Chapter 3 Solubility Equilibrium Answer Key

Transcription

1 BC Science Chemitry 12 Chapter 3 Solubility Equilibrium Anwer Key September 20, The Concept of Solubility Warm Up 1. Potaium phophate Xenon hexafluoride Phophoric acid Sulphur dioxide Ammonium ulphide ionic molecular 2. Ionic compound will conduct electricity. Quick Check 1. No. Some alt are weak electrolyte. They do not diolve to a ignificant extent. 2. a) molecular b) ionic c) ionic d)molecular e) ionic 3. Sample A i tap water. It ha a low electrical conductivity due to low concentration of diolved alt in tap water. Sample B i carbon tetrachloride. It i molecular and doe not conduct electricity. Sample C i eawater. It ha a ignificant amount of diolved ion preent o will conduct electricity well. Quick Check 1. CaCl 2 (aq) Ca 2+ (aq) + 2Cl - (aq) 2. (NH 4 ) 2 C 2 O 4 (aq) 2NH 4 + (aq) + C 2 O 4 2- (aq) 3. Na 3 PO 4 (aq) 3Na + (aq) + PO 4 3- (aq) Edvantage Interactive

2 Quick Check 1. a) Na 2 C 2 O 4 (aq) 2Na + (aq) + C 2 O 2-4 (aq) 0.45 M 0.90 M 0.45 M total [ion] = 0.90 M M = 1.35 M b) AlCl 3 (aq) Al 3+ (aq) + 3Cl - (aq) 2.5 M 2.5 M 7.5 M total [ion] = 2.5 M M = 10.0 M 2. ioh(aq) i + (aq) + OH - (aq) 1.0 M 1.0 M 1.0 M total ion concentration = 2.0 M CaCl 2 (aq) Ca 2+ (aq) + 2Cl - (aq) 0.8 M 0.8 M 1.6 M total ion concentration = 2.4 M The CaCl 2 will have the greater conductivity becaue it ha a greater ion concentration. 3. AlBr 3 (aq) Al 3+ (aq) + 3Br - (aq) According to the diociation equation, 1 mole Al 3+ : 3 mole Br -. If [Br - ] = 0.15 M, then [Al 3+ ] = M. Practice Problem: Calculating Ion Concentration 1. a) [(NH 4 ) 3 PO 4 ] = 0.35 g x 1 mol = 3.6 x 10-3 M g [NH 4 + ] = 3(3.6 x 10-3 ) = 1.1 x 10-2 M [PO 4 3- ] = 3.6 x 10-3 M b) [HCl] = 6.0 mol x = 2.6 M [H + ] = 2.6 M [Cl - ] = 2.6 M 2. Al 2 (SO 4 ) 3 (aq) 2Al 3+ (aq) + 3SO 4 2- (aq) 0.15 M 0.45 M ma of Al 2 (SO 4 ) 3 = x 0.15 mol x g = 13 g 1 1 mol Edvantage Interactive

3 3. [(NH 4 ) 2 SO 4 ] = 0.22 mol x = 0.15 M [NH 4 + ] = 0.30 M, [SO 4 2- ] = 0.15 M [(NH 4 ) 2 S] = 0.45 mol x = 0.14 M [NH + 4 ] = 0.28 M, [S 2- ] = 0.14 M total [NH 4 + ] = 0.58 M, [SO 4 2- ] = 0.15 M [S 2- ] = 0.14 M Practice Problem: Converting Between Unit of Solubility 1. mol MnS = 4.7 x 10-6 g x 1 mol = 5.4 x 10-2 M g 2. Ma Pb(IO 3 ) 2 = x 4.5 x 10-5 mol x g = 7.5 x 10-3 g 1 1 mol 3. mol CaCO 3 = 7.1 x 10-4 g x 1 mol = 7.1 x 10-5 M g Quick Check 1. ZnCO 3 () Zn 2+ (aq) + CO 3 2- (aq) 2. Al(OH) 3 () Al 3+ (aq) + 3OH - (aq) 3. Fe 2 S 3 () 2Fe 3+ (aq) + 3S 2- (aq) 3.1 Review 1. a) any oluble alt or trong acid or bae: NaCl, HCl, NaOH. Many anwer poible. b) any molecular ubtance not including weak acid: CCl 4, SO 2, XeF 6 2. Both are acid with the ame concentration, but HClO 4 i a trong acid that ionize completely. H 3 PO 4 i a weak acid, and remain largely in molecular form. Since there are more ion in the HClO 4, it will have a greater electrical conductivity. 3. a) Mg(ClO 4 ) 2 (aq) 2Mg 2+ (aq) + 2ClO 4 - (aq) b) CaCr 2 O 7 (aq) Ca 2+ (aq) + Cr 2 O 7 2- (aq) c) Cu(CH 3 COO) 2 (aq) Cu 2+ (aq) + 2CH 3 COO - (aq) d) Mn(SCN) 2 (aq) Mn 2+ (aq) + 2SCN - (aq) e) Al(HC 2 O 4 ) 3 (aq) Al 3+ (aq) + 3HC 2 O 4 - (aq) f) Ba(OH) 2 8H 2 O(aq) Ba 2+ (aq) + 2OH - (aq) + 8 H 2 O(l) Edvantage Interactive

4 4. mol BaSO 4 = g x 1 mol = 1.0 x 10-5 M g 5. ma CaCO 3 = x 7.1 x 10-5 mol x g = 1.8 x 10-3 g 1 1 mol 6. [Na 2 Cr 2 O 7 ] = 0.50 g x 1 mol = M g Na 2 Cr 2 O 7 (aq) 2Na + (aq) + Cr 2 O 7 2- (aq) M 2(0.0126) M M [Na + ] = M [Cr 2 O 7 2- ] = M 7. [NaCl] = 5.0 mol x 60. = 3.3 M [Na + ] = 3.3 M [Cl - ] = 3.3 M [MgCl 2 ] = 2.4 mol x 30. = 0.80 M [Mg 2+ ] = 0.80 M [Cl - ] = 1.6 M [Mg 2+ ] = 0.80 M, total [Cl - ] = 4.9 M, [Na + ] = 3.3 M 8. Meaure out 1.0 of water and add olid NaCl until there i undiolved olid remaining. Pour off until 1.0 olution remain. 9. aagbro 3 () Ag + (aq) + BrO 3 - (aq) b) Al 2 (CrO 4 ) 3 () 2Al 3+ (aq) + 2CrO 4 2- (aq) c) Mg(OH) 2 () Mg 2+ (aq) + 2OH - (aq) d) PbSO 4 () Pb 2+ (aq) + SO 4 2- (aq) e) Cu 3 (PO 4 ) 2 () 3Cu 2+ (aq) + 2PO 4 3- (aq) 3.2 Qualitative Analyi Identifying Unknown Ion Warm Up 1. mol AgCl = 1.4 x 10-3 g x 1 mol = 9.8 x 10-6 M g Edvantage Interactive

5 2. mol NaCl = 359 g x 1 mol = 6.14 M g 3. AgCl ha low olubility, and NaCl i oluble. Practice Problem: Predicting the Solubility of Salt Uing the Solubility Table 1. a) low olubility b) low olubility c) oluble d) oluble e) oluble d) oluble 2. (many anwer poible) a) Na2CO3 b) SrSO4 c) Ag + d) NO3-3. The odium hydroxide contain odium ion and hydroxide ion. Sodium ion form oluble compound, o any precipitate mut contain the hydroxide anion. The two compound poible for the precipitate are Mg(OH)2 and Sr(OH)2. According to the Solubility Table, Mg(OH)2 ha low olubility and Sr(OH)2 i oluble. The ample mut contain trontium nitrate ince no precipitate wa oberved. Quick Check 1. a) BaSO4 b) Fe(OH)2 c) Zn3(PO4)2 Edvantage Interactive

6 d) CrCO3 e) Mn(OH)2 2. a) Sr(OH)2(aq) + 2AgNO3(aq) 2AgOH() + Sr(NO3)2(aq) Sr 2+ (aq) + 2OH - (aq) + 2Ag + (aq) + 2NO3 - (aq) 2AgOH() + Sr 2+ (aq) + 2NO3 - (aq) Ag + (aq) + OH - (aq) AgOH() b) MgS(aq) + ZnCl2(aq) ZnS() + MgCl2(aq) Mg 2+ (aq) + S 2- (aq) + Zn 2+ (aq) + 2Cl - (aq) ZnS() + Mg 2+ (aq) + 2Cl - (aq) Zn 2+ (aq) + S 2- (aq) ZnS() c) Na2CO3(aq) + BaS(aq) BaCO3() + Na2S(aq) 2Na + (aq) + CO3 2- (aq) + Ba 2+ (aq) + S 2- (aq) BaCO3() + 2Na + (aq) + S 2- (aq) Ba 2+ (aq) + CO3 2- (aq) BaCO3() Practice Problem: Selective Precipitation (other anwer poible) 1. Firt add NaCl. Filter out the AgCl() Then add Na2S. Filter out the BeS() Finally add Na2SO4. Filter out the BaSO4() 2. Firt add Ca(NO3)2. Filter out the CaSO4() Then add Cu(NO3)2. Filter out the CuS() Finally add AgNO3. Filter out the AgBr() 3. Edvantage Interactive

7 Firt add Sr(NO3)2. Filter out the Sr3(PO4)2() Then add Mg(NO3)2. Filter out the Mg(OH)2() Finally add AgNO3. Filter out the Ag2S() Quick Check 1. - The color of the olution (for example, olution containing Cu 2+ are blue) - flame tet color - olubility amount of precipitate formed, the peed at which a precipitate form, or the ability of a precipitate to re- diolve on the addition of NH3 or HNO3. 2. Firt add NaCl. A precipitate of both PbCl2 and AgCl may form. Filter out the precipitate, dry it and meaure it ma. Put the precipitate into a econd beaker. To the original olution, add NaOH. If a precipitate form, filter out the Cu(OH)2(). To the econd beaker, add 6 M NH3. Any AgCl re- diolve. Filter out the PbCl2(). Dry the PbCl2() and meaure it ma. If the ma had decreaed, the preence of AgCl i confirmed. 3. Firt add HNO3. The formation of bubble confirm the preence of CO3 2- ion. Next, add CaCl2. If a precipitate form, it i Ca3(PO4)2(). Filter out the precipitate. Activity: Uing Titration to Calculate the Unknown [Cl - ] In A Sample of Seawater Quetion: What i the concentration of Cl - in eawater? Procedure: 4. Procedure 1-3 were repeated in 2 more trial, and the data below wa recorded. Trial #1 Trial #2 Trial #3 Edvantage Interactive

8 Initial burette reading (m) Final burette reading (m) Volume AgNO3 added (m) Average volume AgNO3 ued (m) X X Reult and Dicuion 1. Write a balanced net ionic equation that repreent the reaction between ilver ion and chloride ion. Ag + (aq) + Cl - (aq) AgCl() 2. Calculate the mole of ilver nitrate reacted in the titration uing the average volume and concentration of the ilver nitrate. mol AgNO3 = x mol = mol AgNO Uing the mole ratio from the balanced net ionic equation, calculate the mole of chloride ion preent in the ample. mol Cl - = mol Ag + x 1 mol Cl - = mol Cl - 1 mol Ag + 4. Uing the volume of the diluted ample and the mole of chloride calculated above, calculate the concentration of chloride ion in the diluted ample. In the diluted ample, the [Cl - ] = mol Cl - = M Calculate the concentration of chloride ion in the original ample of eawater. Edvantage Interactive

9 The original ample wa diluted 10x, o in the original ample, the [Cl - ] = M. 6. The level of alt in eawater varie acro the planet. The average amount of chloride ion in eawater i 21.2 g/. How doe your ample compare? To convert M Cl - to g/: g = mol x g = 19.6 g 1 mol Our ample ha lightly le chloride in it than the average. 7. Silver ion will form a precipitate with both chloride and chromate ion. Ue a reference book to look up the olubility of AgCl and Ag2CrO4. Calculate the concentration of Ag + in a aturated olution of each. Ue thi information to explain why AgCl precipitate before Ag2CrO4. The olubility of AgCl = 1.3 x 10-5 M The olubility of Ag2CrO4 = 6.5 x 10-5 M When ilver ion are added in the titration, AgCl will precipitate firt becaue it ha a lower olubility than Ag2CrO4. Review Quetion: 1. a) oluble b) low olubility c) low olubility d) oluble e) oluble Edvantage Interactive

10 f) oluble 2. Ye there are ilver and ulphate ion in olution. A low olubility imply mean that very little will diolve; not that it i inoluble. 3. mol AgCH3COO= 11.1 g x 1 mol = M g M < 0.1 M o AgCH3COO would have a low olubility. 4. A formula equation i written with ubtance a compound and tate hown. A complete ionic equation i written with oluble ionic compound and trong acid in diociated form, and compound with low olubility undiociated. State are alo hown. A net ionic equation only how the ubtance that take part in the reaction. Spectator ion are not hown. State are alo hown. 5. A pectator ion i one that doe not take part in the reaction. Na + ion and NO3 - ion are often pectator. 6. a) (NH4)2S(aq) + FeSO4(aq) (NH4)2SO4(aq) + FeS() 2NH4 + (aq) + S 2- (aq) + Fe 2+ (aq) + SO4 2- (aq) 2NH4 + (aq) + SO4 2- (aq) + FeS() Fe 2+ (aq) + S 2- (aq) FeS() b) H2SO3(aq) + CaCl2(aq) 2HCl(aq) + CaSO3() H2SO3(aq) + Ca 2+ (aq) + 2Cl - (aq) 2H + (aq) + 2Cl - (aq) + CaSO3() H2SO3(aq) + Ca 2+ (aq) + SO3 2- (aq) 2H + (aq) + CaSO3() c) CuSO4(aq) + CaS(aq) CuS() + CaSO4() Cu 2+ (aq) + SO4 2- (aq) + Ca 2+ (aq) + S 2- (aq) CuS() + CaSO4() Cu 2+ (aq) + SO4 2- (aq) + Ca 2+ (aq) + S 2- (aq) CuS() + CaSO4() 7. Na + and K + are alkali ion. Compound containing alkali ion are oluble in water. 8. Firt add Na2SO4 to precipitate CaSO4: Ca 2+ (aq) + SO4 2- (aq) CaSO4() Filter out the precipitate Secondly, add Na2S to precipitate Cr2S3: 2Cr 3+ (aq) + 3S 2- (aq) Cr2S3 () Filter out the precipitate Edvantage Interactive

11 Finally add NaOH to precipitate Mg(OH)2: Mg 2+ (aq) + 2OH - (aq) Mg(OH)2() Filter out the precipitate 9. Firt add Mg(NO3)2 to precipitate Mg3(PO4)2: 3Mg 2+ (aq) + 2PO4 3- (aq) Mg3(PO4)2 () Filter out the precipitate Secondly, add Fe(NO3)2 to precipitate FeS: Fe 2+ (aq) + S 2- (aq) FeS() Filter out the precipitate Finally add AgNO3 to precipitate AgCl: Ag + (aq) + Cl - (aq) AgCl() Filter out the precipitate 10. Nitrate are oluble in water. If a precipitate form when a nitrate i added, the precipitate mut contain the cation of the nitrate. 11. AgCl can be diolved in 6M NH3, while AgI cannot. Thi can be ued to eparate AgCl from AgI or AgBr. Carbonate will diolve on the addition of a trong acid uch a HNO a) 2Ag + (aq) + CO3 2- (aq) Ag2CO3() b) Ag2CO3() + 2H + (aq) CO2(g) + H2O(l) + 2Ag + (aq) Ag2CO3() + 4NH3(aq) 2Ag(NH3)2 + (aq) + CO3 2- (aq) 13. a) Hard water contain Ca 2+ ion, Fe 2+ ion and/or Mg 2+ ion. Scale i CaCO3 that collect inide water heater, pipe, kettle and boiler. b) CaCO3() + 2HCl(aq) H2O(l) + CO2(g) + CaCl2(aq) CaCO3() + 2H + (aq) + 2Cl - (aq) H2O(l) + CO2(g) + Ca 2+ (aq) + 2Cl - (aq) CaCO3() + 2H + (aq) H2O(l) + CO2(g) + Ca 2+ (aq) c) Many anwer are poible: Na2CO3 d) Water oftener remove Ca 2+ ion and Mg 2+ ion from hard water. Soap cum can t form without thee ion. 3.3 The Solubility Product Contant - K p Edvantage Interactive

12 Warm Up 1. a) the maximum amount of olute that can diolve in a given volume of olvent at a particular temperature. b) the anwer when number are multiplied together c) a number that doe not change 2. a) PbCl 2 (aq) Pb 2+ (aq) + 2Cl - (aq) b) mol = 4.4 g x 1 mol = M g c) PbCl 2 (aq) Pb 2+ (aq) + 2Cl - (aq) M M M Quick Check 1. a) SrCO 3 () Sr 2+ (aq) + CO 3 2- (aq) b) Mg(OH) 2 () Mg 2+ (aq) + 2OH - (aq) c) Ca 3 (PO 4 ) 2 () 3Ca 2+ (aq) + 2PO 4 3- (aq) 2. a) K p = [Sr 2+ ][CO 3 2- ] b) K p = [Mg 2+ ][OH - ] 2 c) K p = [Ca 2+ ] 3 [PO 4 3- ] 2 3. The olubility i the maximum amount of olute that can be diolved in a particular volume of olvent. The olubility product contant i the product of the ion concentration raied to the power of the coefficient from the equilibrium. Practice Problem 1. a) CaCO 3 () Ca 2+ (aq) + CO 3 2- (aq) 6.1 x 10-5 M 6.1 x 10-5 M 6.1 x 10-5 M K p = [Ca 2+ ][CO 3 2- ] Edvantage Interactive

13 K p = (6.1 x 10-5 )( 6.1 x 10-5 ) K p = 3.7 x 10-9 b) Mn(OH) 2 () Mn 2+ (aq) + 2OH - (aq) 3.6 x 10-5 M 3.6 x 10-5 M 7.2 x 10-5 M K p = [Mn 2+ ][OH - ] 2 K p = (3.6 x 10-5 )( 7.2 x 10-5 ) 2 K p = 1.9 x c) Solubility of BaCrO 4 : mol = 2.8 x 10-3 g x 1 mol = 1.1 x 10-5 M g BaCrO 4 () Ba 2+ (aq) + CrO 2-4 (aq) 1.1 x 10-5 M 1.1 x 10-5 M 1.1 x 10-5 M K p = [Ba 2+ ][CrO 2-4 ] K p = (1.1 x 10-5 )( 1.1 x 10-5 ) K p = 1.2 x d) Solubility of Ag 2 C 2 O 4 : mol = g x 1 mol = 1.1 x 10-4 M g Ag 2 C 2 O 4 () 2Ag + (aq) + C 2 O 2-4 (aq) 1.1 x 10-4 M 2.2 x 10-4 M 1.1 x 10-4 M K p = [Ag + ] 2 [C 2 O 2-4 ] K p = (2.2 x 10-4 ) 2 (1.1 x 10-4 ) K p = 5.2 x Edvantage Interactive

14 2. Solubility of Mg(OH) 2 : mol = 5.5 x 10-5 mol = 1.1 x 10-4 M Mg(OH) 2 () Mg 2+ (aq) + 2OH - (aq) 1.1 x 10-4 M 1.1 x 10-4 M 2.2 x 10-4 M K p = [Mg 2+ ][OH - ] 2 K p = (1.1 x 10-4 )( 2.2 x 10-4 ) 2 K p = 5.2 x Solubility of MgC 2 O 4 : mol = 0.16 g x 1 mol = 9.5 x 10-3 M g MgC 2 O 4 () Mg 2+ (aq) + C 2 O 2-4 (aq) 9.5 x 10-3 M 9.5 x 10-3 M 9.5 x 10-3 M K p = [Mg 2+ ][C 2 O 2-4 ] K p = (9.5 x 10-3 ) (9.5 x 10-3 ) K p = 9.0 x 10-5 Practice Problem: 1. a) AgCl() Ag + (aq) + Cl - (aq) K p = [Ag + ][Cl - ] 1.8 x = 2 = 1.3 x 10-5 M b) Edvantage Interactive

15 FeS() Fe 2+ (aq) + S 2- (aq) K p = [Fe 2+ ][ S 2- ] 6.0 x = 2 = 7.7 x M g = 7.7 x mol x g x 1 = 6.8 x g m 1 1 mol 1000 m m c) Pb(IO 3 ) 2 () Pb 2+ (aq) + 2IO 3 - (aq) 2 K p = [Pb 2+ ][IO - 3 ] x = ()(2) 2 = = 3.7 x = 9.25 x = 4.5 x 10-5 M d) SrF 2 () Sr 2+ (aq) + 2F - (aq) 2 K p = [Sr 2+ ][F - ] x 10-9 = ()(2) 2 = = 4.3 x 10-9 = 1.1 x = 1.0 x 10-3 M g = 1.0 x 10-3 mol x g = 6.8 x g 1 1 mol 2. Edvantage Interactive

16 Fe(OH) 3 () Fe 3+ (aq) + 3OH - (aq) 3 K p = [Fe 3+ ][ OH - ] x = ()(3) 3 = = 2.6 x = 9.6 x = 9.9 x M [OH - ] = 3 = 3(9.9 x ) = 3.0 x M 3. CaC 2 O 4 () Ca 2+ (aq) + C 2 O 4 2- (aq) K p = [Ca 2+ ][ C 2 O 2-4 ] 2.3 x 10-9 = 2 = 4.8 x 10-5 M ma (g) CaC 2 O 4 = x 4.8 x 10-5 mol x g = 4.0 x 10-3 g 1 1 mol Activity: Experimentally Determining the K p of CaCO 3 Quetion: What i the K p of CaCO 3? Procedure: Thi i only one method. Other anwer are poible. Thi procedure i done at room temperature. If another temperature i ued, either heat up or cool down the ditilled water to the required temperature. Reagent: 150 m ditilled water Edvantage Interactive

17 Calcium carbonate olid Apparatu: m beaker 1 hot plate 1 coopula 1 tirring rod filter paper funnel m graduated cylinder 1 cale 1 thermometer Procedure: 1. Prepare a aturated olution of CaCO 3 by meauring out 150 m of ditilled water into the firt beaker. 2. Continue to add olid CaCO 3 and tir until no more CaCO 3 diolve. You hould ee un-diolved olid in the olution. 3. Meaure and note the temperature of the olution in the data table. 4. Filter the aturated olution of CaCO 3 into the econd beaker. 5. Uing the graduated cylinder, meaure 100. m of the filtered olution. 6. Meaure and record the ma of the third beaker in your data table. 7. Pour the 100. m of filtered olution into the third beaker. 8. Heat the beaker on the hot plate until all water ha evaporated. Allow the CaCO 3 to cool. 9. Meaure and record the ma of the beaker and olid CaCO 3. Data Table: Edvantage Interactive

18 Temperature of aturated olution: Ma of beaker: Ma of beaker and CaCO 3 (): Reult and Dicuion: 1. Decribe how you would analyze the data and perform calculation to determine the K p. Calculate the ma of CaCO 3 diolved in 100. m of aturated olution by ubtracting the Ma of beaker from the Ma of beaker and CaCO 3 (). Calculate the molar olubility uing: = ma of CaCO 3 (g) x 1 mol molar ma CaCO 3 (g) Calculate the olubility product contant by ubtituting in for : CaCO 3 () > Ca 2+ (aq) + CO 3 2- (aq) K p = [Ca 2+ ][ CO 3 2- ] = 2 2. What would be ome ource of error? Not removing all of the olid calcium carbonate when filtering the aturated olution. Not completely drying the olid calcium carbonate before weighing it. 3. Compare your procedure to that of another group. Wa their procedure the ame? Another method may be ued uch a removing the diolved calcium ion or carbonate ion by elective precipitation, and then meauring the ma of precipitate formed. Edvantage Interactive

19 3.3 Review Quetion 1. a) Al(OH) 3 () Al 3+ (aq) + 3OH - (aq) K p = [Al 3+ ][ OH - ] 3 b) Cd3(AO4)2 () 3Cd 2+ (aq) + 2 AO4 3- (aq) K p = [Cd 2+ ] 3 [ AO4 3- ] 2 c) BaMoO4 () Ba 2+ (aq) + MoO4 2- (aq) K p = [Ba 2+ ][MoO4 2- ] d CaSO 4 () Ca 2+ (aq) + SO 4 2- (aq) K p = [Ca 2+ ][ SO 4 2- ] e) Pb(IO 3 ) 2 () Pb 2+ (aq) + 2IO 3 - (aq) K p = [Pb 2+ ][ IO 3 - ] 2 f) Ag 2 CO 3 () 2Ag + (aq) + CO 3 2- (aq) K p = [Ag + ] 2 [CO 3 2- ] 2. a) BaSO 3 () Ba 2+ (aq) + SO 3 2- (aq) b) The olubility i the maximum amount of olute diolved in a given volume of olvent. The olubility product contant i the olubility quared ( 2 ) 3. ZnCO 3 () Zn 2+ (aq) + CO 3 2- (aq) 1.1 x 10-5 M 1.1 x 10-5 M K p = [Zn 2+ ][ CO 3 2- ] = (1.1 x 10-5 ) 2 = 1.2 x Edvantage Interactive

20 4. Solubility = g x 1 mol = 4.3 x 10-5 M g Ag 3 PO 4 () 3Ag + (aq) + PO 4 3- (aq) 4.3 x 10-5 M 1.3 x 10-4 M 4.3 x 10-5 M K p = [Ag + ] 3 [PO 3-4 ] = (1.3 x 10-4 ) 3 ( 4.3 x 10-5 M) = 9.5 x CaSO 4 () Ca 2+ (aq) + SO 4 2- (aq) K p = [Ca 2+ ][ SO 4 2- ] 9.1 x 10-6 = 2 = 3.0 x 10-3 M ma gypum (g) = x 3.0 x 10-3 mol x g = 0.26 g 1 1 mol 6. Calcite: CaCO 3 () Ca 2+ (aq) + CO 3 2- (aq) K p = [Ca 2+ ][ CO 3 2- ] 3.4 x 10-9 = 2 = 5.8 x 10-5 M Edvantage Interactive

21 g = 5.8 x 10-5 mol x g x 1 = 5.8 x 10-6 g m 1 1 mol 1000 m m Argonite: CaCO 3 () Ca 2+ (aq) + CO 3 2- (aq) K p = [Ca 2+ ][ CO 3 2- ] 6.0 x 10-9 = 2 = 7.7 x 10-5 M g = 7.7 x 10-5 mol x g x 1 = 7.7 x 10-6 g m 1 1 mol 1000 m m7. Solubility of Pb3(AO4)2 : mol = 3.0 x 10-5 g x 1 mol = 3.3 x 10-8 M g Pb3(AO4)2 () 3Pb 2+ (aq) + 2AO 3-4 (aq) 3.3 x 10-8 M 1.0 x 10-7 M 6.7 x 10-8 M K p = [Pb 2+ ] 3 [ AO 3-4 ] 2 K p = (1.0 x 10-7 ) 3 ( 6.7 x 10-8 ) 2 K p = 4.4 x CdCO 3 () Cd 2+ (aq) + CO 3 2- (aq) K p = [Cd 2+ ][ CO 3 2- ] 1.0 x = 2 = 1.0 x 10-6 M Edvantage Interactive

22 Cd(OH) 2 () Cd 2+ (aq) + 2OH - (aq) 2 K p = [Cd 2+ ][OH - ] x = ()(2) 2 = 4 3 = 1.2 x 10-5 M Diagree. Even though the K p value for CdCO 3 i greater than the K p for Cd(OH) 2, it olubility i le. 9 AgCl() Ag + (aq) + Cl - (aq) K p = [Ag + ][ Cl - ] 1.8 x = 2 = 1.3 x 10-5 M Ag 2 CrO 4 () 2Ag + (aq) + CrO 4 2- (aq) 2 K p = [Ag + ] 2 [ CrO 4 2- ] 1.1 x = (2) 2 () = 4 3 = 6.5 x 10-5 M The olubility of AgCl i le than the olubility of Ag 2 CrO 4. All thing being equal, the precipitate of AgCl will form firt. 10. Edvantage Interactive

23 Ag 2 CO 3 () 2Ag + (aq) + CO 3 2- (aq) 2 K p = [Ag + ] 2 [ CO 2-3 ] 8.5 x = (2) 2 () = 4 3 = 1.3 x 10-4 M ma Ag 2 CO 3 (g) = 2.5 x 1.3 x 10-4 mol x g = 9.0 x 10-2 g 1 1 mol 3.4 Precipation Formation and the Solubility Product K p Warm Up 1. Add olid AgCl until no more olid diolve. There mut be ome undiolved olid in the olution. [Ag + ] = [Cl - ]. 2. a) [AgNO 3 ] = 0.10 mol x = M = [Ag + ] [NaCl] = mol x = M = [Cl - ] b) [Ag + ] i not the ame a [Cl - ] becaue they came from different ource. The [Ag + ] came from the AgNO 3 wherea the [Cl - ] came from the NaCl. c) Ag + (aq) + Cl - (aq) AgCl() Practice Problem: Predicting Whether a Precipitate Will Form 1. Pb(NO 3 ) 2 + 2NaIO 3 Pb(IO 3 ) 2 () + 2NaNO 3 Pb(IO 3 ) 2 () Pb 2+ (aq) + 2IO 3 - (aq) Edvantage Interactive

24 K p = [Pb 2+ ] [IO 3 - ] 2 = 3.7 x [Pb(NO 3 ) 2 ] = 6.3 x 10-2 mol x = 5.3 x 10-4 M = [Pb 2+ ] [NaIO 3 ] = 1.2 x 10-3 mol x 1.0 = 1.2 x 10-3 M = [IO 3 - ] TIP = [Pb 2+ ] [IO 3 - ] 2 = (5.3 x 10-4 )( 1.2 x 10-3 ) 2 = 7.6 x TIP > K p o a precipitate form 2. NH 4 BrO 3 + AgNO 3 AgBrO 3 () + NH 4 NO 3 AgBrO 3 () Ag + (aq) + BrO 3 - (aq) K p = [Ag + ] [BrO 3 - ] = 5.3 x 10-5 [NH 4 BrO 3 ] = 4.5 x 10-3 mol x = 5.5 x 10-5 M = [BrO 3 - ] [AgNO 3 ] = 2.5 x 10-3 mol x = 2.5 x 10-3 M = [Ag + ] TIP = [Ag + ] [BrO 3 - ] = (2.5 x 10-3 )( 5.5 x 10-5 ) = 1.4 x 10-7 TIP < K p o no precipitate form 3. 2NaF + Cd(NO 3 ) 2 CdF 2 () + 2NaNO 3 CdF 2 () Cd 2+ (aq) + 2F - (aq) K p = [Cd 2+ ] [F - ] 2 Edvantage Interactive

25 [NaF] = 0.17 mol x = M = [F - ] [Cd(NO 3 ) 2 ] = 0.22 mol x = 0.15 M = [Cd 2+ ] TIP = [Cd 2+ ] [F - ] 2 = (0.15)( 0.052) 2 = 4.1 x 10-4 If no precipitate form, then TIP < K p. K p mut be greater than 4.1 x Practice Problem: Forming a Precipitate In Solution 1. a) SrF 2 () Sr 2+ (aq) + 2F - (aq) K p = [Sr 2+ ] [F - ] x 10-9 = [Sr 2+ ] (0.045) 2 [Sr 2+ ] = 2.1 x 10-6 M b) 2 SrCO 3 () Sr 2+ (aq) + CO 3 2- (aq) K p = [Sr 2+ ] [CO 2-3 ] 5.6 x = [Sr 2+ ] (2.3 x 10-4 ) [Sr 2+ ] = 2.4 x 10-6 M c) SrSO 4 () Sr 2+ (aq) + SO 4 2- (aq) K p = [Sr 2+ ] [SO 2-4 ] 3.4 x 10-7 = [Sr 2+ ] (0.011) [Sr 2+ ] = 3.1 x 10-5 M 2. Edvantage Interactive

26 MgCO 3 () Mg 2+ (aq) + CO 3 2- (aq) K p = [Mg 2+ ] [CO 2-3 ] 6.8 x 10-6 = (3.2 x 10-3 ) [CO 2-3 ] [CO 2-3 ] = 2.1 x 10-3 M ma Na 2 CO 3 (g) = 10.0 x 2.1 x 10-3 mol x g = 2.2 g 1 mol 3. PbI 2 () Pb 2+ (aq) + 2I - (aq) 2 K p = [Pb 2+ ] [I - ] x 10-9 = ()(2) 2 = 4 3 = 1.3 x 10-3 M [I - ] = 2 = 2.6 x 10-3 M AgI() Ag + (aq) + I - (aq) K p = [Ag + ] [I - ] 8.5 x = [Ag + ] (2.6 x 10-3 ) [Ag + ] = 3.3 x M Quick Check: 1. Edvantage Interactive

27 a) Only changing temperature change the value of K p. b) Changing the temperature or the preence of certain ion in the olution can change the olubility of AgCl. For example, if there are Pb 2+ ion in the olution, the olubility of AgCl would be increaed. 2. Mg(OH) 2 () Mg 2+ (aq) + 2OH - (aq) To decreae the olubility, we need more Mg(OH) 2 (), o we need to caue the equilibirum ytem to hift left. The [Mg 2+ ] or [OH - ] could be increaed by adding Mg(NO 3 ) 2 or NaOH to the olution. 3. it 2 ubtance that would increae the olubility of Mg(OH) 2. Ue e Chatelier Principle and a K p expreion to explain each. Mg(OH) 2 () Mg 2+ (aq) + 2OH - (aq) To increae the olubility, we need more Mg(OH) 2 () to diolve, o we need to caue the equilibirum ytem to hift right. The [Mg 2+ ] or [OH - ] could be decreaed by adding a ubtance to precipitate one of them out uch a Ca(NO 3 ) 2 (to precipitate out the OH - ). Additionally, we could add an acid (uch a HCl) to neutralize the OH - ion. K p = [Mg 2+ ][ OH - ] 2 Since the value of K p doe not change, a decreae in the [Mg 2+ ] (or [OH - ]) caue the equilibrium above to hift right and the concentration of the other ion to increae. Practice Problem: Calculating Solubility With a Common Ion Preent 1 AgIO 3 () > Ag + (aq) + IO 3 - (aq) I C - +x +x E - x x K p = [Ag + ] [IO - 3 ] 3.2 x 10-8 = (x)(0.12) x = 2.7 x 10-7 M = [Ag + ] = [AgIO 3 ] diolved 2. PbI 2 () > Pb 2+ (aq) + 2I - (aq) I Edvantage Interactive

28 C - +x +2x E - x x K p = [Pb 2+ ] [I - ] x 10-9 = (x)(0.10) 2 x = 8.5 x 10-7 M = [Pb 2+ ] = [PbI 2 ] diolved 3. BaSO 4 () > Ba 2+ (aq) + SO 2-4 (aq) I C - +x +x E x x K p = [Ba 2+ ] [SO 4 2- ] 1.1 x = (0.050)(x) x = 2.2 x 10-9 M = [Ba 2+ ] = [BaSO 4 ] diolved ma of BaSO 4 (g) = 2.2 x 10-9 mol x g = 5.1 x 10-7 g/ 1 mol Activity: Experimentally Determining the K p Of Copper(II) Iodate Quetion: What i the approximate value of K p for copper(ii) iodate? Reult and Dicuion: 1. Write a balanced formula, complete ionic and net ionic equation for thi reaction. Cu(NO 3 ) 2 (aq) + 2NaIO 3 (aq) 2NaNO 3 (aq) + Cu(IO 3 ) 2 () Cu 2+ (aq) + 2NO - 3 (aq) + 2Na + (aq) + 2IO - 3 (aq) + 2Na + (aq) + 2NO - 3 (aq) + Cu(IO 3 ) 2 () Cu 2+ (aq) + 2IO - 3 (aq) Cu(IO 3 ) 2 () Edvantage Interactive

29 2. Calculate the [Cu 2+ ] in each of the mixture. Mixture 1: [Cu(NO 3 ) 2 ] = mol x = M = [Cu 2+ ] Mixture 2: [Cu(NO 3 ) 2 ] = mol x = M = [Cu 2+ ] Mixture 3: [Cu(NO 3 ) 2 ] = mol x = M = [Cu 2+ ] Mixture 4: [Cu(NO 3 ) 2 ] = mol x = M = [Cu 2+ ] Mixture 5: [Cu(NO 3 ) 2 ] = mol x = M = [Cu 2+ ] Calculate the [IO 3 - ] in each of the mixture. Mixture 1: [NaIO 3 ] = mol x = M = [Cu 2+ ] Mixture 2: [NaIO 3 ] = mol x = M = [Cu 2+ ] Mixture 3: [NaIO 3 ] = mol x = M = [Cu 2+ ] Mixture 4: [NaIO 3 ] = mol x = M = [Cu 2+ ] Mixture 5: [NaIO 3 ] = mol x = M = [Cu 2+ ] Write the equation for the equilibrium involving the precipitate, and the K p expreion. Cu(IO 3 ) 2 () > Cu 2+ (aq) + 2IO 3 - (aq) Edvantage Interactive

30 K p = [Cu 2+ ] [IO 3 - ] 2 5. Calculate a TIP value for each mixture. Mixture 1: TIP = (0.0050)(0.010) 2 = 5.0 x 10-7 Mixture 2: TIP = (0.0040)(0.0080) 2 = 2.6 x 10-7 Mixture 3: TIP = (0.0030)(0.0060) 2 = 1.1 x 10-7 Mixture 4: TIP = (0.0020)(0.0040) 2 = 3.2 x 10-8 Mixture 5: TIP = (0.0010)(0.0020) 2 = 4.0 x 10-9 (ppt) (ppt) (ppt) (no ppt) (no ppt) 6. State the K p a a range of value from thi data. 3.2 x 10-8 < K p < 1.1 x Compare your range to the tated K p value on the K p table. K p = 6.9 x 10-8 which doe fit into the range tated in quetion 6. Review Quetion: 1. a) FeS() Fe 2+ (aq) + S 2- (aq) K p = [Fe 2+ ] [S 2 ] b) Mg(OH) 2 () Mg 2+ (aq) + 2OH - (aq) K p = [Mg 2+ ] [OH - ] 2 c) Ag 2 CrO 4 () 2Ag + (aq) + CrO 4 2- (aq) K p = [Ag + ] 2 [CrO 4 2- ] 2. a) Edvantage Interactive

31 CaSO 4 b) CaSO 4 () Ca 2+ (aq) + SO 4 2- (aq) K p = [Ca 2+ ] [SO 4 2- ] c) Not necearily. The Ca 2+ came from the CaCl 2 and the SO 4 2- came from the H 2 SO 4. The concenration of ion depend on the original concentration of the CaCl 2 and the H 2 SO CuI 2 (aq) Cu I M 0.015M M PbI 2 () > Pb 2+ (aq) + 2I - (aq) K p = [Pb 2+ ] [I - ] x 10-9 = (x)(0.030) 2 x = 9.4 x 10-6 M = [Pb 2+ ] 4. CaC 2 O 4 () Ca 2+ (aq) + C 2 O 4 2- (aq) K p = [Ca 2+ ] [C 2 O 4 2- ] 2.3 x 10-9 = (5 x 10-3 )[ C 2 O 4 2- ] [ C 2 O 4 2- ] = 5 x 10-7 M 5. AgBr() Ag + (aq) + Br - (aq) K p = [Ag + ] [Br - ] 5.4 x = 2 = 7.3 x 10-7 M = [Ag + ] Edvantage Interactive

32 Ag 2 CO 3 () 2Ag + (aq) + CO 3 2- (aq) K p = [Ag + ] 2 [CO 3 2- ] 8.5 x = (7.3 x 10-7 ) 2 [CO 3 2- ] [CO 3 2- ] = 16 M 6. [AgNO 3 ] = 0.20 mol x = 4.0 x 10-4 M = [Ag + ] AgCl() Ag + (aq) + Cl - (aq) K p = [Ag + ] [Cl - ] 1.8 x = (4.0 x 10-4 ) [Cl - ] [Cl - ] = 4.5 x 10-7 M ma NaCl (g) = x 4.5 x 10-7 mol x g = 2.6 x 10-6 g 1 mol 7. Sr(NO 3 ) 2 + ZnSO 4 SrSO 4 () + Zn(NO 3 ) 2 SrSO 4 () Sr 2+ (aq) + SO 4 2- (aq) K p = [Sr 2+ ] [SO 4 2- ] = 3.4 x 10-7 [Sr(NO 3 ) 2 ] = mol x = 9.2 x 10-5 M = [Sr 2+ ] Edvantage Interactive

33 [ZnSO 4 ] = mol x 1.5 = 1.1 x 10-2 M = [SO 4 2- ] TIP = [Sr 2+ ] [SO 4 2- ] = (9.2 x 10-5 )( 1.1 x 10-2 ) = 1.0 x 10-6 TIP > K p o precipitate form 8. Pb(NO 3 ) 2 + 2NaCl PbCl 2 () + 2NaNO 3 PbCl 2 () Pb 2+ (aq) + 2Cl - (aq) K p = [Pb 2+ ] [Cl - ] 2 = 1.2 x 10-5 [Pb(NO 3 ) 2 ] = g x 1 mol = 1.0 x 10-4 M = [Pb 2+ ] g [NaCl] = M = [Cl - ] TIP = [Pb 2+ ] [Cl - ] 2 = (1.0 x 10-4 )( 0.080) = 8.2 x 10-6 TIP < K p o no precipitate form 9. a) AgCl() Ag + (aq) + Cl - (aq) K p = [Ag + ] [Cl - ] = 1.8 x AgI() Ag + (aq) + I - (aq) K p = [Ag + ] [I - ] = 8.5 x b) See above c) AgI Edvantage Interactive

34 d) AgI() Ag + (aq) + I - (aq) K p = [Ag + ] [I - ] 8.5 x = [Ag + ] (0.020) [Ag + ] = 4.3 x M e) AgCl() Ag + (aq) + Cl - (aq) K p = [Ag + ] [Cl - ] 1.8 x = [Ag + ] (0.020) [Ag + ] = 9.0 x 10-9 M f) 4.3 x M < [Ag + ] < 9.0 x 10-9 M g) AgI() Ag + (aq) + I - (aq) K p = [Ag + ] [I - ] 8.5 x = (9.0 x 10-9 ) [I - ] [I - ] = 9.4 x 10-9 M h) % I - remaining = 9.4 x 10-9 M x 100% = % M o = % precipitated out. 10. MgCO 3 () Mg 2+ (aq) + CO 3 2- (aq) [Mg 2+ ] = 12 mg x 1 g x 1 mol = 4.9 x 10-4 M mg g Edvantage Interactive

35 K p = [Mg 2+ ] [CO 3 2- ] 6.8 x 10-6 = (4.9 x 10-4 ) [CO 3 2- ] [CO 3 2- ]= 1.4 x 10-2 M ma Na 2 CO 3 10H 2 O (g) = 1.0 x 1.4 x 10-2 mol x g = 4.0 g 1 mol 11. PbI 2 () > Pb 2+ (aq) + 2I - (aq) To decreae the olubility, the equilibrium mut be hifted left. By adding Pb(NO 3 ) 2 or KI, the common ion (Pb 2+ or I - ) caue an increae in their repective concentration. 12. BaSO 4 () > Ba 2+ (aq) + SO 4 2- (aq) In Na 2 SO 4, the common ion SO 4 2- exit, o equilibrium hift left and olubility i decreaed. In water, there are no Ba 2+ or SO 4 2- ion. 13. Fe(OH) 3 () > Fe 3+ (aq) + 3OH - (aq) In HCl, the H + ion will neutralize the OH - ion, cauing the [OH - ] to decreae. The equilibrium will hift right and the olubility of Fe(OH) 3 will increae. It i more oluble in HCl than in water. Extenion 14. a) BaSO 4 () Ba 2+ (aq) + SO 4 2- (aq) K p = [Ba 2+ ] [SO 4 2- ] 1.1 x = 2 = 1.0 x 10-5 M ma BaSO 4 (g) = x 1.0 x 10-5 mol x g = 4.7 x 10-4 g Edvantage Interactive

36 1 mol b) BaSO 4 () Ba 2+ (aq) + SO 4 2- (aq) I C +x +x E x x K p = [Ba 2+ ] [SO 4 2- ] 1.1 x = (x) (0.10) x = 1.1 x 10-9 M = [Ba 2+ ] ma BaSO 4 (g) = x 1.1 x 10-9 mol x g = 5.1 x 10-8 g 1 mol Edvantage Interactive