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Chapter 19: Problem 52

Use thermodynamic data from Appendix D to calculate values of \(K_{\mathrm{sp}}\) for the following sparingly soluble solutes: (a) \(\operatorname{AgBr} ;\) (b) \(\operatorname{CaSO}_{4} ;\) (c) \(\operatorname{Fe}(\text { OH })_{3}\). [Hint: Begin by writing solubility equilibrium expressions.

Short Answer

Expert verified
To calculate the \(K_{sp}\) for \(AgBr\), \(CaSO_4\) and \(Fe(OH)_3\), write the solubility equilibrium expressions for them, and calculate \(\Delta G^\circ\) using Gibbs Free energy of formation values from Appendix D. Substitute \(\Delta G^\circ\) into the relationship equation with \(K_{sp}\) to find their values. The values would depend on the specific \(\Delta G_f^\circ\) values given in the Appendix D.

Step by step solution

01

Write Solubility Equilibrium Expressions

For a general solute \(AB\), the dissociation can be represented as \( AB \leftrightarrow A^+ + B^-\).\nHence, for each solute:\n(a) \(AgBr \leftrightarrow Ag^+ + Br^-\)\n(b) \(CaSO_4 \leftrightarrow Ca^{2+}+ SO_4^{2-}\)\n(c) \(Fe(OH)_3 \leftrightarrow Fe^{3+}+ 3OH^-\)
02

Express \(K_{sp}\) in terms of Activities

The solubility product expression is the product of the activities of the ions raised to their respective stoichiometric coefficients. Hence,\n(a) \(K_{sp, AgBr} = a_{Ag^+} \cdot a_{Br^-} \)\n(b) \(K_{sp, CaSO_4} = a_{Ca^{2+}} \cdot a_{SO_4^{2-}}\)\n(c) \(K_{sp, Fe(OH)_3} = a_{Fe^{3+}} \cdot (a_{OH^-})^3 \). At infinite dilution, the activity of a species can be replaced with its concentration.
03

Use Thermodynamics to Calculate \(K_{sp}\)

The relationship between \(\Delta G^\circ\) and \(K_{sp}\) is given by the equation \( \Delta G^\circ = -RT \ln K_{sp} \). Re-arranging this equation gives \( K_{sp} = \exp(-\Delta G^\circ/RT) \). Here, \(R\) is the gas constant (8.314 J/mol.K) and \(T\) is the absolute temperature (normally 298.15 K). Appendix D provides the standard Gibbs Free energy of formation, \(\Delta G_f^\circ\) for different species. The \(\Delta G^\circ\) for the dissociation reaction is the sum of \(\Delta G_f^\circ\) of products minus \(\Delta G_f^\circ\) of reactants. This value can be substituted into the equation above to calculate \(K_{sp}\) for each solute.

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Most popular questions from this chapter

The Gibbs energy available from the complete combustion of 1 mol of glucose to carbon dioxide and water is $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta G^{\circ}=-2870 \mathrm{kJ} \mathrm{mol}^{-1} \end{array}$$ (a) Under biological standard conditions, compute the maximum number of moles of ATP that could form from ADP and phosphate if all the energy of combustion of 1 mol of glucose could be utilized. (b) The actual number of moles of ATP formed by a cell under aerobic conditions (that is, in the presence of oxygen) is about \(38 .\) Calculate the efficiency of energy conversion of the cell. (c) Consider these typical physiological conditions. $$\begin{array}{l} P_{\mathrm{CO}_{2}}=0.050 \mathrm{bar} ; P_{\mathrm{O}_{2}}=0.132 \mathrm{bar} \\\ {[\mathrm{glucose}]=1.0 \mathrm{mg} / \mathrm{mL} ; \mathrm{pH}=7.0} \\ {[\mathrm{ATP}]=[\mathrm{ADP}]=\left[P_{\mathrm{i}}\right]=0.00010 \mathrm{M}} \end{array}$$ Calculate \(\Delta G\) for the conversion of 1 mol ADP to ATP and \(\Delta G\) for the oxidation of 1 mol glucose under these conditions. (d) Calculate the efficiency of energy conversion for the cell under the conditions given in part (c). Compare this efficiency with that of a diesel engine that attains \(78 \%\) of the theoretical efficiency operating with \(T_{\mathrm{h}}=1923 \mathrm{K}\) and \(T_{1}=873 \mathrm{K} .\) Suggest a reason for your result. [ Hint: See Feature Problem 95.]

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

For each of the following reactions, indicate whether \(\Delta S\) for the reaction should be positive or negative. If it is not possible to determine the sign of \(\Delta S\) from the information given, indicate why. (a) \(\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(2 \mathrm{HgO}(\mathrm{s}) \longrightarrow 2 \mathrm{Hg}(1)+\mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NaCl}(1) \longrightarrow 2 \mathrm{Na}(1)+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{g})\) (e) \(\operatorname{Si}\left(\text { s) }+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SiCl}_{4}(\mathrm{g})\right.\)

Explain why you would expect a reaction of the type $\mathrm{AB}(\mathrm{g}) \longrightarrow \mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})$ always to be spontaneous at high rather than at low temperatures.

At \(298 \mathrm{K},\) for the reaction \(2 \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \stackrel{\text { Lest }}{\longrightarrow}\) \(2 \mathrm{POCl}_{3}(1), \Delta H^{\circ}=-620.2 \mathrm{kJ}\) and the standard molar entropies are \(\mathrm{PCl}_{3}(\mathrm{g}), 311.8 \mathrm{JK}^{-1} ; \mathrm{O}_{2}(\mathrm{g}), 205.1 \mathrm{JK}^{-1}\) and \(\mathrm{POCl}_{3}(1), 222.4 \mathrm{JK}^{-1} .\) Determine (a) \(\Delta G^{\circ}\) at \(298 \mathrm{K}\) and (b) whether the reaction proceeds spontaneously in the forward or the reverse direction when reactants and products are in their standard states.
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