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Chemistry

Chemistry. Module 11, part 2. What affects solubility?. Temperature Solids in a liquid Liquids in a liquid Gases in a liquid. What affects Solubility. Pressure Gases: Increasing pressure increases the solubility Pressure has no affect on the solubility of either liquids or solids.

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Chemistry

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  1. Chemistry Module 11, part 2

  2. What affects solubility? Temperature Solids in a liquid Liquids in a liquid Gases in a liquid

  3. What affects Solubility • Pressure Gases: Increasing pressure increases the solubility Pressure has no affect on the solubility of either liquids or solids

  4. Energy in Solutions Exothermic: a process that releases heat Potassium permanganate and glycerine reaction Endothermic: a process that absorbs heat Barium Hydroxide Octahydrate with Ammonium Thiocyanate - YouTube

  5. Applying Stoichiometry The following reaction is performed in a lab: 3 K2CO3 (aq) + 2 Al(NO3)3 (aq) ------------------- Al2(CO3)3 (s) + 6 KNO3 (aq) If 191 mL of 1.25 M aluminum nitrate is added to an excess of potassium carbonate, how many grams of aluminum carbonate will be produced? Convert mL to L: 191mL Al(NO3)3 X 0.001 L = .191 L Al(NO3)3 1 mL Now we need to get into number of moles: Remember Molarity (M) is the same thing as Moles / Liter So…… 1.25 M = 1.25 Moles / Liter 1.25 Moles Al(NO3)3 X .191 L Al(NO3)3 = .239 Moles Al(NO3)3 1 L

  6. 3 K2CO3 (aq) + 2 Al(NO3)3 (aq) ------------------- Al2(CO3)3 (s) + 6 KNO3 (aq) 0.239 Moles Al(NO3)3 X 1 Mole Al2(CO3)3 = .120 Moles Al2(CO3)3 2 Moles Al(NO3)3 Now convert Moles to Grams using the molecular mass .120 Moles Al2(CO3)3 X 234.0 grams Al2(CO3)3 = 28.1 grams Al2(CO3)3 1 Mole Al2(CO3)3

  7. Molality • Another unit of concentration • Abbreviated with lowercase m • m= # moles solute # kg solvent

  8. Molality Example What is the molality of a solution in which 50.0 g of Mg(NO3)2 are dissolved in 500.0 grams of water? m= # moles of solute # kg solvent 50.0 g of Mg(NO3)2X 1 mole Mg(NO3)2 = 0.337 moles Mg(NO3)2 148.3 g of Mg(NO3)2 Now we need to convert grams to kg 500.0 g x 1 kg = 0.5000 kg 1000 g Now we can calculate molality m= 0.337 moles Mg(NO3)2 = 0.674 m Mg(NO3)2 0.5000 kg

  9. Freezing Point Depression Δ T = -i · Kf· m ΔT = Change in Temperature Tfinal – Tinitial m= molality Kf= freezing-point depression constant i= the number of molecules (or ions) the solute splits into when it dissolves

  10. Kf for water = 1.86˚ C/m How do you figure out i? Ionic examples: NaCl ------- Na+ + Cl- Al(NO3)3------ Al+3 + 3NO3- Covalent examples: C6H12O6 ------- C6H12O6

  11. Freezing Point Depression Example What is the freezing point of a solution that is made by dissolving 10.0 grams of KF in 100.0 grams of water? What do we need to know? Δ T = -i · Kf· m Kf water = 1.86 ˚C/m (This will be given to you) m i

  12. First, let’s calculate m for KF m=# moles KF # kg water 10.0 g KF X 1 mole KF = 0.172 moles KF 58.1 g KF 100.0 g water = 0.1000kg water m= 0.172 moles KF = 1.72 m 0.1000 kg water

  13. Let’s find I Is KF Ionic or Covalent? KF ------- K+ + I- so i = 2 Δ T = -i · Kf· m ΔT = (-2) (1.86˚C/m) (1.72 m) = -6.40 ˚ C Remember, this is how much the temperature changes. Because the freezing point of water is 0˚C, the new temperature is -6.40 ˚C.

  14. Think about: Would covalent or ionic compounds have the most impact on freezing-point depression? Why? Would NaCl or Al(NO3)3 have the most impact on freezing-point depression? Why?

  15. Boiling-Point Elevation Δ T = i· Kb · m • In this case, i is positive, not negative • Kb is the Boiling-point elevation constant. This will be given to you. • ΔT tells you the change in temperature. This would be added to the boiling point of the substance.

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