1. Mixtures and Solutions

2. 14.1 Types of Mixtures Objectives: 1. Compare the properties of suspensions, colloids, and solutions 2. Identify types of colloids and types of solutions 3. Describe the electrostatic forces in colloids

3. Suspensions, Colloids and Solutions Recall that a mixture is a combination of two or more substances in which each substance retains its chemical properties Heterogeneous mixtures do not blend smoothly throughout Suspensions and colloids are heterogeneous Def: a suspension is a mixture containing particles that settle out if left undisturbed Suspensions can be filtered to pull out the suspended particles

4. Electrostatic Forces Def: a colloid is a heterogeneous mixture of intermediate sized particles An example is homogenized milk in which the components cannot be separated by settling or filtration If you stir an electrolyte into a colloid, the dispersed particles clump together destroying the colloid Heat also destroys a colloid because it gives energy to colliding particles Def: Brownian motion is the erratic random movement of colloid particles

5. Types of Solutions Def: the Tyndall effect is the phenomenon where colloidal particles scatter light even if the solution appears clear to the eye Recall that a solution is a homogeneous mixture that contain a solute and a solvent The solute is the substance that dissolves The solvent is the dissolving medium Solutions can be solid, liquid or gas though most exist as liquids Def: soluble substance dissolves in a solvent Def: miscible liquids dissolve into each other in any proportion Def: insoluble substances do not dissolve

6. 14.2 Solution Concentration Objectives: 1. Describe concentration using different units 2. Determine the concentration of solutions 3. Calculate the molarity of a solution

7. Units of Concentration Def: the concentration of a solution is a measure of how much solute is dissolved in a specific amount of solvent or solution Concentration is qualitative if you use words such as dilute or concentrated Most solutions are described quantitatively using percent by mass (% mass), percent by volume (% vol), molarity (mol/L) and molality (mol/kg)

8. How to Calculate Concentration Percent by mass = mass of solute x 100 mass of solution Percent by volume = vol of solute x 100 vol of solution Molality (m) = moles of solute kg of solvent ***Molarity (M) = moles of solute liters of solution We will focus on molarity

9. Molarity Use this equation: M = moles of solute liters of soln Q: What is the molarity of an aqueous solution containing 40 g of glucose (C 6 H 12 O 6 ) in 1.5 L of solution 1. find the number of mol of glucose 2. divide mol by liters

10. Q: How would you prepare 1L of 1.5 M solution of NaCl? 1. “of” means multiply, so 1 L x 1.5 mol = 1.5 mol L 2. change mol to grams, so 1.5 mol x 58g = 85 g mol 3. so take 85 g NaCl, place it in a 1 L volumetric flask and fill to the line with H 2 0

11. Since moles of solute = M x liters, we have a new equation: M 1 V 1 = M 2 V 2 where M 1 and V 1 are the molarity and volume of stock solution & M 2 and V 2 are the molarity and volume of dilute solution This equation lets you take stock solution and make less concentrated dilutions Q: What volume of 3.0 M KI stock solution would you use to make 0.3L of 1.25 M solution?

12. It is very important to keep your 1s and 2s straight when plugging into this equation “of” means multiply, so… 1. (0.3L)(1.25 mol/L) X (3.0M)(x) 2. solve Challenge: If 0.5L of 5.00 M stock solution of HCl is diluted to make 2.0 L of solution, how much HCl, in grams, is in the solution?

13. Def: mole fraction is the ratio of the number of moles of solute or solvent in solution to the total number of moles of solute & solvent X A = n A X B = n B n A + n B n A + n B Mole fraction is dimensionless: there are no units in the answer because everything cancels out Q: What is the mole fraction of NaOH in an aqueous solution that contains 28% of NaOH by mass? (Hint: assume 100 g, change g to mol and go from there)

14. 14.3 Factors Affecting Solvation Objectives: 1. Describe how intermolecular forces affect solvation 2. Define solubility 3. Understand what factors affect solubility

15. Intermolecular Forces Attractive forces exist among the particles of all substances When a solid solute is placed in a solvent, the solvent particles completely surround the surface of the solid solute If attractive forces between the solute and the solvent particles are greater than the attractive forces holding the solute particles together, the solvent particles pull the solute particles apart and surround them

16. The surrounded solute particle moves away from the solid solute and out into solution Def: solvation is the process of surrounding solute particles with solvent particles to form a solution “like dissolves like” is the general rule to determine whether solvation will occur in a specific solvent Bonding and polarity play a role Agitation (stirring or shaking), surface area (greater surface area dissolves faster) & temperature (higher temp increases solvation) all affect solvation

17. Solubility Solubility depends on the nature of the solute and solvent Def: an unsaturated solution is one that contains less dissolved solute for a given temperature and pressure than a saturated solution More solute can be dissolved in an unsaturated solution

18. Factors that Affect Solubility Def: a saturated solution contains the maximum amount of dissolved solute for a given amount of solvent at a given temperature and pressure Solubility is affected by raising the temperature For gases, solubility decreases with increased temp For liquids, solubility increases with increased temp due to greater energy and increased particle collisions

19. Def: a supersaturated solution contains more dissolved solute than a saturated solution at the same temperature Not every substance can become supersaturated To make a supersaturated solution, add solute at a high temperature and cool it slowly allowing excess solute to remain in solution Henry’s law states that at a given temp, the solubility (S) is directly proportional to the pressure (P) S1 = S2 where S is given in g/L & P in any units P1 P2