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Solubility Equilibria End of Chapter 17. Solubility Equilibria 16.6 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product.

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Presentation on theme: "Solubility Equilibria End of Chapter 17. Solubility Equilibria 16.6 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product."— Presentation transcript:

1 Solubility Equilibria End of Chapter 17

2 Solubility Equilibria 16.6 AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ]K sp is the solubility product constant MgF 2 (s) Mg 2+ (aq) + 2F - (aq) K sp = [Mg 2+ ][F - ] 2 Ag 2 CO 3 (s) 2Ag + (aq) + CO (aq) K sp = [Ag + ] 2 [CO ] Ca 3 (PO 4 ) 2 (s) 3Ca 2+ (aq) + 2PO (aq) K sp = [Ca 2+ ] 3 [PO ] 2 Dissolution of an ionic solid in aqueous solution: Q = K sp Saturated solution Q < K sp Unsaturated solution No precipitate Q > K sp Supersaturated solution Precipitate will form

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 What is the solubility of silver chloride in g/L ? AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] Initial (M) Change (M) Equilibrium (M) s+s +s+s ss K sp = s 2 s = K sp s = 1.3 x [Ag + ] = 1.3 x M [Cl - ] = 1.3 x M Solubility of AgCl = 1.3 x mol AgCl 1 L soln g AgCl 1 mol AgCl x = 1.9 x g/L K sp = 1.6 x

6

7 If 2.00 mL of M NaOH are added to 1.00 L of M CaCl 2, will a precipitate form? 16.6 The ions present in solution are Na +, OH -, Ca 2+, Cl -. Only possible precipitate is Ca(OH) 2 (solubility rules). Is Q > K sp for Ca(OH) 2 ? [Ca 2+ ] 0 = M [OH - ] 0 = 4.0 x M K sp = [Ca 2+ ][OH - ] 2 = 8.0 x Q = [Ca 2+ ] 0 [OH - ] 0 2 = 0.10 x (4.0 x ) 2 = 1.6 x Q < K sp No precipitate will form

8 Qualitative Analysis of Cations 16.11


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