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Solubility Equilibria

Solubility Equilibria. End of Chapter 16. Solubility Equilibria. AgCl ( s ) Ag + ( aq ) + Cl - ( aq ). MgF 2 ( s ) Mg 2+ ( aq ) + 2F - ( aq ). Ag 2 CO 3 ( s ) 2Ag + ( aq ) + CO 3 2 - ( aq ). Ca 3 (PO 4 ) 2 ( s ) 3Ca 2+ ( aq ) + 2PO 4 3 - ( aq ).

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Solubility Equilibria

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  1. Solubility Equilibria End of Chapter 16

  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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