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Solutions

Solutions. Honors Chemistry Snarrenberg 2012-2013. Catalyst: Tuesday 5/28. Think back to when you did the polarity demo in lab with Styrofoam and eco-peanuts… Think back to Chapter 3… What is a solution? What do you call the substance you’re dissolving?

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Solutions

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  1. Solutions Honors Chemistry Snarrenberg 2012-2013

  2. Catalyst: Tuesday 5/28 Think back to when you did the polarity demo in lab with Styrofoam and eco-peanuts… Think back to Chapter 3… • What is a solution? • What do you call the substance you’re dissolving? • What do you call the substance you’re dissolving it in? • How do you tell which is which? • What is an aqueous solution? • How do ionic compounds dissolve in water? • How do covalent compounds dissolve in water? • Can you think of a substance that doesn’t dissolve in water? What causes them to not dissolve?

  3. soap • After you eat fried chicken for dinner, (or pizza or steak or… anyone hungry yet?), and you’re asked to do the dishes, why do you have to use soap? • Why can’t you just use water? • But then how does the soap come off?

  4. Let’s look at the vocab  • Define the following using the words ‘solute’ and ‘solvent’ • Saturated solution • Unsaturated solution • Concentrated solution • Dilute solution • Demo time dudes!

  5. An aqueous solution containing 30 g of KCl in 50 g of water at 55°C is… • Unsaturated • Saturated • Supersaturated • An aqueous solution containing 5 g of NaNO3 in 10 g of water at STP is… • Unsaturated • Saturated • Supersaturated • An aqueous solution containing 25 g of NaNO3 in 50 g of water at72°C is… • Unsaturated • Saturated • Supersaturated

  6. Demo questions • Is the solution saturated or unsaturated? • Is the solution concentrated or dilute? • What happens as the solution is heated? • ..but wait, wasn’t it saturated already? • What happens as the solution cools? • What happens when a crystal of solid is added?

  7. Comparing flasks • Draw a qualitative picture to represent the concentration in each flask. • How could you quantify the concentration in the flasks?

  8. Which salt has the highest solubility at STP? • Which salt has the lowest solubility at STP? • If you have 50 g of water, how many grams of lead (II) nitrate will make a saturated solution?

  9. Catalyst: Thu 5/30 HW Tonight: 38, 40, 50 • How many grams of sodium acetate can dissolve in 100 g of water at 25°C? • How many grams of sodium acetate can dissolve in 50 g of water at 25°C? • At around what temperature can I dissolve 140 g of sodium acetate to make a saturated solution?

  10. Quantifying it! • How do you quantify the concentration of a solution? • Molarity! (M) – mol/L • 1.0 molar solution, written as 1.0 M, contains 1.0 mols of solute per liter of solution

  11. Calculating molarity • What is the molarity of a solution containing 11.5 g of NaOH with a volume of 1.50 L? • M = moles of solute liters of solution • M = 0.288 molNaOH = 0.192 M NaOH 1.50 L solution

  12. Preparation of a standard solution Standard solution: a solution whose concentration is accurately known

  13. Calculating Molarity • 0.105 M • 0.00061 M • 0.61 M • Calculate the molarity of a solution prepared by dissolving 15.6 g of solid KBr in enough water to make 1.25 L of solution • Calculate the molarity of a solution prepared by dissolving 250.0 g of solid Ca(NO3)2 in enough water to make 2500 mL of solution

  14. POGIL Question A student correctly determines that 17.1 grams of sucrose are needed to make 50 mL of a 1 M sucrose solution. When making this solution in lab, the student measured 50 mL of water and massed 17.1 grams of sucrose. The student then mixed the two together into a new beaker. The student’s resulting solution was not 1 M sucrose as intended. After the student mixed sucrose and water, the resulting solution was poured into a graduated cylinder and it read 62 mL. • What part of the molarity equation did the student overlook when mixing the solution? • What steps should the student take in lab to correctly prepare the 1 M solution?

  15. Note on Molarity • How many mols of NaCl are in a 1.0 M NaCl solution? • …actually, there are no NaCl units in a NaCl solution, only Na+ ions and Cl- ions • 1.0 mols Na+ === 1.0 M Na+ • 1.0 molsCl- === 1.0 M Cl- • Net ionic equations

  16. Concentration of ions in solution • 2.40 M Na+ • 1.20 M SO42- • 1.50 M K+ • 0.750 M CrO42- • 1.20 M Na2SO4 • 0.750 M K2CrO4

  17. Calculating # mols from molarity • Molarity has a complex unit (mol/L) and can be used as a conversion factor! (You must be so excited XD) • How many mols of Ag+ ions are in 25 mL of a 0.75 M AgNO3 solution? • 0.75 M AgNO3 = 0.75 M Ag+

  18. You try  • 1.47 x 10-2mol Na+ • 4.41 x 10-2mol Na+ • How many moles of Na+ ions are present in 42.0 mL of a 0.350 M NaCl solution? • How many mols of Na+ ions are present in 42.0 mL of a 0.350 M Na3PO4 solution?

  19. Calculating mass from molarity • 31.9 g • 41.3 g • What mass of NaNO3 is required to make 2.50 L of 0.150 M solution of NaNO3? • What mass of (NH4)2SO4 is required to make 1.25 L of a 0.250 M solution of (NH4)2SO4?

  20. HW Check • #54: the new concentration of solute is half the original • #58: 0.541 L • #60: 0.13 M • #62: 10.3 mL • Dilutions Review: key words to watch out for that indicate a dilutions problem with M1V1=M2V2 • A change in molarity of the solutions with the same solute • Concentrated / Dilute • New solution / original solution

  21. HW Review

  22. Catalyst: Monday 6/2 • Draw a stoichiometry map (remember, the center link is always Mol A <--> Mol B) to include molarity. Think about all the units you can come from and go to. • HW: 66, 67, 68 (Ch. 15) • Quiz Thursday on Solubility (graph), Molarity, Dilutions, and Solutions Stoich

  23. Stoichiometry Review Propane, used in gas barbecue grills, burns in oxygen according to the following balanced chemical equation: C3H8(g) + 5O2(g)  3CO2 (g) 4H2O Calculate the mass in grams of water vapor produced if 3.11 mol of propane is burned. ANSWER: 224 g

  24. Catalyst: Monday 6/2 • Draw a stoichiometry map (remember, the center link is always Mol A <--> Mol B) to include molarity. Think about all the units you can come from and go to. • What if you started with L of a solution or you needed to go to L or of a solution? Now add a new bridge. • If we are dealing with reactions where the reactants are aqueous solutions, what TYPE of reaction are we talking about? (oooh throw back)

  25. Solutions Stoichiometry#1silver nitrate + potassium chloride  ?

  26. Solutions Stoichiometry #1 • If I mixed 2 mL of 1.5 M silver nitrate with excess potassium chloride, how many grams of solid silver chloride was I able to produce? Answer: 0.43 g

  27. Solutions Stoichiometry #3 Limiting Reactant Problem! Calculate the mass of the white solid CaCO3 that forms when 25.0 mL of a 0.100 M Ca(NO3)2 solution is mixed with 20.0 mL of a 0.150 M Na2CO3 solution. Answer: 0.250 g CaCO3

  28. Announcements • Quiz Thursday on Solubility (graph), Molarity, Dilutions, and Solutions Stoich • HW: 15.7 Notes& #65

  29. Solutions Stoich #4aluminum + copper (II) chloride  ? HW Answers: • #66 – 0.523 g • #67 – 0.976 g PbCrO4 • #68 – 0.300 g

  30. Brainstorm • What pieces of information do you need to find the molarity of the copper (II) chloride solution? • What tools in your toolbox can you use to get those pieces of information? (Hint: one you can find directly, the other must be found indirectly, i.e. through calculation)

  31. Some info… • Volume of CuCl2 solution used: 20.0 mL • Mass of Al before the reaction: 0.59 g • Aluminum is the excess reactant!

  32. % Error Calculation • The actual molarity of the CuCl2 solution was 0.65. What was the percent error?

  33. Try this one… Solutions Stoich #5 One way to determine the amount of chloride ion in a a water sample is to titrate the sample of standard AgNO3 solution to produce solid AgCl. Ag+(aq) + Cl-(aq)  AgCl(s) If a 25.0 mL water sample requires 27.2 mL of 0.104 M AgNO3 in such a titration, what is the concentration of Cl- in the sample? Answer: 0.113 M

  34. Solutions Stoichiometry #2 • Calculate the mass of sodium iodide that must be added to 425.0 mL of a 0.100 M lead (II) nitrate solution to precipitate all of the lead (II) ions as lead (II) iodide. Answer: 12.7 g NaI

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