4)2

Given the balanced chemical equation for the dissolution of Ca3(PO4)2: \[ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightarrow 3\mathrm{Ca}^{2+}(aq) + 2\mathrm{PO}_4^{3-}(aq) \] Let the molar solubility of Ca3(PO4)2 be 'p'. Hence, at equilibrium: \[ [\mathrm{Ca}^{2+}] = 3p \] \[ [\mathrm{PO}_4^{3-}] = 2p \] The Ksp expression for Ca3(PO4)2 is given by: \[ K_{sp} = [\mathrm{Ca}^{2+}]^3[\mathrm{PO}_4^{3-}]^2 \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 1.3 \times 10^{-32}=(3p)^3(2p)^2 \] Solve for 'p', which gives the molar solubility of Ca3(PO4)2.
04

Solving for molar solubility of FePO4

Given the balanced chemical equation for the dissolution of FePO4: \[ \mathrm{FePO}_{4}(s) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{PO}_4^{3-}(aq) \] Let the molar solubility of FePO4 be 'q'. Hence, at equilibrium: \[ [\mathrm{Fe}^{3+}] = q \] \[ [\mathrm{PO}_4^{3-}] = q \] The Ksp expression for FePO4 is given by: \[ K_{sp} = [\mathrm{Fe}^{3+}][\mathrm{PO}_4^{3-}] \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 1.0 \times 10^{-22} = (q)(q) \] Solve for 'q', which gives the molar solubility of FePO4. Upon calculating the molar solubility of Ca3(PO4)2(s) and FePO4(s), we can determine which one has the smaller molar solubility.

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Most popular questions from this chapter

A solution is prepared by adding \(0.10 \mathrm{~mol} \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to \(0.50 \mathrm{~L}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3} .\) Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text {owerall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{\mathrm{x}}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Describe how you could separate the ions in each of the following groups by selective precipitation. a. \(\mathrm{Ag}^{+}, \mathrm{Mg}^{2+}, \mathrm{Cu}^{2+}\) c. \(\mathrm{Pb}^{2+}, \mathrm{Bi}^{3+}\) b. \(\mathrm{Pb}^{2+}, \mathrm{Ca}^{2+}, \mathrm{Fe}^{2+}\)

For which salt in each of the following groups will the solubility depend on pH? a. \(\mathrm{AgF}, \mathrm{AgCl}, \mathrm{AgBr}\) c. \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Sr}\left(\mathrm{NO}_{2}\right)_{2}\) b. \(\mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}\) d. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}\)

A mixture contains \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Mn}^{2+}\) and is saturated with \(0.10 M \mathrm{H}_{2} \mathrm{~S}\). Determine a \(\mathrm{pH}\) where CuS precipitates but MnS does not precipitate. \(K_{\text {op }}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and \(K_{\mathrm{s}}\) for \(\mathrm{MnS}=2.3 \times 10^{-1.3}\)

The solubility of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\) in a \(0.20 \mathrm{M} \mathrm{KIO}_{3}\) solution is \(4.4 \times\) \(10^{-8} \mathrm{~mol} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\).
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