1. The Chemistry of Acids and Bases

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3. Acids and Bases: Molar Solutions At the conclusion of our time together, you should be able to: 1. Define molarity 2. Make a solution with a given molarity 3. Determine the molarity of a given solution

4. Some Definitions A solution is a homogeneous mixture of 2 or more substances in a single phase. The larger constituent is usually regarded as the SOLVENT and the others as SOLUTES.

5. Some Definitions The units are moles per Liter. Specifically, it’s moles of solute per Liter of solution. Abbreviated as mol/L or M. The word Molarity.

6. Some Definitions Moles measure amount of solute usually in grams Molarity measures moles of solute per Liter of solution.

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8. Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution

9. Concentration of Solute Therefore: x both sides by Liters = MV = moles = grams/molar mass Molarity (M) = moles solute liters of solution

10. PROBLEM: Using grams/molar mass = MV. Handout #1 = 58.44 g NaCl x g NaCl 58.44 g NaCl = 1 M x 1 L

11. PROBLEM: Using grams/molar mass = MV. Handout #2 = 42.5 g NaNO 3 x g NaNO 3 85.00 g NaNO 3 = 0.5 M x 1 L

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13. Acids and Bases: Molar Solutions Let’s see if you can: 1. Define molarity 2. Make a solution with a given molarity

14. PROBLEM: Using grams/molar mass = MV. Backside #1 = 245 g H 2 SO 4 x g H 2 SO 4 98.09 g H 2 SO 4 = 2.50 M x 1 L

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16. Acids and Bases: Molar Solutions At the conclusion of our time together, you should be able to: 1. Define molarity 2. Make a solution with a given molarity

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18. PROBLEM: Using grams/molar mass = MV. Backside #4 = 0.741 g Ca(OH) 2 x g Ca(OH) 2 74.10 g Ca(OH) 2 = 0.100 M x 0.100 L

19. Preparing Solutions Determine the mass of solute. Determine the mass of solute. Place in the appropriate volumetric flask. Place in the appropriate volumetric flask. Add deionized water and swirl until solute is dissolved. Add deionized water and swirl until solute is dissolved. Add water to the mark on the neck of the flask. Add water to the mark on the neck of the flask. Stopper and mix thoroughly. Stopper and mix thoroughly.

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21. Stoichiometry: Molar Solutions Let’s see if you can: 1. Define molarity 2. Make a solution with a given molarity

22. Concentration of Solute Therefore: MV = moles = grams/molar mass Molarity (M) = moles solute liters of solution

23. Acids and Bases: Molar Solutions At the conclusion of our time together, you should be able to: 1. Define molarity 2. Make a solution with a given molarity 3. Determine the molarity of a given solution

24. Concentration of Solute Therefore: MV = moles = grams/molar mass Now let’s use this formula to solve another type of problem: Molarity (M) = moles solute liters of solution

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26. PROBLEM: Mr. T accidentally dropped 15.14 g of silver (I) nitrate into 100.0 mL of deionized water. Rather than throw the solution away, help him by determining its molar concentration so that he can label it and still use it.. = 0.8912 M AgNO 3 15.14 g AgNO 3 x 1 mol AgNO 3 169.88 g AgNO 3 = M x 0.100 L

27. PROBLEM: Molarity Problems #2 = 3 M KI 249 g KIx 1 mol KI 166.00 g KI = M x 0.5 L

28. How Would You Make this 3M Solution?? Measure 249 g of KI Measure 249 g of KI Place in a 0.500 L volumetric flask. Place in a 0.500 L volumetric flask. Add deionized water and swirl until solute is dissolved. Add deionized water and swirl until solute is dissolved. Add water to the mark on the neck of the flask. Add water to the mark on the neck of the flask. Stopper and mix thoroughly. Stopper and mix thoroughly.

29. Most Caring Person!! An elderly gentleman had recently lost his wife. Upon seeing the man cry, a little 4 year old neighbor boy went into the old gentleman's yard, climbed onto his lap, and just sat there. When his mother asked him what he had said to the neighbor, the little boy just said, "Nothing, I just helped him cry."

30. Acids and Bases: Molar Solutions Let’s see if you can: 1. Define molarity 2. Make a solution with a given molarity 3. Determine the molarity of a given solution

31. How We Doing?? Need Help????

32. Exit Quiz #1 1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water. 2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution.

33. 1. Calculate the molarity if 0.75 mol of NaCl in placed in 300.0 mL of water. 2. Calculate the number of moles and grams of HCl if there is 12.2 mL of 2.45 M HCl solution. 0.75 mol NaCl = X M x 0.3000 L = 2.5 M NaCl X mol HCl = 2.45 M x 0.0122 L = 0.0299 mol HCl 0.0299 mol HClx 36.46 g HCl 1 mol HCl = 1.09 g HCl

35. Entrance Quiz #2 1. Calculate the molarity if 0.50 mol of KBr in placed in 750 mL of water. 2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl in solution.

36. 1. Calculate the molarity if 0.50 mol of KBr in placed in 750 mL of water. 2. Calculate the number of moles and molarity of HCl if there is 12.2 mL with 2.45 g of HCl. 0.50 mol KBr = X M x 0.75 L = 0.67 M KBr 0.0672 mol HCl = X M x 0.0122 L = 5.51 M HCl 2.45 g HClx 1 mol HCl 36.46 g HCl = 0.0672 mol HCl

37. Stoichiometry: Molar Solutions At the conclusion of our time together, you should be able to: 1. Define molarity 2. Determine the molarity of a given solution 3. Make a solution with a given molarity 4. Dilute a given solution to a new molarity

38. If the Amount (moles) of Solute #1 = #2 M 1 V 1 = moles and M 2 V 2 = moles Therefore if moles of solute are constant: M 1 V 1 = M 2 V 2 Molarity (M) = moles solute liters of solution And Using the Formula:

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40. PROBLEM #1: Using M 1 V 1 = M 2 V 2 = 0.125 M NaOH 0.150 M NaOHx 0.125 L = M x 0.150 L

41. PROBLEM #5: Using M 1 V 1 = M 2 V 2 = 1.20 L Therefore: 0.700 L needs to be added 2.40 M KClx 0.500 L = 1.00 M KCl x X L

42. Stoichiometry: Molar Solutions Let’s see if you can: 1. Define molarity 2. Determine the molarity of a given solution 3. Make a solution with a given molarity 4. Dilute a given solution to a new molarity

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44. How much water do I need to add to 250 mL of 3.0 M HCl to dilute it to 1.0 M HCl? = 0.75 L Total, therefore 0.50 L 3.0 M HClx 0.250 L = 1.0 M HCl x L

45. Name ________Class Period _____ Clicker Number "Making Molar Solutions A1" (10 points) Make 50.00 mL of a 0.100M BaCl 2  2H 2 O solution. __________ grams mass of solute needed __________ Instructor initials (one point)

46. What are the products of the reaction? Balance the equation. Write the balanced molecular, complete ionic and net ionic equations below. Place the precipitate on the 3 rd line of the first row of lines. (5 points) ________ + ________  ________ + ________ ___ ____ + ___ ____  ______ + ___ ____ ____ + ___  _____ What is this the best molar ratio based on the stoichiometry? Should you have put all 50.00 mL of each reactant together to form the most product? Circle yes or no. (2 points) ____________ : ____________ Yes No

47. Preparing Solutions Determine the mass of solute. Determine the mass of solute. Place in the appropriate volumetric flask. Place in the appropriate volumetric flask. Add deionized water and swirl until solute is dissolved. Add deionized water and swirl until solute is dissolved. Add water to the mark on the neck of the flask. Add water to the mark on the neck of the flask. Stopper and mix thoroughly. Stopper and mix thoroughly.

49. PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate Molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M

50. USING MOLARITY moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution? X g H 2 C 2 O 4 x 1 mol H 2 C 2 O 4 90.04 g H 2 C 2 O 4 = 0.0500 x 0.250 L = 1.13 g H 2 C 2 O 4

51. Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? moles = MV mol = 3.0 M x 0.400 L mol = 1.2 48 g NaOH 1.2 mol NaOHx 40.00 g NaOH 1 mol NaOH

52. PROBLEM: Mr. T needs to make 0.500 L of a 0.100 M solution of lead (II) nitrate for a future lab. Please help him by calculating the amount of solute needed and by outlining in detail the steps he would need to take to make this molar solution. = 16.6 g Pb(NO 3 ) 2 X g Pb(NO 3 ) 2 x 1 mol Pb(NO 3 ) 2 331.22 g Pb(NO 3 ) 2 = 0.100 M x 0.500 L

53. PROBLEM # 4: Grams to make 100. mL of 0.100 M calcium hydroxide. = 0.741 g Ca(OH) 2 X g Ca(OH) 2 x 1 mol Ca(OH) 2 74.10 g Ca(OH) 2 = 0.100 M x 0.100 L

54. PROBLEM # 5: 4.00 moles of nitric acid in 1.50 L of solution = what M. = 2.67 M HNO 3 4.00 mol HNO 3 = X M x 1.50 L