Solutions in Chemistry

1. Solutions in Chemistry

2. You are responsible for taking notes from this powerpoint! In class you may work with your group to do calculations and answer questions You are expected to read through the powerpoint and appropriate pages in the book and take notes If you are not prepared for class, you will have to work alone

3. Solutions are Mixtures What is a mixture? How are mixtures different than compounds? What is a homogeneous mixture? Suspensions and Colloids If the particles in a mixture are too big, a suspension or colloid forms Suspension: particles will separate out over time (oil and water) Colloid: particles stay suspended but large enough to reflect light (milk)

4. Solutions: Made of particles so small the parts of the mixture cannot be differentiated • Solute: substance in lesser amount of a solution - said to dissolve in the solvent • Solvent: substance in greater amount of a solution - said to have solute dissolved in it • Examples: • Solid: alloys (brass is copper and zinc) • Liquid: alcohols, vinegar • Gas: air • Gas/liquid: soda • Solid/liquid: coffee • Solid/gas: pollution particles in air

5. Go to the following site • Solutions • Answer the following: • How do you know if a substance is a solution? • How would you separate a solution? • The video also reviews molarity..you may want to watch before moving on!

6. Solvation • Solvation (solute dissolving in solution) occurs when there is an attractive force between the solute and solvent • “like dissolves like” Polar and ionic substances attract to each other through dipole-dipole interactions, hydrogen bonding and ion-dipole attraction Substances with dispersion forces attract to each other, but not to substances with stronger attractive forces

7. Solvation con’t • Generally, the force of attraction between the solvent and solute must be stronger than the attraction between the solute particles themselves for a substance to dissolve If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal.

8. salt dissolving in water

9. Electrolytes • Many ionic substances dissociate when dissolved in water • This solution is conductive due to the charged particles in solution • NaCl (s) → Na+ (aq) + Cl-(aq) • The attraction for the water molecule must be strong enough to overcome the electrostatic attraction between the ions

10. Types of solution • Saturated solution: no more solute may be dissolved in solvent (temperature dependent) • How do you know when a solution is saturated? • Unsaturated solution: more solute may be dissolve in solution • Supersaturated solution: more solute is dissolve in solution than expected at a given temperature

11. Factors affecting solubility • Temperature – why? • For most solids, the solubility of a solid in water increases with temperature: more interaction between solvent and solute particles and more energy to separate solute particles • For gases, solubility decreases with temperature: more KE, more volume and pressure of gas • Pressure – does not affect solid dissolution, but does affect gases – why? • Gases are more soluble with increasing pressure – need more pressure to come out of solution (soda)

12. Solubility curves • 1. Which solid shows the greatest increase of solubility with temperature? • 2. Why is the NH3 line falling with temperature?

13. Questions? • Write down any questions that you have at this point and bring them to class

14. Concentration: • Molarity: What is it?? What mass of NaCl is needed to make a 1.65 M NaCl solution? What is the concentration of IONS in solution?

15. Concentration of ions 34.5g BaCl2 is dissolved in 500.ml water. What is the concentration of each ion in solution? - BaCl2(s) → Ba+2(aq) + 2Cl-(aq) -Calculate moles using molar ratio above - Calculate molarity using given volume

16. Dilution • The key to dilution problems is to remember that the amount of moles or grams does not change – just the volume changes • Example: 250.ml of a .340M NaCl solution is diluted by adding 100.ml of water. What is the new concentration? • Determine the number of moles • .250L x .340 mol/L = .0850 mol • Divide by the new volume • .0850mol/.350L = .243M

17. Questions? • Write down any questions that you have at this point and bring them to class

18. Colligative properties Properties of solutions that are determined by ratio of solute particles to solvent particles, not on the kind of particle present. Vapor pressure Boiling point and freezing point Molality: Concentration term : “m” Molality = moles solute/kg solvent Gives direct relationship between moles of solute to moles of solvent (kg may be converted to moles)

19. Calculating molality What is the molality of a solution made by mixing 45.0g sucrose in 500.ml H2O? 45.0g sucrose x 1mole/342 g = .132 moles 500.ml H2O x 1.00g/1ml x 1kg/1000ml = .500kg .132mole/.500kg = .263m

20. Freezing point depression and Boiling point elevation Adding a solute to solvent will change the physical properties of the solvent When freezing, solute particles interfere with crystallization process and lower temperature needed to crystallize Freezing temperature is lowered When boiling, fewer solvent particles will be at surface (interacting with solute particles) and vapor pressure is lowered. Boiling temperature is raised

21. Calculating changes in freezing and boiling points For any solution, the number of solute particles in solution will determine how much the freezing and boiling point is changed The temperature change from the normal freezing/boiling point will be in constant proportion to the molality of the solution: ΔT =(Kf) (m) or (Kb) (m) For water: Kf = 1.86oC/m (freezing) or Kb = 0.512oC/m (boiling)

22. Calculating freezing point depression and boiling point elevation Example: What will be the freezing point of a .263m sucrose solution? ΔT = (Kf) (m) = (1.86oC/m) (.263m)= .489oC ΔT =.489oC Freezing point = 0oC - .489oC = -.489oC

23. What questions do you have?? Write down any questions and bring them to class