Equilíbrio de Solubilidade

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Chapter 17 Aqueous Ionic Equilibrium 373 Assume that S 5%, the assumption is not valid. Because expanding the expression will give a third-order polynomial, that is not easily solved directly. Solve by successive approximations. Substitute S = 7.83 X 10⁻³ M for the S term that is part of a sum [i.e., the one in (0.10 + S)]. Thus, 2.45 X 10⁻⁵ = (0.10 + 7.83 X 10⁻³) and S = 7.53 X 10⁻³ M. Substitute this new S value again. Thus, 2.45 X 10⁻⁵ = (0.10 + 7.53 X 10⁻³) and S = 7.55 X 10⁻³ M. Substitute this new S value again. Thus, 2.45 X 10⁻⁵ = (0.10 + 7.55 X 10⁻³) and S = 7.55 X 10⁻³ M. So the solution has converged and S = 7.55 X 10⁻³ M. (c) Given: BaF₂ Find: molar solubility (S) in 0.15 M NaF Other: = 2.45 X 10⁻⁵ Conceptual Plan: M NaF MF⁻ then M F⁻, S NaF(s) + ICE table Solution: Because one F⁻ ion is generated for each NaF, = 0.15 M. BaF₂(s) + Initial 0.00 0.15 Change S 2S Kₛₚ(BaF₂) = [F⁻]² = 2.45 X 10⁻⁵ = (S)(0.15 + Equil S Because 2S 0.15, 2.45 X 10⁻⁵ = and S = 1.09 X 10⁻³ M. Confirm that the assumption is 2 (1.09 X 10⁻³) valid. 0.15 X 100% = 1.5%