Ch19_Precip - 1 + Ag PRECIPITATION REACTIONS Chapter 19 Ag+...

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Unformatted text preview: 1 + Ag PRECIPITATION REACTIONS Chapter 19 Ag+ Pb2+ Hg22+ AgCl PbCl2 Hg2Cl2 AgCl(s) AgCl(s) Analysis of Silver Group Ag+(aq) + Cl-(aq) Ag aq) aq) When solution is SATURATED , expt. shows SATURATED expt. that [Ag+] = 1.67 x 10-5 M. [Ag M. This is equivalent to the SOLUBILITY of AgCl. SOLUBILITY of AgCl. What is [Cl-]? [Cl This is also equivalent to the AgCl solubility. 2+ Pb Hg2 Analysis of Silver Group 2+ AgCl PbCl2 Hg2Cl2 4 Analysis of Silver Group AgCl PbCl2 Hg2Cl2 AgCl(s) AgCl(s) + Cl-(aq) aq) Saturated solution has [Ag+] = [Cl-] = 1.67 x 10 -5 M [Cl Use this to calculate Kc Kc = [Ag+] [Cl-] [Cl Ag+ Pb2+ Hg22+ AgCl PbCl2 Hg2Cl2 All salts formed in this experiment are said to be INSOLUBLE and and form when mixing moderately concentrated solutions of the metal ion with chloride ions. Ag+ Pb2+ Hg22+ Ag+(aq) Ag aq) 2 Analysis of Silver Group Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) Ag+(aq) + Cl-(aq) AgCl(s) Ag aq) aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED . 5 Ag+ Pb2+ Hg22+ AgCl PbCl2 Hg2Cl2 AgCl(s) AgCl(s) Kc = [Ag+] Analysis of Silver Group Ag+(aq) + Cl-(aq) Ag aq) aq) [Cl-] = 2.79 x 10-10 [Cl Because this is the product of “ solubilities”, solubilities”, we call it = (1.67 x 10 -5)(1.67 x 10 -5) Ksp = solubility product constant solubility = 2.79 x 10 -10 See Table 19.2 and Appendix J Page 1 3 6 7 Lead(II) Chloride Solubility of Lead(II) Iodide 8 Consider PbI 2 dissolving in water dissolving PbI2(s) √ Pb2+(aq) + 2 I -(aq) Pb aq) aq) Calculate Ksp if solubility = 0.00130 M if Solubility of Lead(II) Iodide Solution Consider PbI 2 dissolving in water dissolving PbI2(s) √ Pb2+(aq) + 2 I -(aq) Pb aq) aq) Calculate Ksp if solubility = 0.00130 M if Solution 1. Solubility = [Pb 2+] = 1.30 x 10 -3 M PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Pb aq) aq) Ksp = 1.9 x 10 -5 1.9 9 2. Ksp = [Pb2+] [I-]2 [Pb = [Pb2+] {2 • [Pb2+]}2 [Pb [I-] = ? [I Ksp = 4 [Pb2+]3 = 4 (solubility) 33 [Pb = 4 (solubility) [I-] = 2 x [Pb2+] = 2.60 x 10 -3 M Ksp = 4 (1.30 x 10 -3))33 = 8.8 x 10 -9 Ksp = 4 ((1.30 x 10 -3 = 8.8 x 10 -9 1.30 8.8 10 Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Hg aq) aq) 11 Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Hg aq) aq) 12 Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Hg aq) aq) Ksp = 1.1 x 10 -18 = [Hg2 2+] [Cl-]2 1.1 [Hg [Cl Ksp = 1.1 x 10 -18 = [Hg2 2+] [Cl-]2 1.1 [Hg [Cl Ksp = 1.1 x 10 -18 = [Hg2 2+] [Cl-]2 1.1 [Hg [Cl If [Hg 22+] = 0.010 M, what [Cl-] is req’d to just [Cl Recognize that Solution Ksp = product of product [Cl-] that can exist when [Hg 22+] = 0.010 M, begin the precipitation of Hg 2Cl2 ? That is, what is the maximum [ Cl-] that can be in solution with 0.010 M Hg 22+ without without forming Hg2Cl 2? maximum ion concs. concs. Precip. begins when product of Precip. ion concs. EXCEEDS the K sp. concs. Page 2 [Cl ] = K sp 0.010 = 1 .1 x 10-8 M If this conc. of Cl- is just exceeded, Hg 2Cl2 conc. is begins to precipitate. Precipitating an Insoluble Salt Hg2Cl2(s) Ksp = 1.1 x 1.1 13 14 Separating Metal Ions Hg22+(aq) + 2 Cl-(aq) Hg aq) aq) A solution contains 0.020 M Ag+ and Pb2+. Add and CrO42- to precipitate red Ag 2CrO4 and yellow to and PbCrO4. Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10 -12 for 9.0 Ksp for PbCrO4 = 1.8 x 10 -14 for 1.8 Solution 2+ + 2+ Cu2+, Ag+, Pb2+ 10 -18 Now raise [Cl-] to 1.0 M. What is the value of [Cl [Hg22+] at this point? Ksp Values Ksp Values AgCl 1.8 x 10-10 AgCl 1.8 x 10-10 PbCl22 1.7 x 10-5 PbCl 1.7 x 10-5 PbCrO44 1.8 x 10-14 1.8 x 10-14 PbCrO 1.8 Solution [Hg22+] = Ksp / [Cl-]2 [Cl = Ksp / (1.0)2 = 1.1 x 10 -18 M (1.0) 1.1 The substance whose Ksp is first is exceeded precipitates first. The ion requiring the lesser amount of CrO 42- ppts. first. ppts. The concentration of Hg 2 2+ has been reduced has by 1016 ! 16 15 Separating Salts by Differences in K sp sp 17 18 Separating Salts by Differences in K sp sp Separating Salts by Differences in K sp sp Separating Salts by Differences in K sp sp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate and to red Ag2CrO4 and yellow PbCrO4. Which precipitates first? and Ksp for Ag2CrO4 = 9.0 x 10-12 for 9.0 Ksp for PbCrO4 = 1.8 x 10-14 for 1.8 A solution contains 0.020 M Ag+ and Pb2+. Add and CrO42- to precipitate red Ag 2CrO4 and yellow to and PbCrO4. PbCrO 4 ppts. first. . ppts Ksp (Ag2CrO4)= 9.0 x 10-12 (Ag Ksp (PbCrO4) = 1.8 x 10-14 (PbCrO How much Pb 2+ remains in solution when Ag+ remains begins to precipitate? A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate and to red Ag2CrO4 and yellow PbCrO4. and Ksp (Ag2CrO4)= 9.0 x 10-12 (Ag Ksp (PbCrO4) = 1.8 x 10-14 (PbCrO How much Pb2+ remains in solution when Ag+ begins to precipitate? remains begins Solution Calculate [CrO 42-] required by each ion. [CrO42-] to ppt. PbCrO 4 = Ksp / [Pb2+] ppt. [Pb = 1.8 x 10 -14 / 0.020 = 9.0 x 10 -13 M 0.020 [CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 ppt. [Ag = 9.0 x 10 -12 / (0.020) 2 = 2.3 x 10 -8 M 9.0 (0.020) 2.3 PbCrO4 precipitates first precipitates Page 3 Solution We know that [CrO 42-] = 2.3 x 10 -8 M to begin to to ppt. Ag2CrO4. ppt. What is the Pb 2+ conc. at this point? conc. [Pb2+] = K sp / [CrO42-] = 1.8 x 10 -14 / 2.3 x 10 -8 M [CrO 2.3 = 7.8 x 10-7 M 7.8 Lead ion has dropped from 0.020 M to < 10 -6 M 19 20 Simultaneous Equilibria Simultaneous Equilibria Section 19.8 • PbCl2(s) + CrO42- • Salt Salt PbCl PbCl2 PbCrO4 PbCrO4 + 2 Cl- PbCl 2(s) Ksp PbCl 2 PbCrO 4 1.8 x 10-14 Complex Ions Add NH3 to light blue [Cu(H 2O)4]2+ ------> to ------> light blue Cu(OH)2 and then deep blue [Cu(NH3)4]2+ and Ksp 1.7 x 10-5 1.8 x 10-14 Pb 2+ + CrO42CrO 1.7 x 10-5 The combination of metal ions with small molecules such as H2O and NH3 ------> ------> PbCrO4 + 2 ClPbCrO Cl Pb2+ + 2 ClPb Cl COMPLEX IONS K1 = Ksp PbCrO4 PbCrO K2 = 1/Ksp 1/K [Cu(NH3)4]2+ [Cu(NH • Knet = K1 • K2 = 9.4 x 108 9.4 • Net reaction is product-favored 22 Dissolving Precipitates by forming Complex Ions Formation of complex ions explains why you can dissolve a ppt. by forming a complex ppt. ion. AgCl(s) ¸ Ag+ + Cl AgCl(s) Ag Cl Ag+ + 2 NH3 --> Ag(NH3)2 NH --> Ag(NH Ksp = 1.8 x 10 -10 1.8 + Kform = 1.6 x 10 7 1.6 ------------------------------------AgCl(s) + 2 NH3 ¸ Ag(NH3)2+ + ClAgCl(s) Ag(NH Knet = Ksp • Kform = 2.9 x 10 -3 2.9 Page 4 21 Section 19.10 Section 19.8 • Add CrO42- to solid PbCl 2. The less to soluble salt, PbCrO 4, precipitates • PbCl2(s) + CrO42- Simultaneous Equilibria and Complex Ions All metal ions form All metal ions form complex ions with water complex ions with water —and are of the type —and are of the type n+ [M(H22O)x]n+where x = 4 [M(H O)x] where x = 4 and 6. and 6. 23 24 Common Ion Effect Adding an Ion “Common” to an Adding an Ion “Common” to an Equilibrium Equilibrium 25 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and in (b) in 0.010 M Ba(NO3)2. Ba(NO Ksp for BaSO4 = 1.1 x 10-10 for 1.1 BaSO4(s) √ Ba2+(aq) + SO42-(aq) Ba aq) aq) Solution Ksp = [Ba2+] [SO42-] = (0.010 + y) (y) [Ba Because y < 1.1 x 10 -5 M (= x, the solubility in (= pure water), this means 0.010 + y is about equal to 0.010. Therefore, Ksp = 1.1 x 10 -10 = (0.010)(y) 1.1 (0.010)(y) y = 1.1 x 10 -8 M = solubility in presence of solubility added Ba2+ ion. ion. 27 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and in (b) in 0.010 M Ba(NO3)2. Ba(NO Ksp for BaSO4 = 1.1 x 10-10 for 1.1 Calculate the solubility of BaSO 4 in (a) pure in water and (b) in 0.010 M Ba(NO3)2. Ba(NO Ksp for BaSO 4 = 1.1 x 10 -10 for 1.1 BaSO4(s) Ba2+(aq) + SO42-(aq) Ba aq) aq) Solution Solubility in pure water = [Ba 2+] = [SO42-] = x Ksp = [Ba2+] [SO42-] = x 2 [Ba x = (Ksp)1/2 = 1.1 x 10-5 M (K 1.1 Solubility in pure water = 1.1 x 10 -5 mol/L mol/L The Common Ion Effect 26 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and in (b) in 0.010 M Ba(NO3)2. Ba(NO Ksp for BaSO4 = 1.1 x 10-10 for 1.1 BaSO4(s) √ Ba2+(aq) + SO42-(aq) Ba aq) aq) BaSO4(s) √ Ba2+(aq) + SO42-(aq) Ba aq) aq) Solution Solution Solubility in pure water = 1.1 x 10 -5 mol/L. mol/L. Now dissolve BaSO 4 in water already in containing 0.010 M Ba 2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO 4 be be less than or greater than in pure water?___ 28 initial change equilib. equilib. 29 The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and in (b) in 0.010 M Ba(NO3)2. Ba(NO Ksp for BaSO4 = 1.1 x 10-10 for 1.1 BaSO4(s) √ Ba2+(aq) + SO42-(aq) Ba aq) aq) SUMMARY Solubility in pure water = x = 1.1 x 10 -5 M Solubility in presence of added Ba 2+ = 1.1 x 10 -8 M 1.1 Le Chatelier’s Principle is followed! Page 5 [Ba2+] 0.010 +y 0.010 + y [SO42-] 0 +y y ...
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