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1. Solutions

2. Objectives Differentiate the different types of solutions. Determine the different factors that can affect solubility Promote the use of beneficial solutions

3. Mixtures Mixture: a combination of two or more substances that do not combine chemically but remain the same individual substances; can be separated by physical means. Two types: Heterogeneous Homogeneous

4. Heterogeneous Mixture “Hetero” means “different” Consists of visibly different substances or phases (solid, liquid, gas) Can be separated by filtering Example:

5. Homogeneous Mixture “Homo” means the same has the same uniform appearance and composition throughout; maintain one phase (solid, liquid, gas) Commonly referred to as solutions Example: Salt Water

6. Solution Solution: a mixture of two or more substances that is identical throughout (homogeneous) can be physically separated composed of solutes and solvents the substance being dissolved the substance that dis solve s the solute Iced Tea (solution) Iced Tea Mix (solute) Water (solvent) Salt water is considered a solution. How can it be physically separated?

7. Solution The solvent is the largest part of the solution, and the solute is the smallest part of the solution S O L V E N T S O L U T E

8. Types of Solutions Gaseous solutions – air = Oxygen + Nitrogen Liquid solutions – drinks = mix + water Solid solutions – alloys = steel, brass, etc

9. Concentration the amount of solute dissolved in a solvent at a given temperature described as dilute if it has a low concentration of solute dissolved described as concentrated if it has a high concentration of solute dissolved

10. Concentration Unsaturated - has a less than the maximum concentration of solute dissolved Saturated - has the maximum concentration of solute dissolved (can see solid in bottom of solution) Supersaturated -contains more dissolved solute than normally possible (usually requires an increase in temperature followed by cooling)

11. Solubility the amount of solute that dissolves in a certain amount of a solvent at a given temperature and pressure to produce a saturated solution

12. Factors affecting solubility of solids Temperature e increased temperature causes solids to dissolve faster Shaking Shaking (agitation) causes solids to dissolve faster Smaller particles dissolve Faster because they have more surface area Particle Size e

13. Miscible liquids can easily dissolve in one another. Immiscible liquids are not soluble in each other.

14. Polarity and Dissolving Chemists use the saying “like dissolves like”: Polar solutes tend to dissolve in polar solvents. Nonpolar solutes tend to dissolve in nonpolar solvents.

15. Solubility Curves Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.

16. To read the graph, find the line for the substance. The amount that dissolves at a given temperature is on the y- axis.

17. How much KNO 3 dissolves in 100g (or 100mL H 2 O at 50 o C? 1. Find the line ( green ) 2. Find the temperature and follow up to the line.( red arrow ) 3. Read across to the y- axis and this is the answer. (blue arrow ) 4. Since it is more than ½-way between 80 and 90, it is 87.

18. A point on the line is a saturated solution. Above the line is supersaturated . Below the line is unsaturated .

19. Using Solubility Curves What is the solubility of NaNO 3 in 100 g of H 2 O at 0°C? How many grams of KNO 3 will dissolve in 200g of H 2 O at 45°C? How much water is needed to dissolve 190g of NaNO 3 at 30°C? 73g NaNO 3 75g = ? 100g H 2 O 2 00g H 2 O = 150 g KNO 3 95g = 190g 100g H 2 O ? g H 2 O = 200 g H 2 O

20. Molarity (M)

21. Mole Fraction (x) A component in a solution is equal to the number of moles of that component divided by the total number of moles of all component present. moles = m= mass MM= molar mass  

22. Chemistry-Borders IPC-Solutions-Borders

23. Chemistry-Borders IPC-Solutions-Borders

24. Molarity Molarity is the concentration of a solution expressed in moles of solute per Liter of solution. Molarity is a conversion factor for calculations Molarity (M) = moles of solute solution in Liters

25. Molality Molality of a solution is the number of moles of solute per kilogram of solvent. Molality = moles of solute mass of solvent (kg)

26. Molarity M = mol (solute) L (solution) Example 1: What is the molarity of a solution that has 2.3 moles of sodium chloride in 0.45 liters of solution?

27. Molarity M = mol (solute) L (solution) Example 2: How many moles of Na 2 CO 3 are there in 10.0 L of 2.0 M solution?

28. Molarity M = mol (solute) L (solution) Example 3: How many moles of KNO 3 are needed to make 450. mL of 1.5 molar solution? 450. mL 1L 1.5 mol KNO 3 1 1000mL 1L = .675 moles KNO 3

29. Molarity M = mol (solute) L (solution) Example 4: How many grams of NaCl are needed to make 3.0 L of 1.5 M solution? 3.0 L 1.5 mol NaCl 58.44 g NaCl 1 1 L 1 mol NaCl = 262.98 g NaCl

30. Molarity M = mol (solute) L (solution) Example 5: How many L of 4.0 M solution can be made with 132g of NaCl ? 132 g NaCl 1 mol NaCl 1 L 1 58.44 g NaCl 4.0 mol NaCl = 0.565 L

31. Dilutions Chemistry-Borders IPC-Solutions-Borders

32. Dilutions and Molarity Use this formula to make a more dilute solution from a concentrated solution Molarity 1 x Volume 1 = Molarity 2 x Volume 2 (Concentrated) (Dilute) (before) = (after) M 1 V 1 = M 2 V 2

33. Example 1 How many liters of 2.5 M HCl are required to make 1.5 L of 1.0 M HCl ? M 1 V 1 = M 2 V 2 M 1 = 2.5 M V 1 = ? M 2 = 1.0 M V 2 = 1.5 L (2.5M) V 1 = (1.0M) (1.5 L) 2.5M 2.5M = 0.60L

34. Example 1 M 1 = 2.5M V 1 = 0.60L M 2 = 1.0 M V 2 = 1.5 L How much water should you add to the volume of 2.5M HCl you calculated above to make the solution? (draw this in your notes) 1 st add .60L of HCl to measuring device.

35. Example 1 M 1 = 2.5M V 1 = 0.60L M 2 = 1.0 M V 2 = 1.5 L How much water should you add to the volume of 2.5M HCl you calculated above to make the solution? Then add enough water to get to 1.5L of solution V 2 – V 1 = Amount of water 1.5L – 0.60L = 0.90L water

36. Example 1 M 1 = 2.5M V 1 = 0.60L M 2 = 1.0 M V 2 = 1.5 L How much water should you add to the volume of 2.5M HCl you calculated above to make the solution? Final solution is 1.5L of 1.0M HCl

37. Example 2 250.0 mL of a 0.500 M HCl solution needs to be made from concentrated HCl . What volume of the concentrated solution is needed if its molarity is 12.0 M? M 1 V 1 = M 2 V 2 M 1 = V 1 = M 2 = V 2 =

38. Example 2 250.0 mL of a 0.500 M HCl solution needs to be made from concentrated HCl . What volume of the concentrated solution is needed if its molarity is 12.0 M? M 1 V 1 = M 2 V 2 M 1 = 12.0M V 1 = 10.4mL M 2 = 0.500M V 2 = 250.0mL How much water would you add to make the final solution? 250.0mL - 10.4mL = 239.6mL

39. Objectives Use different ways of expressing concentration of solutions. Calculate the solution by percent by mass, by volume, and by mass-volume

40. Mass Percent Chemistry-Borders IPC-Solutions-Borders

41. Mass Percent Solutions can also be represented as percent of solute in a specific mass of solution. For a solid dissolved in water, you use percent by mass which is Mass Percent. % by mass = mass solute x 100 mass of solution **Mass of solution = solute mass + solvent mass

42. Example 1 If a solution that has a mass of 800.0 grams contains 20.0 grams of NaCl , what is the concentration using Percent by Mass ?

43. Example 2 If 10.0 grams of NaCl is dissolved in 90.0 grams of water , what is the concentration using Percent by Mass ?

44. Example 3 How many grams of sodium bromide are in 200.0g of solution that is 15.0% sodium bromide by mass? % by mass = mass solute x 100 mass of solution   % by mass = ? g NaBr x 100 = 15.0%NaBr 200.0g solution   g NaBr = 200.0 x 15.0 100 = 30 g NaBr

45. PERCENT BY VOLUME It involves the volume of both the solute and the solution. percent by volume = volume of solute X 100 volume of solution

46. Example 1 Determine the volume/volume percent solution made by combining 25 mL of ethanol with enough water to produce 200 mL of the solution.

47. Example 2 A wine contains 12% alcohol by volume. Calculate the volume (in ml) of alcohol in 350 ml of the wine.

48. Percent by mass-volume Relationship between the mass of the solution and the volume of the solution Percent by mass-volume= mass of solute in grams X 100 volume of solution in ml

49. Example 1 Calculate the mass-volume percent of a 762.5 ml solution that is prepared by dissolving 289.15 g of calcium nitrate, Ca(NO3)2, in water.

50. Example 2 A 50 ml of 12% by mass-volume solution was used in an experiment. How many grams of solute does the solution contain?

51. Objectives understand the concept of stoichiometry perform stoichiometric calculations

52. Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure

53. stoichiometric coefficient or stoichiometric number is the number of molecules that participate in the reaction. 3Fe(s) + 4H2O(l) ⇾ Fe3O4 (s)+ 4H2 (g)

54. Limiting Reagent In a chemical reaction, it is possible that one of the reactants is present in excess amount. excess reactant- the reaction stops immediately as soon as one of the reactants is totally consumed N2 + 3H2 ➝ 2NH3

55. Balancing chemical reactions are frequently written as an equation, using chemical symbols. 2Na(s)+2HCl(aq)→2NaCl(aq)+H2(g)

56. Solution Stoichiometry Chemistry-Borders IPC-Solutions-Borders reactants products

57. Solution Stoichiometry When we previously did stoichiometry for a reaction to determine theoretical yield, we only worked with GRAMS and MOLES Ex/ How many MOLES of HCl are required to react with 13 GRAMS of zinc? Zn + 2 HCl  ZnCl 2 + H 2

58. Solution Stoichiometry But we may be given something OTHER than grams and moles We can use stoichiometry to solve for ANY unit. We just need to make sure units cancel out and we end up with the unit we are trying to solve for! The mole ratio using coefficients from the balanced chemical equation is the key to switching between compounds

59. Solution Stoichiometry Ex/ How many LITERS of 12 M HCl are required to react with 13.0 GRAMS of zinc? Zn + 2 HCl  ZnCl 2 + H 2 13.0g Zn 1 mole Zn 2 mol HCl 1L HCl 1 65.38g Zn 1 mol Zn 12 mol HCl Remember – Molarity (M) is a conversion Factor = 0.03 L HCl

60. Solution Stoichiometry Ex/ How many grams of NaOH would be required to react with 1.50 L of 3.75M sulfuric acid? H 2 SO 4 + NaOH  Na 2 SO 4 + H 2 O 1.50L 1 H 2 SO 4 3.75 mole H 2 SO 4 2 mol Na OH 40.01g Na OH 1 1 L H 2 SO 4 1 mole H 2 SO 4 1 mol Na OH = 450. g Na OH

61. Fermentation is a complex chemical process of making wine by converting glucose into ethanol and carbon dioxide: C6H12O6(s)  2 C2H5OH (l) + 2 CO2(g) a. Calculate the mass of ethanol produced if 500.0 grams of glucose reacts completely.

62. Fermentation is a complex chemical process of making wine by converting glucose into ethanol and carbon dioxide: C6H12O6(s)  2 C2H5OH (l) + 2 CO2(g) Calculate the volume of carbon dioxide gas produced at STP if 100.0 grams of glucose reacts.

63. Fermentation is a complex chemical process of making wine by converting glucose into ethanol and carbon dioxide: C6H12O6(s)  2 C2H5OH (l) + 2 CO2(g) A. If 17.5 moles of ethanol were produced, how many moles of glucose were there in the beginning?

64. Consider the reaction of zinc metal with hydrochloric acid, HCl( aq ). A. Write the balance equation for this reaction. Zn(s) + HCl( aq )  H2(g) + ZnCl2( aq ) B. Calculate the moles of HCl needed to react completely with 8.25 moles of zinc.

65. Consider the reaction of zinc metal with hydrochloric acid, HCl( aq ). Zn(s) + 2 HCl( aq )  H2(g) + ZnCl2( aq ) C. Calculate the grams of zinc chloride produced if 0.238 grams of zinc react completely