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Chapter 17: Problem 29

Write balanced equations and solubility product expressions for the solubility equilibria of these compounds: (a) \(\mathrm{CuBr},\) (b) \(\mathrm{ZnC}_{2} \mathrm{O}_{4},\) (c) \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (e) \(\mathrm{AuCl}_{3}\) (f) \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Short Answer

Expert verified
The solubility product expressions are \[K_{sp}=[\mathrm{Cu}^{+}][\mathrm{Br}^{-}]\], \[K_{sp}=[\mathrm{Zn}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]^{2}\], \[K_{sp}=[\mathrm{Ag}^{+}]^{2}[\mathrm{CrO}_{4}^{2-}]\], \[K_{sp}=[\mathrm{Hg}_{2}^{2+}][\mathrm{Cl}^{-}]^{2}\], \[K_{sp}=[\mathrm{Au}^{3+}][\mathrm{Cl}^{-}]^{3}\] and \[K_{sp}=[\mathrm{Mn}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\].

Step by step solution

01

Establishing the General Equilibrium Equation for Solubility

The general equation for solubility equilibrium is: \(\mathrm{AB}_{(s)} \rightleftharpoons \mathrm{A}^{+}_{(aq)} + \mathrm{B}^{-}_{(aq)}\). Balancing is not needed here since every reactant is reacting in a ratio of 1:1.
02

Solubility Equilibrium for CuBr

Here's the balanced solubility equation for \(\mathrm{CuBr}\): \(\mathrm{CuBr}_{(s)} \rightleftharpoons \mathrm{Cu}^{+}_{(aq)} + \mathrm{Br}^{-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Cu}^{+}][\mathrm{Br}^{-}]\].
03

Solubility Equilibrium for ZnC2O4

The balanced solubility equation for \(\mathrm{ZnC}_{2} \mathrm{O}_{4}\) is: \(\mathrm{ZnC}_{2} \mathrm{O}_{4(s)} \rightleftharpoons \mathrm{Zn}^{2+}_{(aq)} + 2\mathrm{C}_{2} \mathrm{O}_{4}^{2-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Zn}^{2+}][\mathrm{C}_{2} \mathrm{O}_{4}^{2-}]^{2}\].
04

Solubility Equilibrium for Ag2CrO4

The balanced solubility equation for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is: \(\mathrm{Ag}_{2} \mathrm{CrO}_{4(s)} \rightleftharpoons 2\mathrm{Ag}^{+}_{(aq)} + \mathrm{CrO}_{4}^{2-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Ag}^{+}]^{2}[\mathrm{CrO}_{4}^{2-}]\].
05

Solubility Equilibrium for Hg2Cl2

The balanced solubility equation for \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) is: \(\mathrm{Hg}_{2} \mathrm{Cl}_{2(s)} \rightleftharpoons \mathrm{Hg}_{2}^{2+}_{(aq)} + 2\mathrm{Cl}^{-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Hg}_{2}^{2+}][\mathrm{Cl}^{-}]^{2}\].
06

Solubility Equilibrium for AuCl3

The balanced solubility equation for \(\mathrm{AuCl}_{3}\) is: \(\mathrm{AuCl}_{3(s)} \rightleftharpoons \mathrm{Au}^{3+}_{(aq)} + 3\mathrm{Cl}^{-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Au}^{3+}][\mathrm{Cl}^{-}]^{3}\].
07

Solubility Equilibrium for Mn3(PO4)2

The balanced solubility equation for \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is: \(\mathrm{Mn}_{3}\left(\mathrm{PO}_{4}\right)_{2(s)} \rightleftharpoons 3\mathrm{Mn}^{2+}_{(aq)} + 2\mathrm{PO}_{4}^{3-}_{(aq)}\). The solubility product expression is: \[K_{sp}=[\mathrm{Mn}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\].

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Most popular questions from this chapter

\(\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)\) has a larger \(K_{\mathrm{sp}}\) value than that of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{\mathrm{sp}}=1.4 \times 10^{-5}\right)\). Does it fol- low that \(\mathrm{CaSO}_{4}\) also has greater solubility \((\mathrm{g} / \mathrm{L}) ?\)

A 0.2688 -g sample of a monoprotic acid neutralizes \(16.4 \mathrm{~mL}\) of \(0.08133 \mathrm{M}\) KOH solution. Calculate the molar mass of the acid.

The \(\mathrm{p} K_{\mathrm{a}}\) of phenolphthalein is \(9.10 .\) Over what \(\mathrm{pH}\) range does this indicator change from \(95 \%\) HIn to \(95 \% \mathrm{In}^{-} ?\)

The solubility product of \(\mathrm{PbBr}_{2}\) is \(8.9 \times 10^{-6} .\) Determine the molar solubility (a) in pure water, (b) in \(0.20 M\) KBr solution, (c) in \(0.20 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with the carbonate and then "back-titrating" the remaining acid with \(\mathrm{NaOH}\). (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a \(0.1022-\mathrm{g}\) sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).
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