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Ch.17 - Applications of Aqueous Equilibria
All textbooksMcMurry 8th EditionCh.17 - Applications of Aqueous EquilibriaProblem 131
Chapter 17, Problem 131

In quantitative analysis, Al3+ and Mg2+ are separated in an NH4+ - NH3 buffer having pH approximately equal to 8. Assuming cation concentrations of 0.010 M, show why Al(OH)3 precipitates but Mg(OH)2 does not.

Verified step by step guidance
1
Step 1: Write the relevant chemical equilibrium reactions for the precipitation of Al(OH)_3 and Mg(OH)_2. For Al(OH)_3, the reaction is: Al^{3+} + 3OH^- \rightleftharpoons Al(OH)_3(s). For Mg(OH)_2, the reaction is: Mg^{2+} + 2OH^- \rightleftharpoons Mg(OH)_2(s).
Step 2: Determine the solubility product constants (K_{sp}) for Al(OH)_3 and Mg(OH)_2. These values are typically found in a chemistry reference table. For example, K_{sp} for Al(OH)_3 is approximately 1.9 \times 10^{-33} and for Mg(OH)_2 is approximately 1.8 \times 10^{-11}.
Step 3: Calculate the hydroxide ion concentration [OH^-] in the buffer solution at pH 8. Use the relationship pH + pOH = 14 to find pOH, and then calculate [OH^-] using [OH^-] = 10^{-pOH}.
Step 4: Use the K_{sp} expressions to determine the conditions for precipitation. For Al(OH)_3, the expression is K_{sp} = [Al^{3+}][OH^-]^3. For Mg(OH)_2, the expression is K_{sp} = [Mg^{2+}][OH^-]^2. Substitute the known concentrations and [OH^-] into these expressions.
Step 5: Compare the calculated ion product for each compound with its K_{sp}. If the ion product exceeds K_{sp}, precipitation occurs. Analyze the results to show that the ion product for Al(OH)_3 exceeds its K_{sp}, indicating precipitation, while the ion product for Mg(OH)_2 does not exceed its K_{sp}, indicating no precipitation.
Related Practice
Textbook Question
Citric acid (H3Cit) can be used as a household cleaning agent to dissolve rust stains. The rust, represented as Fe(OH)3, dissolves because the citrate ion forms a soluble complex with Fe3+ (a) Using the equilibrium constants in Appendix C and Kf = 6.3 x 10^11 for Fe(Cit), calculate the equilibrium constant K for the reaction. (b) Calculate the molar solubility of Fe(OH)3 in 0.500 M solution of H3Cit.
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Open Question
Hard water contains alkaline earth cations such as Ca2+, which react with CO32- to form insoluble deposits of CaCO3. Will a precipitate of CaCO3 form if a 250 mL sample of hard water having [Ca2+] = 8.0 x 10^-4 M is treated with the following? (a) 0.10 mL of 2.0 x 10^-3 M Na2CO3 (b) 10 mg of solid Na2CO3
Open Question
The pH of a sample of hard water having [Mg2+] = 2.5 x 10^-4 M is adjusted to pH 10.80. Will Mg(OH)2 precipitate?
Textbook Question
Can Fe2+ be separated from Sn2+ by bubbling H2S through a 0.3 M HCl solution that contains 0.01 M Fe2+ and 0.01 M Sn2+? A saturated solution of H2S has [H2S] ≈ 0.10 M. Values of Kspa are 6 x 10^2 for FeS andd 1 x 10^-5 for SnS.
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Open Question
Can Co2+ be separated from Zn2+ by bubbling H2S through a 0.3 M HCl solution that contains 0.01 M Co2+ and 0.011 M Zn2+? A saturated solution of H2S has [H2S] approximately equal to 0.10 M. Values of Ksp are 3 for CoS and 3 x 10^-2 for ZnS.
Textbook Question
Will FeS precipitate in a solution that is 0.10 M in Fe(NO3)2, 0.4 M in HCl, and 0.10 M in H2S? Will FeS precipitate if the pH of the solution is adjusted to pH 8 with an NH4+ - NH3 buffer? Kspa = 6 x 10^2 for FeS.
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