AP Ksp - -5 [Ag + ] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M...

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Solubility Equilibria End of Chapter 16
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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 3 2 - ( aq ) K sp = [Ag + ] 2 [CO 3 2 - ] Ca 3 (PO 4 ) 2 ( s ) 3Ca 2+ ( aq ) + 2PO 4 3 - ( aq ) K sp = [Ca 2+ ] 3 [PO 3 3 - ] 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
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16.6
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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
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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 ) 0.00 + s 0.00 + s s s K sp = s 2 s = K sp s = 1.3 x 10
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Unformatted text preview: -5 [Ag + ] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M Solubility of AgCl = 1.3 x 10-5 mol AgCl 1 L soln 143.35 g AgCl 1 mol AgCl x = 1.9 x 10-3 g/L K sp = 1.6 x 10-10 16.6 16.6 If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 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 &gt; K sp for Ca(OH) 2 ? [Ca 2+ ] = 0.100 M [OH-] = 4.0 x 10-4 M K sp = [Ca 2+ ][OH-] 2 = 8.0 x 10-6 Q = [Ca 2+ ] [OH-] 2 = 0.10 x (4.0 x 10-4 ) 2 = 1.6 x 10-8 Q &lt; K sp No precipitate will form Qualitative Analysis of Cations 16.11...
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AP Ksp - -5 [Ag + ] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M...

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