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Page 1: Solutions in Chemistry

Solutions in Chemistry

Page 2: Solutions in Chemistry

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

Page 3: Solutions in Chemistry

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)

Page 4: Solutions in Chemistry

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

Page 5: Solutions in Chemistry

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!

Page 6: Solutions in Chemistry

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

Page 7: Solutions in Chemistry

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.

Page 9: Solutions in Chemistry

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

Page 10: Solutions in Chemistry

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

Page 11: Solutions in Chemistry

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)

Page 12: Solutions in Chemistry

Solubility curves1. Which solid

shows the greatest

increase of solubility with temperature?

2. Why is the NH3 line falling

with temperature?

Page 13: Solutions in Chemistry

Questions?

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

Page 14: Solutions in Chemistry

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?

Page 15: Solutions in Chemistry

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

Page 16: Solutions in Chemistry

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

Page 17: Solutions in Chemistry

Questions?

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

Page 18: Solutions in Chemistry

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)

Page 19: Solutions in Chemistry

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

Page 20: Solutions in Chemistry

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

Page 21: Solutions in Chemistry

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)

Page 22: Solutions in Chemistry

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

Page 23: Solutions in Chemistry

What questions do you have??

• Write down any questions and bring them to class


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