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Dilution Review

Dilution Review

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Dilution Review

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  1. Dilution Review • Take 5 minutes with your partner to solve the scenario below. • Oops! I spilled 100mL of water into my 300mL of 6 M HCl. Is my solution too weak to use if my lab requires a 1M to 1.5 M HCl solution? • How do you know?

  2. Colligative Properties

  3. Defining Colligative Properties • PHYSICAL properties of the solvent that change when a solute is added • The degree of change is dependent upon the number of solute particles added to the solvent • The more solute particles = the greater the change in colligative property

  4. Number of Particles of Solute • Think back to what you’ve learned about ionic and covalent compounds dissolving in water. • Ionic compounds DISSOCIATE as they dissolve. • Covalent compounds do NOT dissociate as they dissolve. • Apply this idea. • How many solute particles will 1 piece of NaCl add to a solvent? • Answer: 2 (Na+ and Cl-)

  5. Applying the Concept Further • How many solute particles will 1 piece of magnesium phosphate add to a solvent? • Mg3(PO4)2 breaks into 3 Mg2+ ions and 2 PO43- ions. Answer: 5 solute particles • How many solute particles will 1 piece of sugar add to a solvent? • Sugar is covalent. Answer: 1 solute particle • Finally, which of the two above will affect a colligative property the most?

  6. Two Colligative Properties • Boiling point • Freezing point

  7. Boiling Point Elevation Boiling point is changed when solute is added to the solvent. • Boiling point elevates when a solute is added to a solvent. • The solution requires more energy to reach boiling. • Example: Salt water will not boil at 100°C. It will boil at a HIGHER temp.

  8. Calculating the NEW Boiling Pt. ∆Tb = Kbm i • ∆Tb = change in boiling pt. • Kb = boiling point constant for the solvent (will be given) • m = molality • i = number of ions present (USE ONLY WITH IONIC SOLUTES!)

  9. Example of Boiling Pt. Calculation • What is the boiling point when 15.0g NaCl is dissolved into 200 mL of water? (Kb of water is 0.52 °C/m) • ∆Tb = Kbm i • You are solving for ∆Tb , and you have the Kb to use. • m =You’ll need to calculate the molality from the info in the question. (change 15.0 g of NaCl to moles and 200 mL to kg and plug in) • i = Finally, is the solute ionic? YES…NaCl is ionic and will give TWO ions when dissolved.

  10. What is the boiling point when 15.0g NaCl is dissolved into 200 mL of water? • You’ll need to calculate molality first. • 15.0g NaCl x 1 mol NaCl = 0.256 moles NaCl 58.5 g NaCl • 200 mL = 200 g = 0.2 kg of water • Molality = 0.256 moles NaCl/ 0.2 kg water = 1.28m • ∆Tb = Kbm i • ∆Tb = (0.52 C/m)(1.28m)(2) • ∆Tb = 1.33 C (note: This is the CHANGE in boiling point. It does NOT answer our question.) • New Boiling Point: 100C + 1.33C = 101.33 • We took the regular boiling point of water and added the change since boiling point ELEVATES with a solute added.

  11. Ten Minute PartnersBoiling Point Calculations Kb of water = 0.52°C/m • What is the boiling point when 30.0g CaCl2 is dissolved into 200 mL of water? • The molality of the above example is similar to the problem that we worked together a moment ago. Explain the difference in boiling temperature. • CHALLENGING: How many grams of NaCl would need to be added to the water to change the boiling temperature of 200 mL to 110°C?

  12. Answers to Boiling Point Calculations 1. ∆Tb= (1.36m)(0.52C/m)(3) = 2.12C 100C + 2.12C = 102.12C 2. The difference is due to the difference in the number of solute particles (ions). • 10C = m(0.52)(2) m = 9.62 9.62m = mole/0.2kg mole = 1.92 mole NaCl 112.2 grams NaCl

  13. Freezing Point Depression Freezing point is changed when solute is added to the solvent. • Freezing point depresses when a solute is added to a solvent. • The solution requires a lower temp to reach freezing. • Example: Salt water will not freeze at 0°C. It will freeze at a LOWER temp.

  14. Calculating a NEW Freezing Pt. ∆Tf = Kfm i • ∆Tf = change in freezing pt. • Kf = freezing point constant for the solvent (will be given) • m = molality • i = number of ions present (USE ONLY WITH IONIC SOLUTES)

  15. Example of Freezing Pt. Calculation • What is the freezing point when 15.0g NaCl is dissolved into 200 mL of water? • (Kf of water is 1.86 °C/m) • ∆Tf = Kfm i • You are solving for ∆Tf , and you have the Kf to use. • m =You’ll need to calculate the molality from the info in the question. (change 15.0 g of NaCl to moles and 200 mL to kg and plug in) • i = Finally, is the solute ionic? YES…NaCl is ionic and will give TWO ions when dissolved.

  16. What is the freezing point when 15.0g NaCl is dissolved into 200 mL of water? • You’ll need to calculate molality first. • 15.0g NaCl x 1 mol NaCl = 0.256 moles NaCl 58.5 g NaCl • 200 mL = 200 g = 0.2 kg of water • Molality = 0.256 moles NaCl/ 0.2 kg water = 1.28m • ∆Tf = Kfm i • ∆Tf = (1.86 C/m)(1.28m)(2) • ∆Tf = 4.76 C (note: This is the CHANGE in freezing point. It does NOT answer our question.) • New Freezing Point: 0C – 4.76C = -4.76C • We took the regular freezing point of water and subtracted the change since freezing point DEPRESSES with a solute added.

  17. Ten Minute PartnersFreezing Point Calculations Kf of water = 1.86°C/m • What is the freezing point when 30.0g CaCl2 is dissolved into 200 mL of water? • The molality of the above example is similar to the problem that we solved together. Explain the difference in freezing temperature. • CHALLENGING: How many grams of NaCl would need to be added to the water to change the freezing temperature of 200 mL to -8°C?

  18. Answers to Freezing Point Calculations • ∆Tf= (0.271m)(1.86C/m)(3) =1.51C 0C - 1.51C = -1.51C 2. The difference is due to the difference in the number of solute particles (ions). • 8C = m(1.86C/m)(2) molality = 2.15 2.15 = mole/0.2kg mole = 0.43 mole NaCl 25.14 grams NaCl

  19. Practice Practice Practice! • The 15-4 Practice Problems in your practice packet are boiling point/freezing point calculations. • The key is posted at the front of the classroom. • PLEASE SOLVE MANY OF THESE PROBLEMS! You’ll see several on Wednesday’s test.