Unit 12/Chapter 12 Notes Solutions, Suspensions and Colloids A. Solutions 1. Soluble = capable of being dissolved. 2. Smaller particles are eventually uniformly distributed until they cannot be seen and mixture has the same properties/composition throughout 3. Solution = a homogeneous mixture of two/more substances in a single phase. 4. Composed of a solute and a solvent. (Solute = substance that is dissolved, typically one in lesser amount) 5. Particles are smaller than 1 nm. 6. Solutions do not separate upon standing/cannot be separated by filtration. 7. When light is shined through, they do not scatter light. B. Suspensions 1. If the solute particles are so large that they settle out unless the mixture is stirred constantly or agitated = suspension. 2. Example = muddy water 3. Particles over 1000 nm in diameter form suspensions 4. The particles in suspension can be separated from the heterogeneous mixtures by passing the mixture through a filter. 5. They can also be separated by standing 6. They look murky or opaque and may scatter light, but not transparent C. Colloids 1. Particles that are intermediate in size between those in solutions and suspensions form mixtures known as colloid dispersions or colloids. 2. Particles between 1 nm and 1000 nm can form colloids. 3. In the muddy water example, the large particles in suspension would settle out, but the colloidal particles remain dispersed and thus the water stays cloudy a. Even with a filter, you cannot remove colloid particles. b. They are small enough to be suspended throughout the solvent by the constant movement of the surrounding molecules. c. Colloids are heterogeneous like suspensions, though they appear homogeneous. 4. Examples of colloids: page 404 5. Tyndall Effect a. Many colloids appear homogenous because the individual particles cannot be seen. b. The particles are, however, large enough to scatter light. c. Example = a headlight is visible on foggy night. d. This effect is called the Tyndall effect – occurs when light is scattered by colloidal particles dispersed in a transparent medium. e. Used to distinguish between a solution and a colloid. Electrolytes vs. Nonelectrolytes A. Substances that dissolve in water are classified according to whether they yield molecules or ions in solution. B. A substance that dissolves in water to give a solution that conducts electric current (ions separate and thus can carry electric current) is called an electrolyte. A solution with this characteristic is an electrolyte solution. 1. NaCl and any soluble ionic compounds are examples 2. Some highly polar molecular compounds also form ions when dissolved in water (HCl for example) C. A substance that dissolves in water to give a solution that does not conduct an electric current is called a nonelectrolyte. Solution is nonelectrolyte solution. 1. Sugar is an example. 2. These substances do not form ions when dissolved. The Solution Process A. Factors affecting the rate of dissolution 1. Increasing the surface area of the solute 2. Agitating a solution 3. Heating a Solvent B. Solubility 1. There is a limit to the amount of solute that can be dissolved -­‐ eventually solute returning to the crystal is the same as the solute dissolving from the crystal. 2. The physical state in which the opposing processes of dissolution and crystallization of a solute occur at equal rates = solution equilibrium. 3. Depends upon the nature of the solute, the nature of the solvent and the temperature – differs from solution to solution. B. Saturated vs. Unsaturated Solutions 1. A solution that contains the maximum amount of dissolved solute is described as a saturated solution. 2. If you add more solute to a saturated solution, it falls to the bottom and does not dissolve. 3. A solution that contains less solute than a saturated solution under the existing conditions = unsaturated solution. C. Supersaturated solutions 1. Typically, when a saturated solution of a solute whose solubility increases with temperature is cooled, the excess solute drops to the bottom, but not always 2. A supersaturated solution is a solution that contains more dissolved solute than a saturated solution contains under the same conditions. 3. Dropping a small crystal of the solute into a supersaturated solution (called seeding) starts the rapid formation of crystals. D. The solubility of a substance is the amount of that substance required to form a saturated solution with a specific amount of solvent at a specified temperature. 1. The solubility of sugar, for example, is 204 g per 100 g of water at 20°C. 2. Temperature must be specified as it affects solubility. 3. In a gas, pressure must also be specified. 4. The rate at which a solid dissolves is unrelated to solubility. E. Solute-­‐Solvent Interactions 1. Why are there differences in solubility? 2. What makes substances dissolve depends on the type of bonding, the polarity or nonpolarity of the molecules, and the intermoleculear forces between the solute and solvent. 3. Typically, “like dissolves like” is a rough, but useful rule! 4. Liquid solutes and solvents a. Liquid solutes and solvents that are not soluable in each other are immiscible. (oil and water) b. Liquids that dissolve freely in one another in any proportion are said to be completely miscible. c. Demonstration/Activity – immiscible and miscible 5. Henry’s Law – The solubility of a gas in a liquid is directly proportional to the partial pressure of that gas on the surface of the liquid a. Applies to gas-­‐liquid solutions at constant temp. b. This is how carbonated beverages are made = increase the solubility of carbon dioxide by increasing the pressure c. Rapid escape of the gas from the liquid (effervescence) occurs when you open the can and reduce the pressure. Molarity and Concentration A. The concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution. B. Many medications are solutions – pharmacists must ensure concentrations are correct. C. Solutions can be referred to as dilute or concentrated. 1. Dilute = small amount of solute in a solvent. 2. Concentrated – large amount of solute in solvent. 3. These terms do not apply to the saturation of the solution. D. Molarity 1. The number of moles of solute in one liter of solution 2. You need to know the molar mass to determine the molarity. 3. 1 M NaOH = one mole of NaOH in every liter of solution. 4. M = amount of solute (mol)/volume of solution (L) 5. How to make solutions: i. Determine the amount of grams needed. ii. Add the amount of grams to some of solvent (not 1 L) iii. Dissolve iv. Add enough more of solvent to reach 1 L v. Mix carefully as instructed. 6. To determine molarity of a solution, first calculate moles from grams, then divide moles by liters. Be sure to convert to liters if volume is not given in liters. E. Molality 1. The number of moles of solution in one kg of solution 2. You need to know the molar mass to determine the molality 3. m = amount of solute (mol)/kg of solvent (kg) 4. To determine molality of a solution, first calculate moles from grams, then divide moles by kilograms. Be sure to convert to kilograms if mass is not given in kg 5. Be sure to correctly determine which is the solute and solvent! F. Converting from one concentration (M) of solution to another (M2) 1. M1V1 = M2V2 is a useful equation for solving problems converting a solution from one molarity to another. 2. The volume calculated is the amount of concentrated solute 3. Remember to subtract from total amount of solution desired for water added. Example #1: You have 3.50 L of a solution that contains 90.0 g of sodium chloride. What is the molarity? Step 1: Begin by converting grams to moles 90 g NaCl X 1 mole NaCo/58.44 g NaCl (periodic table) = 1.54 mol NaCl Step 2: Divide moles by liters to determine molarity 1.54 mol NaCl/ 3.50 L solution = 0.440 M Example #2: What is the molarity of an aqueous solution with a volume of 450. mL that contains 200 g of iron (II) chloride? Step 1: Convert grams to moles 200 g FeCl2 X 1 mole/126.8 g = 1.58 moles FeCl2 Step 2: Convert mL to L 450 mL = .450 L Step 3: Divide moles by liters to determine molarity 1.58/.450 = 3.51 M Example #3: If you want to make 500. mL of a .250 M solution of KOH, how many grams of KOH would be needed? Step 1: Convert mL to L 500. mL = .5 L Step 2: Determine the moles of KOH .5 L X .250 mole/ 1 L = .125 mol KOH Step 3: Convert moles to grams .125 mol KOH X 56.1 grams KOH/1 mol = 7.01 grams KOH Example: #4: You are given 146.3 g of NaCl and asked to make a 3.0 M solution of NaCl. You should add water until the solution reaches what volume? Step 1: Convert grams to moles 146.3 g NaCl X I mole/58.44 grams = 2.5 moles Step 2: Determine liters from moles and molarity 2.50 moles X 1.0 L/3.0 moles = 0.83 L Example #5: What is the molality of a solution with 16.0 g of NaCl in 125 g of water? Step 1: Convert solute grams to moles 16.0g NaCl X 1 mole/58.5 grams = 0.274 moles Step 2: Convert solvent grams to kg 125 grams X 1 kg/1000 g = .125 kg Step 3: Determine molality by dividing moles by kg 0.274/.125 = 2.19 m Example #6: What volume (mL) of concentrated HF (16.5 M) should be used to prepare
125.0 mL of a 0.350 M solution? What volume of water is used?
Step 1: Use equation M1V1 = M2V2 to determine volume of HF
16.5M (V1) = 0.350M (125.0 mL)
V1 = 0.350(125.0)/16.5 = 2.65 mL of concentrated HF
Step 2: Volume of water used = subtract concentrated HF from total volume
Water = 125.0 mL - 2.65 mL = 122.4 mL of water