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Chapter 18: Problem 1

Write \(K_{\text {sp }}\) expressions for the following equilibria. For example, for the reaction \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\) \(\mathrm{Cl}^{-}(\mathrm{aq}), K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\). (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (b) \(\operatorname{Ra}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ra}^{2+}(\mathrm{aq})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) \rightleftharpoons 3 \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})\) (d) \(\mathrm{PuO}_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{PuO}_{2}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})\)

Short Answer

Expert verified
The \(K_{sp}\) expressions for the given equilibria are: (a) \(K_{sp}=[Ag^+]^2[SO_{4}^{2-}]\), (b) \(K_{sp}=[Ra^{2+}][IO_{3}^{-}]^2\), (c) \(K_{sp}=[Ni^{2+}]^3[PO_{4}^{3-}]^2\), (d) \(K_{sp}=[PuO_{2}^{2+}][CO_{3}^{2-}]\)

Step by step solution

01

Write expression for Ag2SO4

The equilibrium reaction for the dissociation of \(Ag_2SO_4\) is: \(Ag_{2}SO_{4}(s) \rightleftharpoons 2Ag^{+}(aq)+SO_{4}^{2-}(aq)\). From this, the \(K_{sp}\) expression can be written as: \(K_{sp}=[Ag^+]^2[SO_{4}^{2-}]\)
02

Write expression for Ra(IO3)2

The equilibrium reaction for the dissociation of \(Ra(IO_3)_2\) is: \(Ra(IO_3)_2(s) \rightleftharpoons Ra^{2+}(aq)+2IO_{3}^{-}(aq)\). The \(K_{sp}\) expression can be written as: \(K_{sp}=[Ra^{2+}][IO_{3}^{-}]^2\)
03

Write expression for Ni3(PO4)2

The equilibrium reaction for the dissociation of \(Ni_3(PO_4)_2\) is: \(Ni_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Ni^{2+}(aq)+2PO_{4}^{3-}(aq)\). Using this information, we can write the \(K_{sp}\) expression as: \(K_{sp}=[Ni^{2+}]^3[PO_{4}^{3-}]^2\)
04

Write expression for PuO2CO3

The equilibrium reaction for the dissociation of \(PuO_2CO_3\) is: \(PuO_{2} CO_{3}(s) \rightleftharpoons PuO_{2}^{2+}(aq)+CO_{3}^{2-}(aq)\). From this, we can write the \(K_{sp}\) expression as: \(K_{sp}=[PuO_{2}^{2+}][CO_{3}^{2-}]\)

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Most popular questions from this chapter

Determine whether \(1.50 \mathrm{g} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (oxalic acid: \(K_{\mathrm{a}_{1}}=\) \(\left.5.2 \times 10^{-2}, K_{\mathrm{a}_{2}}=5.4 \times 10^{-5}\right)\) can be dissolved in \(0.200 \mathrm{L}\) of \(0.150 \mathrm{M} \mathrm{CaCl}_{2}\) without the formation of \(\mathrm{CaC}_{2} \mathrm{O}_{4}(\mathrm{s})\left(K_{\mathrm{sp}}=1.3 \times 10^{-9}\right)\).

Can the following ion concentrations be maintained in the same solution without a precipitate forming: \(\left[\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]^{3-}\right]=0.048 \mathrm{M},\left[\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right]=0.76 \mathrm{M},\) and \(\left[\mathrm{I}^{-}\right]=2.0 \mathrm{M} ?\)

A solution is \(0.010 \mathrm{M}\) in both \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{SO}_{4}^{2-}\). To this solution, \(0.50 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\text { aq })\) is slowly added. (a) Which anion will precipitate first from solution? (b) What is \(\left[\mathrm{Pb}^{2+}\right]\) at the point at which the second anion begins to precipitate? (c) Are the two anions effectively separated by this fractional precipitation?

Adding \(1.85 \mathrm{g} \mathrm{Na}_{2} \mathrm{SO}_{4}\) to \(500.0 \mathrm{mL}\) of saturated aqueous \(\mathrm{BaSO}_{4}:\) (a) reduces \(\left[\mathrm{Ba}^{2+}\right] ;\) (b) reduces \(\left[\mathrm{SO}_{4}^{2-}\right]\); (c) increases the solubility of \(\mathrm{BaSO}_{4} ;\) (d) has no effect.

The electrolysis of \(\mathrm{MgCl}_{2}(\mathrm{aq})\) can be represented as \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) The electrolysis of a 315 mL sample of \(0.185 \mathrm{M} \mathrm{MgCl}_{2}\) is continued until \(0.652 \mathrm{L} \mathrm{H}_{2}(\mathrm{g})\) at \(22^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) has been collected. Will \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) precipitate when electrolysis is carried to this point? [Hint: Notice that \(\left[\mathrm{Mg}^{2+}\right]\) remains constant throughout the electrolysis, but \(\left.\left[\mathrm{OH}^{-}\right] \text {increases. }\right]\)
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