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(Chapter 13 Andrew Meade Mitchell Korotkin Kayla Devin)

(Chapter 13 Andrew Meade Mitchell Korotkin Kayla Devin). Solubility Rules:. -The presence of certain ions may cause a compound to be soluble or insoluble. - The solubility rules are a list of ions and whether they make a compound soluble or insoluble.

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(Chapter 13 Andrew Meade Mitchell Korotkin Kayla Devin)

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  1. (Chapter 13 Andrew Meade Mitchell Korotkin Kayla Devin)

  2. Solubility Rules: -The presence of certain ions may cause a compound to be soluble or insoluble. - The solubility rules are a list of ions and whether they make a compound soluble or insoluble. -If there are no solids (all products are soluble), then there is no reaction

  3. Example: Potassium chromate- K2CrO4 1. Look at solubility rules for chromate. Normally, chromates tend to be insoluble. However, one of the exceptions is the IA elements, and because potassium(K) is an IA element, Potassium chromate is soluble. Answer: SOLUBLE

  4. Example: Use the solubility rules to predict the reaction products: AuCl (aq) + NaNO3 (aq) AuNO3(aq) + NaCl(aq) **There is no reaction because both of the products are aqueous

  5. Complete Ionic Equations ** Notice the Spectator Ions circled- these are ions that do not take part in a chemical reaction and are found on both sides of the reaction **

  6. Net Ionic Equations -Include only the compounds that undergo a chemical change in a reaction in an aqueous solution -To do this simply omit the spectator ions from the complete ionic equation

  7. Scaffolded Example Write the complete and net ionic equation for the molecular equation: HCl(aq) + NaOH (aq) NaCl(aq) + H2O(l) 1. Split up aqueous reactants and products into ions. (complete ionic equation) H+ + Cl- + Na+ + OH- Na+ +Cl- + H2O 2. Cancel out spectator ions H+ + Cl- + Na+ + OH- Na+ +Cl- + H2O Net ionic equation: H+ + OH- H2O

  8. PROBLEM 1 Write the complete and net ionic equation for the molecular equation: NaCl(aq) + AgNO3 (aq) AgCl(s) + NaNO3(aq)

  9. Problem 1: ANSWER Write the complete and net ionic equation for the molecular equation: NaCl(aq) + AgNO3 (aq) AgCl(s) + NaNO3(aq) 1. Split up aqueous reactants and products into ions. (complete ionic equation) Na+ + Cl- + Ag+ + NO3- AgCl + Na+ + NO3- 2. Cancel out spectator ions Na+ + Cl- + Ag+ + NO3- AgCl(s) + Na+ + NO3- Net ionic equation: Ag+ + Cl- AgCl(s)

  10. Problem 2 Write the complete and net ionic equation for the molecular equation: KI(aq) + AgClO3 (aq) AgI(s) + KClO3(aq)

  11. Problem 2: ANSWER Write the complete and net ionic equation for the molecular equation: KI(aq) + AgClO3 (aq) AgI(s) + KClO3(aq) 1. Split up aqueous reactants and products into ions. (complete ionic equation) K+ + I- + Ag+ + ClO3- AgI + K+ + ClO3- 2. Cancel out spectator ions K+ + I- + Ag+ + ClO3- AgI(s) + K+ + ClO3- Net ionic equation: Ag+ + I- AgI(s)

  12. Colligative Properties - The physical properties of a solution are different from those of the pure solvent. - Some of these differences are due to the mere presence of solute particles in the solution. - Colligative properties depend on the number of particles dissolved in a given mass of solvent. - They do not depend on the chemical nature of the solute or solvent The 4 Colligative Properties Are: • Vapor Pressure Lowering • Freezing Point Depression • Boiling Point Elevation • Osmotic Pressure

  13. Vapor Pressure Lowering -Vapor pressure occurs because some molecules of a pure liquid leave the liquid state and enter the gaseous state (vaporization) -At the same time, molecules from the gaseous state return to the liquid state (condensation) -Equilibrium is established when the rate of vaporization and condensation becomes equal -The gas pressure resulting from the vapor molecules over the liquid is the vapor pressure -Vapor Pressure of a solvent containing a nonvolatile solute is lower than the vapor pressure of the pure solvent

  14. Osmotic Pressure - Solvents are able to move through a semipermeable membrane, where they move from a high to a low concentration. -Through the movement of the solute, the levels of the solution becomes uneven. -The osmosis of the solvent will stop when the pressure difference becomes large and the levels are very uneven. -The pressure difference is called osmotic pressure.

  15. Boiling Point Elevation -The molal boiling point constant (Kb) is the boiling point elevation of the solvent in a 1-molal solution of a nonvolatile, nonelectrolyte solute -The boiling point elevation is the difference between the boiling points of the pure solvent and a non electrolyte solution of that solvent and is directly proportional to the molal concentration of the solution -Can be calculated using the formula: Δ T b = K b (i) (m) Molality Boiling Point Elevation Boiling Constant Van't Hoff Constant

  16. Boiling Point Elevation Explanation Boiling point is essentially the amount of energy it takes to make the vapor pressure of a solution equal to the external pressure. With normal water, the amount of energy it takes for the vapor pressure to equal the external pressure is reached at 100 ºC. 100 ºC However, when a solute like NaCl is added, the vapor pressure of the solution is lowered, and it takes more energy to equalize the vapor and external pressure, causing a higher boiling point. Boiling Point Vapor pressure

  17. Scaffolded Example ΔTb = Kb (m)(i) Kb = .52 ºC/m m = i = 1. What is the boiling point elevation when 35.0 g of NaCl is dissolved in 750 g of water? (Kb for water is .52 ºC/m) .799 m STEP 1: FIND MOLALITY STEP 3: Plug in values STEP 2: Van't Hoff Factor 2.00 1 mole NaCl 58.43 g NaCl NaCl 35.0g NaCl X = .599 moles NaCl ΔTb = Kb (m)(i) 1 kg H2O 1000 g H2O 750g H2O X = .750 kg H2O Step 1: Find molality Step 1: Find molality Step 3: Plug in values into the formula ΔTb = .52ºC/m(.799m)(2.00) Step 2: Determine Van't Hoff Factor Step 2: Determine Van't Hoff Factor Cl- Na+ molality = moles solute kg solvent m= .599 moles NaCl .750 kg H2O Step 3: Plug in values into the formula ΔTb = .831 ºC molality = .799 m 2 ions = Van't Hoff Factor of 2

  18. Practice 1 1. What is the boiling point elevation when 15 g of ammonia (NH3) is dissolved in 250 g of H2O? (Kb for water is .52 ºC/m) Δ Tb = Kb(i)(m)

  19. Practice 1: Answer Δ Tb = Kb (i)(m) • 15 g NH3 x • m= • Δ Tb = .52ºC/m(1)(3.52m) • Δ Tb = 1.8 ºC 1 mole NH3 17.024 g NH3 = .88 mole NH3 .88 moles NH3 .25 kg H20 3.52 m

  20. Practice 2 1. What is the new boiling point when 23 g of ammonia (NH3) is dissolved in 180 g of H2O? (Kb for water is .52 ºC/m) Δ Tb = Kb(i)(m)

  21. Practice 2: Answer Δ Tb = Kb (i)(m) • 23.0 g NH3 x • m= • Δ Tb = .52ºC/m(1)(7.5m) • Δ Tb = 3.5ºC • 100ºC+3.5ºC = 103.5ºC 1 mole NH3 17.024 g NH3 = 1.35 mole NH3 1.35 moles NH3 .18 kg H20 7.5 m

  22. Freezing Point Depression -Freezing point depression is the ability of a dissolved solute to lower the freezing point of its solution. -When a solute is added to a solvent, more kinetic energy must be withdrawn from the solution for it to solidify. The equation for freezing point depression is: ΔTf = Kf (m)(i) Freezing Point Depression Van't Hoff Constant Freezing Constant Molality

  23. Freezing Point Depression Explanation Free moving molecules in liquid water. With pure water, the water molecules move freely around, and it is easy for the water to freeze into a lattice structure as solid ice. Water freezing Water and NaCl molecules However, when a solute like NaCl is added, the additional movement from the solute particles impairs the freezing process and makes the freezing process require more energy, lowering the freezing point. Water partially freezing

  24. Scaffolded Example ΔTf = Kf (m)(i) Kf = -1.86 ºC/m m = i = 1. What is the freezing point depression when 65.0 g of NaCl is dissolved in 1250 g of water? (Kf for water is -1.86 ºC/m) .888 m STEP 1: FIND MOLALITY STEP 3: Plug in values STEP 2: Van't Hoff Factor 2.00 1 mole NaCl 58.43 g NaCl NaCl 65.0g NaCl X = 1.11 moles NaCl ΔTf = Kf (m)(i) 1 kg H2O 1000 g H2O 1250g H2O X = 1.250 kg H2O Step 1: Find molality Step 3: Plug in values into the formula Step 1: Find molality ΔTf = -1.86ºC/m(.888m)(2.00) Step 2: Determine Van't Hoff Factor Step 2: Determine Van't Hoff Factor Cl- Na+ molality = moles solute kg solvent m= 1.11 moles NaCl 1.250 kg H2O Step 3: Plug in values into the formula ΔTf = -3.31 ºC 2 ions = Van't Hoff Factor of 2 molality = .888 m

  25. PRACTICE 1 1. What is the new freezing point when 90.0 g of oxygen gas is dissolved in 1750 g of water? (Kf for water is -1.86 ºC/m) ΔTf = Kf (m)(i)

  26. PRACTICE 1: ANSWERS ΔTf = Kf (m)(i) • 90.0 g O2 x • m= • Δ Tf = -1.86ºC/m(1.60m)(1) • Δ Tf = -2.98 ºC • 0 - 2.98 = -2.98 ºC 1 mole O2 32.00 g O2 = 2.80 moles O2 2.80 moles O2 1.75 kg H20 1.60 m

  27. PRACTICE 2 1. What is the freezing point depression when 48.0 g of oxygen gas is dissolved in 2300 g of water? (Kf for water is -1.86 ºC/m) ΔTf = Kf (m)(i)

  28. PRACTICE 2: ANSWERS ΔTf = Kf (m)(i) • 48.0 g O2 x • m= • Δ Tf = -1.86ºC/m(.652m)(1) • Δ Tf = -1.21ºC 1 mole O2 32.00 g O2 = 1.50 moles O2 1.50 moles O2 2.30 kg H20 .652 m

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