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Solubility Equilibria in Chemistry

This text explores solubility equilibria, Ksp constants, and dissolution of ionic solids in aqueous solutions. Learn about saturated, unsaturated, and supersaturated solutions, and calculate molar and mass solubility. Practice qualitative cation analysis.

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Solubility Equilibria in Chemistry

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  1. Solubility Equilibria End of Chapter 17

  2. Solubility Equilibria AgCl (s) Ag+(aq) + Cl-(aq) MgF2(s) Mg2+(aq) + 2F-(aq) Ag2CO3(s) 2Ag+(aq) + CO32-(aq) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Ksp= [Ag+][Cl-] Ksp is the solubility product constant Ksp= [Mg2+][F-]2 Ksp= [Ag+]2[CO32-] Ksp= [Ca2+]3[PO33-]2 Dissolution of an ionic solid in aqueous solution: No precipitate Unsaturated solution Q < Ksp Q = Ksp Saturated solution Precipitate will form Q > Ksp Supersaturated solution 16.6

  3. 16.6

  4. Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility (g/L)is the number of grams of solute dissolved in 1 L of a saturated solution. 16.6

  5. AgCl (s) Ag+(aq) + Cl-(aq) Initial (M) Change (M) Equilibrium (M)  s = Ksp 1.3 x 10-5 mol AgCl 1 L soln x 143.35g AgCl 1 mol AgCl What is the solubility of silver chloride in g/L ? Ksp= 1.6 x 10-10 0.00 0.00 Ksp= [Ag+][Cl-] Ksp= s2 +s +s s s s = 1.3 x 10-5 [Cl-] = 1.3 x 10-5M [Ag+] = 1.3 x 10-5M = 1.9 x 10-3g/L Solubility of AgCl = 16.6

  6. 16.6

  7. 2 Q = [Ca2+]0[OH-]0 If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form? The ions present in solution are Na+, OH-, Ca2+, Cl-. Only possible precipitate is Ca(OH)2 (solubility rules). Is Q > Kspfor Ca(OH)2? [OH-]0 = 4.0 x 10-4M [Ca2+]0 = 0.100 M = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8 Ksp = [Ca2+][OH-]2 = 8.0 x 10-6 Q < Ksp No precipitate will form 16.6

  8. Qualitative Analysis of Cations 16.11

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