Solution stoichiometry.pptx

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1. Solution stoichiometry

2. STOICHIOMETRY IN SOLUTIONS Identify the compound/element present in the combined solution. Write the balanced net ionic equation. Calculate the moles of reactants. Calculate the moles of product(s). Convert to grams or other units.

3. 3 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl ( aq )  AlCl 3 ( aq ) + H 2 (g) 2 6 2 3 Gram to Gram Conversions

4. 4 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl ( aq )  AlCl 3 ( aq ) + H 2 (g) 2 6 2 3 Now let’s get organized. Write the information below the substances. 3.45 g ? grams Gram to Gram Conversions

5. 5 Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl ( aq )  AlCl 3 ( aq ) + H 2 (g) 2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles. Now use the molar ratio. Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

6. 6 Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M) When working problems, it is a good idea to change M into its units.

7. 7 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. What type of problem(s) is this? Molarity followed by dilution. Solutions

8. 8 A solution is prepared by dissolving 3.73 grams of AlCl 3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. 1 st : = mol L 3.73 g 200.0 x 10 - 3 L 0.140 2 nd : M 1 V 1 = M 2 V 2 (0.140 M)(10.0 mL) = (? M)(100.0 mL) 0.0140 M = M 2 molar mass of AlCl 3 dilution formula final concentration Solutions

9. 9 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. First write a balanced Equation. ____ NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1

10. 10 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound. ____ NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL Since 1 L = 1000 mL, we can use this to save on the number of conversions Our Goal

11. 11 Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now let’s get to work converting. ____ NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL = mL NaOH H 2 SO 4 35.0 mL H 2 SO 4 0.125 mol 1000 mL H 2 SO 4 NaOH 2 mol 1 mol H 2 SO 4 1000 mL NaOH 0.102 mol NaOH 85.8 Units Match Solution Stoichiometry: shortcut

12. 12 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH) 2 ? 1st write out a balanced chemical equation Solution Stoichiometry

13. 13 What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH) 2 ? 2HCl(aq) + Ba(OH) 2 (aq)  2H 2 O(l) + BaCl 2 0.40 M 47.1 mL 0.75 M ? mL = mL HCl Ba(OH) 2 47.1 mL 1 mol Ba(OH) 2 HCl 2 mol 0.40 mol HCl HCl 1000 mL 1 7 6 Units match Solution Stoichiometry

14. 14 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? First write a balanced chemical reaction. ____ HCl ( aq ) + ____Ba(OH) 2 ( aq )  ____H 2 O(l) + ____BaCl 2 ( aq ) 2 1 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L

15. 15 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3 . Determine the molarity of the Ca(OH) 2 solution. We must first write a balanced equation. Solution Stochiometry Problem:

16. 16 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3 . Determine the molarity of the Ca(OH) 2 solution. Ca (OH) 2 ( aq ) + HNO 3 ( aq )  H 2 O(l) + Ca (NO 3 ) 2 ( aq ) 2 2 48.0 mL 19.2 mL 0.385 M = mol (Ca(OH) 2) L (Ca(OH) 2 ) 19.2 mL HNO 3 48.0 x 10 -3 L ? M units match! 0.0770 Solution Stochiometry Problem:

17. 17 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO 2 , is used in rebreathing gas masks to generate oxygen. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) a. How many moles of O 2 can be produced from 0.15 mol KO 2 and 0.10 mol H 2 O? b. Determine the limiting reactant. 4KO 2 (s) + 2H 2 O(l)  4KOH(s) + 3O 2 (g) 0.15 mol 0.10 mol ? moles Two starting amounts? Where do we start? Hide one

18. 18 Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO 3 ) 3 in 455 mL of solution. After you have worked the problem, click here to see setup answer Try this problem (then check your answer):

19. Factors affecting solubility

20. Polar liquids tend to dissolve readily in polar solvents. Pairs of liquids that mix in all proportions are said to be miscible and liquid that do not mix are immiscible . Hydrogen bonding interactions between solute and solvent may lead to high solubility. The solubility of alcohol decreases as the number of carbons increases. like dissolves like Non polar substances are soluble in non polar solvents. Nature of the solute and solvent

21. Temperature The solubility of most solid solutes in water increases as the temperature of the solution increases. In contrast, the solubility of gases in water decreases with increasing temperature.

22. Pressure The solubilities of solids and liquids are not appreciably affected by pressure. The solubility of a gas in any solvent increases as the pressure of the gas over the solvent increases. E.g. Carbonated Beverages.

23. ASSESSMENT How does the following factors affect the solubility of a solution: a . Nature of Solute and Solvent b . Temperature (solid, liquid, gas) c . Pressure (solid, liquid, gas)

24. Colligative properties

25. Colligative Properties Solution properties that depend on the number of solute particles present . Among colligative properties are Vapor pressure lowering Boiling point elevation Melting point depression Osmotic pressure

26. Vapor Pressure As solute molecules are added to a solution, the solvent become less volatile . Solute-solvent interactions contribute to this effect.

27. Vapor Pressure Therefore, the vapor pressure of a solution is lower than that of the pure solvent.

28. Raoult’s Law P A = X A P  A where X A is the mole fraction of compound A P  A is the normal vapor pressure of A at that temperature NOTE: This is one of those times when you want to make sure you have the vapor pressure of the solvent .

29. SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr . PRACTICE EXERCISE The vapor pressure of pure water at 110°C is 1070 torr . A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?

30. SAMPLE EXERCISE 13.8 Calculation of Vapor-Pressure Lowering Glycerin (C 3 H 8 O 3 ) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr (Appendix B). Solution Analyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute. Plan: We can use Raoult’s law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, X A , is the ratio of the number of moles of solvent (H 2 O) to total solution (moles C 3 H 8 O 3 + moles H 2 O). Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C 3 H 8 O 3 and H 2 O:

31. PRACTICE EXERCISE The vapor pressure of pure water at 110°C is 1070 torr . A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer:  0.290 SAMPLE EXERCISE 13.8 continued The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water. We now use Raoult’s law to calculate the vapor pressure of water for the solution:

32. Boiling Point Elevation and Freezing Point Depression Solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

33. Boiling Point Elevation The change in boiling point is proportional to the molality of the solution:  T b = K b  m where K b is the molal boiling point elevation constant, a property of the solvent.  T b is added to the normal boiling point of the solvent.

34. Freezing Point Depression The change in freezing point can be found similarly:  T f = K f  m Here K f is the molal freezing point depression constant of the solvent.  T f is subtracted from the normal freezing point of the solvent.

35. Boiling Point Elevation and Freezing Point Depression In both equations,  T does not depend on what the solute is , but only on how many particles are dissolved.  T b = K b  m  T f = K f  m

36. Colligative Properties of Electrolytes Because these properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) show greater changes than those of nonelectrolytes. e.g. NaCl dissociates to form 2 ion particles; its limiting van’t Hoff factor is 2.

37. Colligative Properties of Electrolytes However, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 M solution of methanol does. It doesn’t act like there are really 2 particles.

38. Osmosis Semipermeable membranes allow some particles to pass through while blocking others. In biological systems, most semipermeable membranes (such as cell walls) allow water to pass through, but block solutes.

39. Osmosis In osmosis, there is net movement of solvent from the area of higher solvent concentration ( lower solute concentration ) to the are of lower solvent concentration ( higher solute concentration ). Water tries to equalize the concentration on both sides until pressure is too high.

40. Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure ,  , is n V  = ( ) RT = MRT where M is the molarity of the solution If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic .

41. Osmosis in Blood Cells If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic . Water will flow out of the cell, and crenation results.

42. Osmosis in Cells If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic . Water will flow into the cell, and hemolysis results.