1. Chapters 4 & 11: Properties of Solutions

2.  Many common chemical reactions occur in water, or aqueous solution. To understand how chemical species interact in solution, we must first understand water, the universal solvent.  Water is an excellent solvent due to:  Its shape; water is a bent molecule.  Electrons aren’t shared evenly (oxygen is more electronegative)  Electrons spend more time close to O than to H  This uneven distribution of charge makes water polar.

3.  Water is held together by covalent bonds  When water surrounds an ionic crystal, the H end attracts the anion and the O end attracts the cation. This process is called hydration.  Hydration causes salts to dissolve. H 2 O also dissolves polar covalent substances such as C 2 H 5 OH.

4.  H 2 O doesn’t dissolve nonpolar covalent substances.  The difference in a substances’ ability to dissolve is due to its interaction with itself, and the solvent solution.

5.  Review:  Solute-the substance that dissolves  Solvent-the substance that does the dissolving  Electrical conductivity-the ability of a solution to conduct an electrical current  Strong electrolyte-a highly ionized solution that easily conducts electrical current  Weak electrolyte-a solution with few ions that does not conduct a current very well  Nonelectrolyte- a solution made of a soluble compound that does not ionize and thus does not conduct an electrical current.

7.  A solute that ionizes completely conducts very well.  He said that the best conductors are: ▪ Soluble salts (like sodium chloride or magnesium nitrate) ▪ Strong acids like HCl(aq), HNO 3 (aq), H 2 SO 4 (aq) ▪ strong bases that contain OH - (have a bitter taste, slippery feel and include NaOH and KOH)

8.  NaC 2 H 3 O 2  Na + + C 2 H 3 O 2 -

9.  HC 2 H 3 O 2  H + + C 2 H 3 O 2 - 99% 1%  Ammonia (NH 3 ) is a weak base. NH 3 + H 2 O  NH 4 + + OH -

11.  In order to perform stoichiometric calculations with solutions, we must know the nature of the solution and the amounts of chemical present. M = moles of solute liters of solution

12. 23.4 g Na 2 SO 4 1 mol Na 2 SO 4 = 0.165 mol Na 2 SO 4 142.06g Na 2 SO 4 0.165 mol = 1.32 M 0.125 L

13. 0.350L 0.50 mol Na 2 SO 4 142.06g Na 2 SO 4 = 24.9g 1 L 1 mol Na 2 SO 4

14. Dilution problem (M 1 V 1 = M 2 V 2 ) (1.000M)(V 1 ) = (0.250M)(500.0mL) V 1 = 125 mL

16.  Solutions can generally be described as dilute (very little solute per volume of solvent) of concentrated (a lot of solute per volume of solvent). Another way to describe a solution is by looking at the mass of the solute in terms of the mass of the entire solution, or mass percent.  We can also examine the percent solute to solvent using the mole fraction of the solution, or χ.  Finally, we can look at the amount of solute in moles per amount of solvent in kg to find molality.

17. Mass % = Mass solute x 100 Mass of solution Mole Fraction χ A = n A. n A + n B Molality = moles solute Kg solvent

18.  10.5g C 2 H 5 OH 1 mol C 2 H 5 OH = 0.228mol C 2 H 5 OH  46.08g C 2 H 5 OH  1,210.5 g soln = 1.2105 L soln  1,200.0g H 2 O/18.02g = 66.59 mol H 2 O  M = 0.228 mol /1.2105 L = 0.188M C 2 H 5 OH  m = 0.228 mol/1.2000 kg = 0.19 m C 2 H 5 OH  χ C2H5OH = (0.228) = 0.00341  (0.228 + 66.59)

19.  When a solution forms, it takes place in three distinct steps:  Expanding the solute- The solute particles must move away from one another. This may form ions or molecules. (ΔH 1 )  Expanding the solvent- The solvent must break any internal bonds so that it can surround and hydrate the solute. (ΔH 2 )  The solute and solvent begin to interact to form the solution. (ΔH 3 )

22.  The heat of solution (ΔH soln ) is equal to the sum of heat produced or consumed in the three steps.  If the heat of solution is very positive, then energy must come into the system to make the solute dissolve, so the substance is insoluble.  If the heat of solution is very negative or only slightly positive, then energy is leaving the system as the solute dissolves and the solute is soluble.  The table below summarizes these interactions.

23. ΔH 1 ΔH 2 ΔH 3 ΔH soln Outcome Polar solute and solvent Large positive value Large negative value Small positive value Solution forms Nonpolar solute, polar solvent Small positive value Large positive value Small positive value Large positive value No solution forms Nonpolar solute and solvent Small positive value Solution forms Polar solute, nonpolar solvent Large positive value Small positive value Large positive value No solution forms

24.  Although we will discuss many of these in other chapters, the following summarizes the reasons for the values of H in the chart above.  Structural effects- the shape of solute molecules affects their polarity, and thus their ability to dissolve in solvents.  Ex. Vitamin A vs. Vitamin C

25.  Pressure effects- as pressure increases on the surface of a solvent, it may increase the rate of dissolution of a solid or liquid solute, but it does not affect the solubility of these solutes. If the solute is a gas, however, increasing pressure on the system will increase the solubility of the gas in the solvent.  Temperature effects- again, as temperature increases in a solvent, the rate of dissolution will increase. Many solid and liquid solutes will become more soluble, but a few solid and liquid solutes, and almost all gaseous solutes, will become less soluble as temperature increases.

26.  When a solution is created, the properties of the solvent are often changed. For example, adding a solution of water and antifreeze will allow your car to keep running, whether the temperature outside is very high or very low.

27.  A volatile compound is one that vaporizes easily due to weak internal bonding (ex. Acetone)  A nonvolatile compound has strong internal bonding and thus does not vaporize easily (ex. water)  When a volatile solute is added to a solvent like water, the volatile compound will vaporize to some extent, adding to the amount of particles above the surface of the solution. In a closed container, this increases the vapor pressure.  Nonvolatile solutes, however, will not change the amount of vaporized particles. These types of solutions are called ideal solutions.

28.  You can see that the more solute is added, the smaller the value for the mole fraction of solvent, and thus the smaller the pressure.  If a solute is added that actually attracts water molecules, it may lower vapor pressure in a closed container by keeping water in a liquid state. Solutes like salt actually bind to water and keep it from vaporizing, therefore reducing the number of water vapor particles above the surface of the solution.

29.  Matter can be separated as shown in the diagram below.  One of the ways we can classify solutions is to look at the particles they are made of.  Homogenous solutions are made up of relatively small particles that are permanently suspended in solvent.  Heterogeneous solutions are made up of relatively large particles that will settle out of solution over time.

30.  Colloids fall into a special category because they are made up of intermediately sized particles that stay suspended due to interactions with the surrounding solvent. ▪ The only way to distinguish between a colloid and a solution is to shine a light through them. Colloid particles are large enough that they bounce light, creating a “beam” you can see (called the Tyndall effect). Solutions will not show a “beam” effect.