11/3/09 Aqueous Solubility of Compounds • Not all compounds dissolve in water. • Solubility varies from compound to compound. Chapter 5: Chemical Reactions Soluble ionic compounds Most molecular dissociate. Ions are compounds stay solvated associated in water. © 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2 Aqueous Solubility of Ionic Compounds Aqueous Solubility of Ionic Compounds © 2008 Brooks/Cole 3 © 2008 Brooks/Cole 4 Aqueous Solubility of Ionic Compounds Exchange Reactions: Precipitation (a) Nitrates (b) Hydroxides (soluble) (insoluble) AgNO Cu(NO ) 3 3 2 Cu(OH)2 AgOH (c) Sulfides CdS Sb2S3 PbS (NH4)2S soluble insoluble © 2008 Brooks/Cole 5 © 2008 Brooks/Cole 6 1 11/3/09 Precipitation Reaction Precipitation Reactions When ionic solutions mix, a precipitate may form: Na2SO4(aq) and Ba(NO3)2(aq) are mixed. Will they react? + 2- Products? Na SO4 Not all ions react 2+ - Ba NO3 KNO3(aq) + NaCl(aq) No reaction NaNO3 and BaSO4 . Insoluble? A reaction occurs if a product is insoluble. The solubility rules help predict reactions. © 2008 Brooks/Cole 7 © 2008 Brooks/Cole 8 Net Ionic Equations Net Ionic Equations Soluble ionic compounds fully dissociate: + - AgNO3(aq) Ag (aq) + NO3 (aq) + - KCl(aq) K (aq) + Cl (aq) On mixing: + - + - Ag (aq) + NO3 (aq) + K (aq) + Cl (aq) + - AgCl(s) + K (aq) + NO3 (aq) Net ionic equation: Ag+(aq) + C -(aq) → AgC (s) l l © 2008 Brooks/Cole 9 © 2008 Brooks/Cole 10 Net Ionic Equations Net Ionic Equations (NH4)2S(aq) + Hg(NO3)2(aq) → HgS + 2 NH4NO3 Product Solubility HgS – insoluble + (All sulfides – except group 1A, 2A and NH4 salts) NH4NO3 – soluble (All ammonium salts; all nitrates) (NH4)2S(aq) + Hg(NO3)2(aq) → HgS + NH4NO3 not balanced ammonium sulfide mercury(II) sulfide mercury(II) nitrate ammonium nitrate © 2008 Brooks/Cole 11 © 2008 Brooks/Cole 12 2 11/3/09 Net Ionic Equations Acids Increase the concentration of H+ ions in water. + 2- 2+ - 2 NH4 (aq) + S (aq) + Hg (aq) + 2 NO3 (aq) + - → HgS(s) + 2 NH4 (aq) + 2 NO3 (aq) • Change the color of pigments (indicators) . Litmus, phenolphthalein… S2-(aq) + Hg2+(aq) → HgS(s) © 2008 Brooks/Cole 13 © 2008 Brooks/Cole 14 Acids Bases Increase the concentration of OH- (hydroxide ion) in water. Bases: • Counteract an acid (neutralize an acid). • Change an indicator’s color (phenolphthalein…). • Have a bitter taste. • Feel slippery. H2O © 2008 Brooks/Cole 15 © 2008 Brooks/Cole 16 Common Acids and Bases Neutralization Reactions acid + base → salt + water. Neutralizations are exchange reactions. © 2008 Brooks/Cole 17 © 2008 Brooks/Cole 18 3 11/3/09 Net Ionic Equations for Acid-Base Reactions Net Ionic Equations for Acid-Base Reactions © 2008 Brooks/Cole 19 © 2008 Brooks/Cole 20 Gas-Forming Exchange Reactions Gas-Forming Exchange Reactions Carbonic acid (unstable) Tums® H2CO3(aq) → H2O(l) + CO2(g) Alka-Seltzer ® often written: CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) © 2008 Brooks/Cole 21 © 2008 Brooks/Cole 22 Gas-Forming Exchange Reactions Gas-Forming Exchange Reactions Sulfurous acid (unstable) H2SO3(aq) → H2O(l) + SO2(g) or Na2SO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + H2O(l) +SO2(g) © 2008 Brooks/Cole 23 © 2008 Brooks/Cole 24 4 11/3/09 Oxidation-Reduction Reactions Oxidation-Reduction Reactions In all cases: • If something is oxidized, something must be reduced. • Redox reactions move e-. +2 e- 2 Ag+(aq) + Cu(s) 2 Ag(s) + Cu2+(aq) -2 e- Here: • Cu changes to Cu2+. • Cu loses 2 e-; each Ag+ gains one e- + • Ag is reduced (ore turned to metal) • Gain of e- = reduction (so, loss of e- = oxidation) (note: H2 is oxidized, so CuO is an oxidizing agent) © 2008 Brooks/Cole 25 © 2008 Brooks/Cole 26 Redox Reactions and Electron Transfer Redox Reactions and Electron Transfer e- M X M loses electron(s) X gains electron(s) M is oxidized X is reduced M is a reducing agent X is an oxidizing agent © 2008 Brooks/Cole 27 © 2008 Brooks/Cole 28 Common Oxidizing and Reducing Agents Displacement Reactions Oxidizing Agent Reaction Product + + - O2 (oxygen) O2 (oxide ion) A XZ AZ X H2O2 (hydrogen peroxide) H2O(l) - - - - Redox: F2, Cl2, Br2, I2 (halogen) F , Cl , Br , I (halide ion) HNO3 (nitric acid) nitrogen oxides (NO, NO2..) Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s) - 3+ Cr2O7 (dichromate ion) Cr (chromium(III) ion) 0 +2 +2 0 - 2+ MnO4 (permanganate ion) Mn (manganese(II) ion) Cu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2 Ag(s) 0 +1 +2 0 Reducing Agent Reaction Product + H2 (hydrogen) H or H2O C CO and CO2 M (metal: Na, K, Fe…..) Mn+ (Na+, K+, Fe3+…..) © 2008 Brooks/Cole 29 © 2008 Brooks/Cole 30 5 11/3/09 lithium Activity Series of Metals potassium Displacement Reactions Li barium K strontium Ba Powerful reducing agents at the top. Sr Displace H from H O (l), calcium 2 2 Ca steam or acid sodium Na magnesium Mg aluminum Higher elements displace lower ones: Al manganese Mn Displace H from steam or 2 zinc Zn(s) + CuSO (aq) ZnSO (aq) + Cu(s) Zn acid chromium 4 → 4 Cr Ease of iron oxidation Mg(s) + 2 HCl(aq) → MgCl (aq) + H (g) Fe nickel 2 2 increases Ni tin Sn Displace H2 from acid lead Pb hydrogen H antimony Sb copper Cu No reaction with H2O, mercury Hg steam or acid silver Ag palladium Pd platinum Pt Au gold © 2008 Brooks/Cole 31 © 2008 Brooks/Cole 32 Displacement Reactions Displacement Reactions Potassium + water 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g) © 2008 Brooks/Cole 33 © 2008 Brooks/Cole 34 Solution Concentration Molarity moles solute mol Relative amounts of solute and solvent. Molarity = = liters of solution L • Shorthand: [NaOH] =1.00 M The brackets [ ] represent “molarity of ” © 2008 Brooks/Cole 35 © 2008 Brooks/Cole 36 6 11/3/09 Molarity Molarity Calculate the molarity of sodium sulfate in a solution 6.37 g of Al(NO3)3 are dissolved to make a 250. mL 3+ that contains 36.0 g of Na2SO4 in 750.0 mL of aqueous solution. Calculate (a) [Al(NO3)3] (b) [Al ] - solution. and [NO3 ]. 36.0 g nNa2SO4 = = 0.2534 mol 142.0 g/mol (a) Al(NO3)3 FM = 26.98 + 3(14.00) + 9(16.00) = 213.0 g mol 0.2534 mol [Na2SO4 ] = Unit change! 0.7500 L (mL to L) 6.37 g 213.0 g/mol [Na2SO4 ] = 0.338 mol/L = 0.338 M © 2008 Brooks/Cole 37 © 2008 Brooks/Cole 38 Molarity Solution Preparation 6.37 g of Al(NO3)3 in a 250. mL aqueous solution. Calculate (a) the molarity of the 3+ - Al(NO3)3 , (b) the molar concentration of Al and NO3 ions in solution. Solutions are prepared either by: 1. Diluting a more concentrated solution. 3+ - 1 Al(NO3)3 ≡ 1 Al 1 Al(NO3)3 ≡ 3 NO3 or… 3+ [Al3+] = 0.120 M Al(NO ) 1 Al = 0.120 M Al3+ 3 3 2. Dissolving a measured amount of solute and 1 Al(NO3)3 diluting to a fixed volume. - - [NO3 ] = 0.120 M Al(NO3)3 = 0.360 M NO3 © 2008 Brooks/Cole 39 © 2008 Brooks/Cole 40 Solution Preparation by Dilution Solution Preparation from Pure Solute Prepare a 0.5000 M solution of potassium MconcVconc = MdilVdil permanganate in a 250.0 mL volumetric flask. Mass of KMnO4 = 0.1250 mol x 158.03 g/mol M V 17.8 M x 75.0 mL Mdil = conc conc = = 1.34 M = 19.75 g Vdil 1000. mL © 2008 Brooks/Cole 41 © 2008 Brooks/Cole 42 7 11/3/09 Solution Preparation from Pure Solute Molarity and Reactions in Aqueous Solution • Rinse all the solid from the weighing dish into the flask. • Fill the flask ≈ ⅓ full. • Swirl to dissolve the solid. nA = [ A ] x V • Fill the flask to the mark on the neck. [product] = n / (total volume). • Shake to thoroughly mix. product © 2008 Brooks/Cole 43 © 2008 Brooks/Cole 44 Molarity and Reactions in Aqueous Solution Molarity and Reactions in Aqueous Solution H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) ? mol 5.850 x 10-3 mol 2 NaOH ≡ 1 H2SO4 -3 1 H2SO4 H2SO4 required = 5.850 x 10 mol NaOH H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l) 2 NaOH = 2.925 x 10-3 mol V needed: 0.234 mol acid n = 0.0250 L = 5.850 x 10-3 mol NaOH L mol H SO 1 L V = 2 4 = 2.925 x 10-3 mol [H2SO4] 0.0875 mol = 0.0334 L = 33.4 mL © 2008 Brooks/Cole 45 © 2008 Brooks/Cole 46 Molarity and Reactions in Aqueous Solution Molarity and Reactions in Aqueous Solution A 4.554 g H C O / NaC mixture was neutralized by 29.58 mL of 0.550M A 4.554 g mixture of oxalic acid, H C O and NaCl 2 2 4 l 2 2 4 NaOH. What was the weight % of oxalic acid in the mixture? was neutralized by 29.58 mL of 0.550M NaOH. What was the weight % of oxalic acid in the mixture? Mass of acid consumed = 8.135 x10-3 mol x (90.04 g/mol acid) = 0.7324 g of oxalic acid weight of acid nNaOH = 0.02958 L x 0.550 mol/L = 0.01627 mol Weight percent = sample weight x 100% 1 H2C2O4 ≡ 2 NaOH Oxalic acid reacted = 0.7324 g Weight percent = x 100% = 16.08% 1 H C O 4.554 g 0.01627 mol NaOH 2 2 4 = 8.135 x10-3 mol 2 NaOH © 2008 Brooks/Cole 47 © 2008 Brooks/Cole 48 8 11/3/09 Molarity and Reactions in Aqueous Solution Molarity and Reactions in Aqueous Solution 25.0 mL of 0.234 M FeCl and 50.0 mL of 0.453 M NaOH are mixed.