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17 Additional Aspects of Solutions to Exercises Aqueous Equilibria = 4.5 10⁻⁹ 5.6 10⁻¹¹ x10⁻¹⁴ = (c) K = 8.036 X 10⁻¹³ = = s³; S = 9.297 10⁻⁵ = 9.3 10⁻⁵ M [OH⁻] = S = 9.297 10⁻⁵ M; pOH = 4.03165 = 4.03; pH = 14 - 4.03 = 9.97 (d) pH = 8.2, pOH = 14 8.2 = 5.8. = 10⁻⁵.⁸ = 1.585 10⁻⁶ = 2 10⁻⁶ M 8.036 10⁻¹³ = 1.585 S = 7.121 10⁻⁴ = 7 10⁻⁴ M (e) pH = 7.5, pOH = 14 7.5 = 6.5. = 10⁻⁶.⁵ = 3.162 10⁻⁷ = 3 10⁻⁷ M 8.036 10⁻¹³ = s²( 3.162 S = 1.549 10⁻³ = 2 10⁻³ M The drop in pH from 8.2 to 7.5 approximately doubles (from 7 10⁻⁴ M to 15 10⁻⁴ M) the molar solubility of CaCO₃(s). 17.98 (a) Hydroxyapatite: = Fluoroapatite: Ksp = (b) For each mole of apatite dissolved, one mole of OH- or F⁻ is formed. Express molar solubility, in terms of and [F⁻]. Hydroxyapatite: = = 5s, = 3s = 6.8 = (s) = 84,375 = 8.059 = 8.1 Use logs to find S. S = 3.509 = 3.5 M Ca₅(PO₄)₃OH. Fluoroapatite: [F⁻] = s, = 5s, = 3s = = 1.185 = 1.2 S = 6.109 10⁻⁸ = 6.1 17.99 Cr(OH)₃(s) + 11 Cr(OH)₃(s) + 11 K K = = (6.7 = 0.536 = 0.5 K = [Cr(OH)₄⁻] ; pOH = pH = = 4.0; [OH⁻] = 1 10⁻⁴ M [Cr(OH)₄⁻] = K [OH⁻] = 0.536(1 = 5.36 10⁻⁵ = 5 10⁻⁵ M (The value for Cr(OH)₃ listed in Appendix D is different from the value given in this exercise.) 17.100 Analyze/Plan. Calculate the solubility of Mg(OH)₂ in 0.50 M Find Ksp for Mg(OH)₂ in Appendix D. is a weak acid, which will increase the solubility of Mg(OH)₂. Combine the various interacting equilibibria to obtain an overall reaction. Calculate K for this reaction and use it to calculate solubility (s) for Mg(OH)₂ in 0.50 M Solve. 542