chapter 16 notes

chapter 16 notes - Chapter 16 More Equilibria in Aqueous...

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©2010 Donald L. Siegel, All Rights Reserved Chapter 16 Chapter 16 More Equilibria in Aqueous Solutions: Slightly Soluble Salts and Complex Ions
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©2010 Donald L. Siegel, All Rights Reserved Solubility Solubility Equilibria Equilibria and the and the Solubility Product Solubility Product W How do we measure solubility? Use an equilibrium constant W Consider CaF 2 (s) Ca 2+ (aq) +2F - (aq) Make usual equilibrium constant [] [ ] ) ( 2 2 2 s CaF F Ca K eq + = 1 [ ][ ] 2 2 + = F Ca K sp
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©2010 Donald L. Siegel, All Rights Reserved Solubility vs. Solubility vs. K K sp sp W K sp is the equilibrium constant for the dissolution of a salt W Solubility is the amount of the salt that dissolves W We can use one to calculate the other
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©2010 Donald L. Siegel, All Rights Reserved Example Example W Calculate the K sp of Bi 2 S 3 if the solubility is 1.0 × 10 -15 M at 25 ºC. W Step 1: Find the equilibrium concentration of each ion Stoichiometry: Bi 2 S 3 (s) 2Bi 3+ (aq) + 3S 2- (aq) [Bi 3+ ] = 2[Bi 2 S 3 ][ S 2- ]=3[Bi 2 S 3 ] [Bi 3+ ] = 2 × 1.0 × 10 -15 M = 2.0 × 10 -15 M [S 2- ]=3±×1.0±×10 -15 M = 3.0 × 10 -15 M W Step 2: Calculate the K sp K sp = (2.0 × 10 -15 ) 2 ( 3.0 × 10 -15 ) 3 K sp = 1.1 × 10 -73
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©2010 Donald L. Siegel, All Rights Reserved Example Example W Calculate the solubility of Cu(IO 3 ) 2 at 25 ºC in water. W Step 1: Look up K sp K sp of Cu(IO 3 ) 2 = 1.4 × 10 -7 W Step 2: Set up equilibrium constant Cu(IO 3 ) 2 (s) Cu 2+ (aq) + 2IO 3 - (aq) Initial 0 0 Change +x +2x Equil x 2x K sp = x(2x) 2 = 4x 3 = 1.4 × 10 -7 x = 3.3 × 10 -3
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©2010 Donald L. Siegel, All Rights Reserved Relative Solubility Relative Solubility W Determining which salt is more soluble is not always obvious W Comparisons reduce to 2 cases Salts that produce the same number of ions Salts that produce different numbers of ions
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©2010 Donald L. Siegel, All Rights Reserved Case 1: Same number of ions Case 1: Same number of ions W Direct comparison of K sp s all that’s needed The relationship between K sp and s will be the same W Examples Ag 3 PO 4 (K sp = 1.8 × 10 -18 ) and ScF 3 (K sp = 4.2 × 10 -18 ) CuI (K sp = 1.1 × 10 -12 ) and SrSO 4 (K sp = 3.2 × 10 -7 ) 4 27 sp K s = sp K s =
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©2010 Donald L. Siegel, All Rights Reserved Case 2: Different Numbers of Ions Case 2: Different Numbers of Ions W Must actually calculate the solubilities W Example CuS (K sp = 8.5 × 10 -45 ), Ag 2 S (K sp = 1.6 × 10 -49 ), Bi 2 S 3 (K sp = 1.1 × 10 -73 ) CuS s= 8.5 × 10 -45 = 9.2 × 10 -23 M, Ag 2 S s=[0.25(1.6 × 10 -49 )] 1/3 = 3.4 × 10 -17 M, Bi 2 S 3 s= [(1.1 × 10 -73 )/(4×27) ] 1/5 = 1.0 × 10 -15 M Bi 2 S 3 >Ag 2 S>CuS
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©2010 Donald L. Siegel, All Rights Reserved Common Ion Effect Common Ion Effect W When 2 reactions have an ion in common, and one of the reactions is the dissolution of a slightly soluble salt, the changed solubility of the salt is called the common ion effect.
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This note was uploaded on 11/09/2011 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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chapter 16 notes - Chapter 16 More Equilibria in Aqueous...

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