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TP The molarity of the NaCl aqueous solution shown in the image is approximately 1. 20 M 2. 20 M 3. 2 M 4. 0.2 M 5. 0.02 M 6. 0.02 M Lecture 19 CH101 A1 (MWF 9:05 am) Friday, October 19, 2018 For today Dispersion postscript: Lone pairs versus cloud size Dissolving ionic solids Solubility guidelines fig 6.28, p 181 memorize Precipitation reactions Concentrations after precipitation Next lecture: Ionization of molecular solutes; Self ionization of water; Acid base reactions: Competition for H Response Counter 10 1 Relative dispersion: lone pairs vs cloud size Relative dispersion: lone pairs vs cloud size Boiling point Boiling point CH 3 CH 3 89 CH 3 CH 3 89 CH 3 CH 2 CH 3 42 Lone pair dispersion CH 3 OCH 3 24 CH 3 CH 2 CH 3 42 Lone pair dispersion CH 3 OCH 3 24 CH 3 CH 2 CH 2 CH 3 0.5 Lone pair dispersion CH 3 CH 2 OCH 3 7.5 3 4 1

Relative dispersion: lone pairs vs cloud size Relative dispersion: lone pairs vs cloud size Boiling point Boiling point CH 3 CH 3 89 CH 3 CH 3 89 CH 3 CH 2 CH 3 42 Lone pair dispersion CH 3 OCH 3 24 CH 3 CH 2 CH 3 42 Lone pair dispersion CH 3 OCH 3 24 CH 3 CH 2 CH 2 CH 3 0.5 Lone pair dispersion CH 3 CH 2 OCH 3 7.5 CH 3 CH 2 CH 2 CH 3 0.5 Lone pair dispersion CH 3 CH 2 OCH 3 7.5 CH 3 CH 2 OCH 2 CH 3 34.6 Cloud bond pair CH 3 CH 2 CH 2 CH 2 CH 3 36 dispersion CH 3 CH 2 OCH 2 CH 3 34.6 Cloud bond pair CH 3 CH 2 CH 2 CH 2 CH 3 36 dispersion CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 69 5 6 Dissolving ionic salts Na and Cl disperse throughout water Demonstration: Conductivity of water Pure water Na + (aq) + Cl (aq) 8 9 2

Dissolving ionic salts Na and Cl disperse throughout water Representing an NaCl aqueous solution If 2 mol of NaCl s is dissolved in 1 L of water, How many moles of NaCl are present? How many moles of NaCl are present? How many moles of Na are present? How many moles of Cl are present? How many moles of H 2 O are present? 10 12 Representing 2 M NaCl aqueous solution Make a sketch of a 2 M NaCl aqueous solution, representing waters as chevrons Na ions as Cl ions as Estimating molarity 13 The concentration of water is 55.5 mol/l. Get molarity from the relative numbers of ions and water molecules. 15 3

TP The molarity of the NaCl aqueous solution shown in the image is approximately 1. 20 M 2. 20 M 3. 2 M 4. 0.2 M 5. 0.02 M 6. 0.02 M Solubility guidelines Response Counter 10 17 20 Soluble ionic compounds, p181 (memorize!) Insoluble ionic compounds, p 181 (memorize!) 26 27 4

Precipitation reactions NaCl and AgNO 3 are both soluble in water, but AgCl is not. If aqueous solutions of NaCl and AgNO 3, are mixed, the Ag and Cl ions can encounter one another. Precipitation reactions When Ag aq and Cl aq ions do encounter one another, they will pair up as a lattice of AgCl 32 33 Precipitation reactions As more pairs form, they will aggregate so much that solid AgCl settles out of the solution, so that no AgCl remains in solution. This process is called precipitation. TP When aqueous solutions of copper II chloride and sodium carbonate are combined, 1. no precipitate will form 2. CuCO 3 will precipitate 3. NaCl will precipitate 4. both CuCO 3 and NaCl will precipitate 34 35 5

Quiz Some solid sodium carbonate and solid potassium nitrate are placed together into pure water. After thorough stirring and allowing things to settle, 1. NaNO 3 will have precipitated 2. K 2 CO 3 will have precipitated 3. Neither Na 2 CO 3 and nor KNO 3 will dissolve, and so will have settled as solids to the bottom of the solution 4. Everything will dissolve, no precipitate will form, and so the solution will be clear Concentrations before and after precipitation 36 37 These solutions are mixed and a precipitate forms. Write the net ionic equation for this process. What are the spectator ions for this process? What is the limiting reactant? NH 4 2 CO 3 2 NH 4 CO 3 BaI 2 Ba 2 I NH 4 Ba CO 3 I 38 39 6

NH 4 2 CO 3 2 NH 4 CO 3 BaI 2 Ba 2 I NH 4 Ba CO 3 I 2 0.20 M 0.150 L 0.060 mol 0.40 M 0.150 L 0.060 mol 0.20 M 0.150 L 0.030 mol 2 0.40 M 0.150 L 0.120 mol Quiz You are given 150 ml of a 0.20 M aqueous solution of ammonium carbonate These solutions are mixed and a precipitate forms. The limiting reagent is. 1. BaI 2 2. NH 4 2 CO 3 3. NH 4 4. Ba 5. CO 3 6. I 40 41 These solutions are mixed and a precipitate forms NH 4 2 CO 3 2 NH 4 CO 3 BaI 2 Ba 2 I Ba CO 3 BaCO 3 Ba aq CO 3 aq BaCO 3 s start 0.060 mol 0.030 mol 0 change end 42 These solutions are mixed and a precipitate forms. Write the net ionic equation for this process. What are the spectator ions for this process? What is the limiting reactant? How many moles of the solid precipitate are formed? After precipitation, what are the molarities M of each ion remaining in solution? 43 7

These solutions are mixed and a precipitate forms NH 4 2 CO 3 2 NH 4 CO 3 BaI 2 Ba 2 I Ba CO 3 BaCO 3 These solutions are mixed and a precipitate forms NH 4 2 CO 3 2 NH 4 CO 3 BaI 2 Ba 2 I Ba CO 3 BaCO 3 Ba aq CO 3 aq BaCO 3 s start 0.060 mol 0.030 mol 0 change end 44 Ba CO 3 BaCO 3 start 0.060 mol 0.030 mol 0 change 0.030 mol 0.030 mol 0.030 mol end 0.030 mol 0 0.030 mol 45 These solutions are mixed and a precipitate forms. Write the net ionic equation for this process. What are the spectator ions for this process? What is the limiting reactant? How many moles of the solid precipitate are formed? After precipitation, what are the molarities M of each ion in solution? Is the final solution electrically neutral? 46 TP You are given 150 ml of a 0.20 M aqueous solution of ammonium carbonate These solutions are mixed and a precipitate forms. After the precipitation, the solution 25% 25% 25% 25% 1. will be positively charged excess total positive charge 2. will be electrically neutral equal total positive and negative charge 3. will be negatively charged excess total negative charge 4. More information needed Ba CO 3 BaCO 3 start 0.060 mol 0.030 mol 0 change 0.030 mol 0.030 mol 0.030 mol end 0.030 mol 0 0.030 mol 47 8

These solutions are mixed and a precipitate forms. After the precipitation, the total electrical charge due to the ions remaining in solution is 0. Moles of ions Moles of charge NH 4 0.060 0.060 I 0.120 0.120 Ba 0.030 0.060 Total 0 48 9