1. Chapter 11 Solutions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

2. Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes.

3. Solutes Spread evenly throughout the solution. Cannot be separated by filtration. Can be separated by evaporation. Are not visible, but can give a color to the solution. Nature of Solutes in Solutions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

4. Examples of Solutions

5. Water Is the most common solvent. Is a polar molecule. Forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings (Intermolecular Forces)

6. Formation of a Solution Na + and Cl - ions, On the surface of a NaCl crystal are attracted to polar water molecules. In solution are hydrated as several H 2 O molecules surround each.

7. When NaCl(s) dissolves in water, the reaction can be written as: H 2 O NaCl(s) Na + (aq) + Cl - (aq) solid separation of ions Equations for Solution Formation

8. Solid LiCl is added to water. It dissolves because: A. The Li + ions are attracted to the 1) oxygen atom (  - ) of water. 2) hydrogen atom (  + ) of water. B. The Cl - ions are attracted to the 1) oxygen atom (  - ) of water. 2) hydrogen atom (  + ) of water. Learning Check

9. Two substances form a solution: When there is an attraction between the particles of the solute and solvent. When a polar solvent, such as water, dissolves polar solutes such as sugar and ionic solutes such as NaCl. When a nonpolar solvent, such as hexane (C 6 H 14 ), dissolves nonpolar solutes such as oil or grease. Like Dissolves Like

10. Which of the following solutes will dissolve in water? Why? 1) Na 2 SO 4 2) gasoline (nonpolar) 3) I 2 4) HCl Learning Check

11. In water, Strong electrolytes produce ions and conduct an electric current. Weak electrolytes produce a few ions. Nonelectrolytes do not produce ions. Solutes and Ionic Charge Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

12. Strong electrolytes, Dissociate in water producing positive and negative ions. Produce an electric current in water. In equations show the formation of ions in aqueous (aq) solutions. H 2 O 100% ions NaCl(s) Na + (aq) + Cl − (aq) H 2 O CaBr 2 (s) Ca 2+ (aq) + 2Br − (aq) Strong Electrolytes

13. A weak electrolyte, Dissociates only slightly in water. In water forms a solution of only a few ions and mostly undissociated molecules. HF(g) + H 2 O(l) H 3 O + (aq) + F - (aq) NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) Weak Electrolytes

14. Solubility Is the maximum amount of solute that dissolves in a specific amount of solvent. Can be expressed as grams of solute in 100 grams of solvent, usually water. g of solute 100 g water Solubility

15. Unsaturated solutions Contain less than the maximum amount of solute. Can dissolve more solute. Saturated solutions Contain the maximum amount of solute that can dissolve. Have undissolved solute at the bottom of the container.

16. Soluble and Insoluble Salts Ionic compounds that Dissolve in water are soluble salts. Do NOT dissolve in water are insoluble salts. ***

17. Equations for Forming Solids A molecular equation shows the formulas of the compounds. Pb(NO 3 )(aq) + 2NaCl(aq) PbCl 2 (s) + 2NaNO 3 (aq) An ionic equation shows the ions of the compounds. Pb 2+ (aq) + 2NO 3 − (aq) + 2Na + (aq) + 2Cl − (aq) PbCl 2 (s) + 2Na + (aq) + 2NO 3 − (aq) NOTE: the “spectator ions” A net ionic equation shows only the ions that form a solid. Pb 2+ (aq) + 2Cl − (aq) PbCl 2 (s)

18. Learning Check A precipitate forms in the following reaction. Write the molecular, ionic and net ionic equations for the reaction. BaCl 2 (aq) + Na 2 SO 4 (aq)

19. The mass percent (%m/m) Concentration is the percent by mass of solute in a solution. mass percent (%m/m) = g of solute x 100 g of solute + g of solvent Is the g of solute in 100 g of solution. mass percent = g of solute x 100 100 g of solution Mass Percent

20. Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute) x 100 = 16.0% (m/m) 50.00 g KCl solution Calculating Mass Percent

21. The volume percent (%v/v) is: Percent volume (mL) of solute (liquid) to volume (mL) of solution. volume % (v/v) = mL of solute x 100 mL of solution Solute (mL) in 100 mL of solution. volume % (v/v) = mL of solute 100 mL of solution Volume Percent

22. Percent Conversion Factors Two conversion factors can be written for each type of % value.

23. Write two conversion factors for each solutions: A. 8.50%(m/m) NaOH B. 5.75%(v/v) ethanol Learning Check

24. How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1 Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaCl STEP 2 g solution g NaCl STEP 3 Write the 10.0% (m/m) as conversion factors. 10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4 Set up using the factor that cancels g solution. 225 g solution x 10.0 g NaCl = 22.5 g NaCl 100 g solution Using Percent Factors

25. Molarity (M) Molarity (M) is: A concentration term for solutions. Gives the moles of solute in 1 L solution. moles of solute liter of solution

26. Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared: By weighing out 58.5 g NaCl (1.00 mol) and Adding water to make 1.00 liter of solution.

27. What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? 1) 0.557 M 2) 1.44 M 3) 1.71 M Learning Check

28. Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions.

29. How many grams of AlCl 3 are needed to prepare 125 mL of a 0.150 M solution? 1) 20.0 g AlCl 3 2) 16.7g AlCl 3 3) 2.50 g AlCl 3 Learning Check

30. How many milliliters of 2.00 M HNO 3 contain 24.0 g HNO 3 ? 1) 12.0 mL 2) 83.3 mL 3) 190. mL Learning Check

31. Dilution In a dilution, Water is added Volume increases Concentration decreases M 1 V 1 = M 2 V 2 initial diluted

32. Learning Check What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? 1) 27.0 mL 2) 60.0 mL 3) 90.0 mL

33. Using Molarity of Reactants How many mL of 3.00 M HCl are needed to completely react with 4.85 g CaCO 3 ? 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 1 Given: 3.00 M HCl; 4.85 g CaCO 3 Need: volume in mL STEP 2 Plan: g CaCO 3 mol CaCO 3 mol HCl mL HCl

34. Using Molarity of Reactants (cont.) 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 3 Equalities 1 mol CaCO 3 = 100.09 g; 1 mol CaCO 3 = 2 mol HCl 1000 mL HCl = 3.00 mol HCl STEP 4 Setup 4.85 g CaCO 3 x 1 mol CaCO 3 x 2 mol HCl x 1000 mL HCl 100.09 g CaCO 3 1 mol CaCO 3 3.00 mol HCl = 32.3 mL HCl required

35. Learning Check If 22.8 mL of 0.100 M MgCl 2 is needed to completely react 15.0 mL of AgNO 3 solution, what is the molarity of the AgNO 3 solution? MgCl 2 (aq) + 2AgNO 3 (aq) 2AgCl(s) + Mg(NO 3 ) 2 (aq) 1) 0.0760 M 2) 0.152 M 3) 0.304 M