1. WATER AND AQUEOUS SOLUTIONS Part I

2. I. Properties of Liquids A. Water 11. Water is a triatomic molecule with covalent H-O bonds. 22. Water experiences high surface tension and hydrogen bonding. 33. Water molecules are polar. 44. Water absorbs more energy than most liquids: a. High boiling point b. High specific heat c. High heat of vaporization B. Surface Tension and Hydrogen Bonding. 11. Almost all solids and liquids expand when the temperature is raised. 22. Liquids can become solids by slowing the kinetic energy of the particles and can become gases by increasing the kinetic energy. 33. Liquids, particularly water, have a property known as surface tension. 44. Surface tension allows the liquid to behave like it has an elastic skin.

3. 55. It is the inward force or pull which tends to minimize surface area. 66. Surface tension is caused by the intermolecular forces within the liquid. aa. The intermolecular forces between the interior liquid molecules are the same in every direction. bb. However, the forces on the surface are not the same. cc. The surface molecules experience forces that attract down or laterally. dd. The difference is what causes surface tension and capillary rise. 77. Liquids that have hydrogen in the molecular structure experience hydrogen bonding. 88. This occurs when the hydrogen of one of the molecules is attracted to another atom on another molecule with unshared pair of electrons. 99. The attractive force of the hydrogen bond is not as strong as a chemical bond.

4. C. Specific Heat Capacity 11. The specific heat of water is 4.184 J/g°C 22. Because of hydrogen bonding, the heat capacity is greater than that of other substances. II. Water Vapor and Ice A. Evaporation and Condensation 11. Because of hydrogen bonding, water absorbs a lot of heat as it evaporates or vaporizes. 22. An extensive network of hydrogen bond tightly holds the molecules in liquid water together. 33. Those bonds must be broken before water changes from a liquid to the vapor state. 44. The reverse of evaporation is condensation.

5. B. Ice 11. Water is denser in its liquid form than in its solid form (ice). 22. Ice expands to an open crystal lattice structure. 33. The molecules are further apart than they are in liquid with empty space between the molecule. III. Aqueous Solutions A. Solvents and Solutes 11. A solution is a uniform, homogeneous mixture of two or more substances. 22. It consist of two parts: aa. solute - the substance that dissolves and is present in small amounts. bb. solvent- the dissolving substance. 33. The formation of a solution is not a chemical interaction, therefore they do not conform to the law of definite proportions.

6.  4 4. A mixture is a solution if: aa. The particles of the solute and solvent are the size of the molecules or smaller. bb. The solute particles are evenly dispersed throughout the solute. cc. The dissolved particles will not come out of solution. dd. The solution is clear and transparent. ee. The solution cannot be separated by filtration. 55. The following should be considered: aa. If a polar solvent and a polar solute are mixed, then a solution is likely. bb. If a polar solvent and a nonpolar solute are mixed, then a solution is not likely. cc. If a nonpolar solvent and a nonpolar solute are mixed, then a solution is likely. dd. If a nonpolar solvent and a polar solute are mixed, then a solution is not likely. 66. The above can be summarized as “ like dissolves like”. 77. Solutions are formed by the process of solvation.

7. C. Electrolytes and Nonelectrolytes 11. Compounds that conduct an electrical charge in an aqueous solution are called electrolyte. 22. All ionic compounds are electrolytes. 33. Compounds that do not conduct an electrical charge in an aqueous solution are called nonelectrolyte. 44. Most molecular or organic compounds are nonelectrolytes. B. The Solution Process 11. The solute particles separate from the solid crystal. 22. The solvent particles move apart. 33 Solute particles are attracted to the solvent particles and are surrounded by the solvent molecules.

8. D. Water of Hydration 11. Crystals that are chemically bonded to water molecules are called hydrated crystals and the water is called water of hydration. ex. CuSO 4 5 H 2 O 22. It is possible effloresce (remove) the water from hydrated crystals by either raising the temperature or lowering the pressure or both. 33. The resulting compound becomes anhydrous meaning without water. 44. Anhydrous substances have a strong attraction to water. 55. Substances that absorb water out of the air are called hygroscopic and are good desiccants (drying agents). 66. When substance absorb enough water to dissolve into liquid form are called deliquescent.

9. IV. Colloids and Suspensions A. Suspensions 11. Suspensions are dispersions of particles larger than 100 nm in a continuous medium. 22. Suspensions settle out solution. 33. Suspensions can be separated by filtration and membrane. 44. Suspensions scatter light. 55. Suspensions can not affect the colligative properties.

10. B. Colloids 11. Colloids are dispersion of particles from 1 nm to 100 nm in at least one dimension in a continuous medium. 22. Colloids do not settle out of solution. 33. Colloids pass unchanged through ordinary filter paper but can be separated by a membrane. 44. Colloid scatter light (called the Tyndall effect). 55. Colloids can not affect the colligative properties. 66. Colloids experience Brownian motion which is continuous random motion caused by the bombardment of the colloid particles by the particle of the medium. 77. Emulsions are colloidal dispersions of liquids in liquids.

11. SOLUTIONS Part 2

12. I. Properties of Solution A. Solution Formation 11. The effect of temperature a. Increase in the rate of solubility b. The addition of heat cause the solvent molecules to move further apart thereby allowing more solute to come into contact with the solvent molecules. 22. The effect of particle size a. Smaller particles increase the rate of solubility. b. The larger surface area of the numerous particles exposes more solute molecules to the solvent than does the surface area of a single large lump. 33. The effect of stirring a. Increase in the rate of solubility. b. Stirring tends to bring the solute particles into contact with all particles of the solvent.

13. B. Solubility 11. A solution is considered to be in a single phase in which a state of dynamic equilibrium exist. 22. Solution equilibrium refers to the point at which the number of particles leaving the crystal surface equals the number returning to the surface. 33. There are three terms referring to the solution equilibrium: a. Saturated b. Unsaturated c. Supersaturated 44. Saturated is a solution in which solvation is equal to desolvation and the solution has some undissolved solid at the bottom. 55. Unsaturated is a solution containing less than the saturated amount of solute for that temperature and the solution is completely clear with no undissolved solid. 66. Supersaturated is a solution that is holding more solute than would normally hold at that temperature and the solution is completely clear but the addition of a single crystal will cause all of the solute to come out of solution. 77. Two liquids that dissolve into each other are miscible. 88. Two liquids that do not dissolve into each other are immiscible.

14. C. Factors Affecting Solubility 11. Solubility of most solids increases with an increase of temperature of the solvent. 22. Solubility of gases increase as the temperature of the solvent decreases. 33. Gas solubility increases as the partial pressure of the gas above the solution increases.

15. DILUTIONS Concentration of Solutions

16. Dilution is the process of preparing a less concentrated solution from a more concentrated one. moles of solute before dilution = moles of solute after dilution

17. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = M1V1M1V1 M2V2M2V2 = 4.5

18. Diluting Solutions  Often once you have made a stock solution, you need to dilute it to a working concentration.  To determine how to dilute the stock solution, use the formula: M 1 V 1 = M 2 V 2 M 1 – concentration of stock M 2 - concentration of diluted solution V 1 – volume needed of stock V 2 – final volume of dilution

19. Example : How many milliliters of a 5 M stock solution of NaCl are needed to prepare 100 ml of a 0.4 M solution? Diluting Solutions M 1 V 1 = M 2 V 2 (5) V 1 = (0.4)(100) V 1 = 8 ml

20. How would you prepare 60.0 mL of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M 1 V 1 = M 2 V 2 M 1 = 4.00 M 2 = 0.200V 2 = 0.06 L V 1 = ? L V 1 = M2V2M2V2 M1M1 = 0.200 x 0.06 4.00 = 0.003 L = 3 mL 3 mL of acid + 57 mL of water= 60 mL of solution

21. Another Dilution Problem If 32 mL stock solution of 6.5 M H 2 SO 4 is diluted to a volume of 500 mL What would be the resulting concentration? M 1 *V 1 = M 2 *V 2 (6.5M) * (32 mL) = M 2 * (500.0 mL) M 2 = 500 mL 6.5 M * 32 mL M 2 = 0.42 M

22. Chemistry Introduction to Molarity

23. Prescription drugs in the correct concentration make you better. In higher concentration they can kill you.

24. Chemists need to make solutions that have precise concentrations

25. General Properties of Aqueous Solutions A solution is a homogenous mixture of two or more substances. The substance present in the largest amount (moles) is referred to as the solvent. The other substances present in smaller amounts are called the solutes. A substance that dissolves in a particular solvent is said to be soluble in that solvent.

26. Solutions solute Solvent When the solvent is water the solution is said to be aqueous

27. Concentration of Solutions Molarity (M), or molar concentration, is defined as the number of moles of solute per liter of solution.

28. The concentration of a solution tells you how much solute that is dissolved in a given amount of solvent (water). The molarity is the concentration of a solution. Molarity=Moles Liter 1. 2.255 moles of NaCl is dissolved in 4.0 L of water, calculate the molarity. Molarity=2.255 moles 4.0 L =0.56 moles/liter =0.56 M (Molar) where M means moles per 1 liter 2. 5.00 g KOH is dissolved in 250. mL of water, calculate the molarity. Molarity=5.00 g x 1 mole 56.11 g --------------------------- 0.250 L =0.356 M

29. 4.5

30. Molarity  To make a 1 molar (1M) solution, first add 1 mole of solute to a 1-L volumetric flask half filled with distilled water. 16.2  Swirl the flask carefully to dissolve the solute. Fill the flask with water exactly to the 1-L mark.

31. 16.2