Chapter 12 Solutions

12.1 What Is a Solution? Solution are made of two components. What are they? How would you describe a solution? What’s in Coke?

12.1 What Is a Solution? What are the three types of solutions? Give an example of each! What’s this?

12.1 What Is a Solution? What are aqueous solutions? Why is water the “universal solvent?

12.2 Formation of Solutions Solute-separation: Dissociation! Describe dissociation of an ionic crystal… Contrast it to that of a molecular substance…

12.4 Solubility, Temperature, and Pressure What is solubility? What factors effect it?

12.4 Solubility, Temperature, and Pressure Solubility Relationships

12.4 Solubility, Temperature, and Pressure What’s a saturated solution? Use the solubility graph to find how much copper sulfate makes a saturated solution in 100g of 20o water: What if you has 200g of solution? Explain…

12.4 Solubility, Temperature, and Pressure What’s an unsaturated solution? How would you make an unsaturated NaCl(aq) at 50oC? How would you make an unsaturated LiCl(aq) at 70oC?

12.6 Concentration What’s concentration? What are some terms that describe concentration? What are some ways to quantify concentration?

12.6 Molarity Molarity: Describe how to form 1M NaCl(aq) What is the relationship between M and concentration? Describe how to dilute a solution of known concentration…

12.6 Dilution Practice problem… Dilution Equation: Illustrate how dilution changes the concentration of a solution:

12.7 Percent Composition Percent composition Amount of solute divided by amount of solution Three main types of percent composition: Percent composition by mass (mass %) Percent composition by volume (vol %) Percent composition by mass/volume (% m/v)

12.7 Percent Composition What is the percent-by-mass concentration of a solution containing 10.0g of sucrose and enough water to make 100g of solution in total? What is the percent-by-volume concentration of a solution made from 25.0mL of liquid ethanol and enough water to give 100.0 mL of solution?

12.9 Colligative Properties of Solutions Normal boiling point Temperature where a liquid boils at one atmosphere Normal freezing point Temperature where a liquid freezes at one atmosphere

12.9 Colligative Properties of Solutions Heating Curve of Water

12.9 Colligative Properties of Solutions (Continued) Colligative property Property that depends on the number of solute particles Does not depend on type of particle, only number of particles Melting point and freezing point are examples because they both change when a solute is dissolved in water. Let’s see…

12.9 Colligative Properties of Solutions Colligative property (continued)

12.9 Colligative Properties of Solutions (Continued) How many moles of dissolved solute particles are present in each of the following beakers?

12.9 Colligative Properties of Solutions (Continued) How would the boiling and freezing points of the two solutions compare?

Complete lab report(s)! End of Chapter 12 Exam 3 = chapters 8,9 and 12 Studying… Pick up an objective sheet Make sure you know how to write formulas! REALLY! Do text problems that go along with the notes Complete lab report(s)! Final EXAM! Use notes, old exams and text problems!

12.6 Molarity

Molarity can be used in conversions! Ex: How many moles of NaCl are there in 500 mL of1.0 M NaCl solution? How many grams?

12.6 Molarity Ex: How many moles of NaCl are there in 500 mL of1.0 M NaCl solution? How many grams? 500mLx 1L/1000mL = 0.5L M= moles solute/Liters solution 1.0M = x moles/0.5L X= 0.5moles NaCl

12.6 Molarity Ex: How many moles of NaCl are there in 500 mL of1.0 M NaCl solution? How many grams? NaCl= 58.443g/mol 0.5moles NaCl x 58.433g/mol= 29.217g

12.6 Molarity (Continued) How many moles of NaCl are there in 200 mL of 1.0 M NaCl solution? How many grams?

12.6 Molarity (Continued) How many mL of a 1.500 M solution of NaCl do you need to obtain 100.0 g of NaCl? NaCl= 58.443g/mol 100.g NaCl / 58.442g/mol x =1.71moles NaC 1.500M= 1.71mols/ X Liters X= 1.14L 1.14lL x 1000ml/L= 1140mL of solution

12.6 Molarity (Continued) Preparing a solution from scratch Calculate the amount of material that is needed. Add material to the flask and dilute to the known volume.

How would you prepare 500.0 mL of a 0.15 M NaCl solution? 12.6 Molarity (Continued) How would you prepare 500.0 mL of a 0.15 M NaCl solution?

12.6 Molarity (Continued) How would you prepare 500.0 mL of a 0.15 M NaCl solution? Then add enough water to form the 500 mL of solution.

12.6 Molarity (Continued) What volume of 0.50 M NaCl stock solution is needed to form 500 mL of 0.15 M NaCl solution?

12.9 Colligative Properties of Solutions (Continued) How many moles of dissolved solute particles are present in each of the following beakers? The two solutions contain equal numbers of dissolved solute particles. Remember that the NaCl dissociates to Na+ and Cl-.

12 Cool Lab Type Solution Problem How many moles of CaF2 are there in 25.0 mL of 0.350 M CaF2(aq)?

12 Cool Lab Type Solution Problem How many moles of CaF2 are there in 25.0 mL of 0.350 M CaF2(aq)?

12 Cool Lab Type Solution Problem What volume of 0.350 M CaF2 solution contains 0.00875 mole of CaF2?

12 Cool Lab Type Solution Problem What volume of 0.350 M CaF2 solution is required to obtain 0.00875 mole of CaF2?

12 Cool Lab Type Solution Problem How would you prepare 9.70 g of PbCl2(s) from a 0.100 M solution of Pb(NO3)2 and a 0.200 M solution of CaCl2?

12 Cool Lab Type Solution Problem How would you prepare 9.70 g of PbCl2(s) from a 0.100 M solution of Pb(NO3)2 and a 0.200 M solution of CaCl2? Step 1: Write the balanced chemical equation: Pb(NO3)2(aq) + CaCl2(aq)  PbCl2(s) + Ca(NO3)2(aq)

12 Cool Lab Type Solution Problem How would you prepare 9.70 g of PbCl2(s) from a 0.100 M solution of Pb(NO3)2 and a 0.200 M solution of CaCl2? Step 2: Convert given product mass to moles:

12 Cool Lab Type Solution Problem How would you prepare 9.70 g of PbCl2(s) from a 0.100 M solution of Pb(NO3)2 and a 0.200 M solution of CaCl2? Step 3: Use molar ratio to determine moles of reactants required

How would you prepare 9. 70 g of PbCl2(s) from a 0 How would you prepare 9.70 g of PbCl2(s) from a 0.100 M solution of Pb(NO3)2 and a 0.200 M solution of CaCl2? Step 4: Convert from moles to desired units:

12.8 Reactions in Solution (Continued) How would you prepare 9.70 g of PbCl2(s) from a 0.100 M solution of Pb(NO3)2 and a 0.200 M solution of CaCl2? So you would combine 349 mL of the lead nitrate solution with 175 mL of the calcium chloride solution, and then use a funnel and filter paper to isolate the 9.70 g of PbCl2 that forms.

Final EXAM! Use notes, old exams and text problems! End of Chapter 12 Exam 3 = chapters 8,9 and 12 Studying… Pick up an objective sheet Make sure you know how to write formulas! REALLY! Do text problems that go along with the notes Complete lab report Final EXAM! Use notes, old exams and text problems!

12.9 Colligative Properties of Solutions (Continued) Calculating changes in freezing point and boiling point: Kf = freezing-point constant Kb = boiling-point constant

12.2 Energy and the Formation of Solutions (Continued) Hydrogen bonding Leads to strong attractions called hydrogen bonds

12.2 Formation of Solutions Dissolution of an ionic salt Water allows for: Solute-separation—freeing of ions from the crystal lattice of the solute Solvent-separation—breaking apart of the water molecules Solvation—moving of the ions into spaces in the solvent All occurs at once as the salt dissolves.

12.2 Energy and the Formation of Solutions (Continued) Solvent-separation step Making room in the solvent for the ions

12.2 Energy and the Formation of Solutions (Continued) Solvation step Formation of attractive forces between solvent particles and the solute particles

12.2 Energy and the Formation of Solutions (Continued) Hydration The surrounding of solute ions by solvent molecules Solvation when water is the solvent

12.2 Energy and the Formation of Solutions A solute dissolves when: the energy released during solvation is larger than the energy needed in the first two steps

12.2 Energy and the Formation of Solutions (Continued) Consider the ionic compound magnesium chloride, MgCl2. Do you think the hydration energy for this compound is greater than, less than, or about equal to that of NaCl?

12.2 Energy and the Formation of Solutions (Continued) Consider the ionic compound magnesium chloride, MgCl2. Do you think the hydration energy for this compound is greater than, less than, or about equal to that of NaCl? Since in MgCl2 the Mg has a +2 charge where Na has a +1 charge, the hydration energy released for MgCl2 would be larger.

12.2 Energy and the Formation of Solutions (Continued) System Combination of solute and solvent Examining energy changes will decide if a solute dissolves E is symbol for energy changes If overall E is positive, the solute will not dissolve. If overall E is negative, the solute will dissolve.

12.2 Energy and the Formation of Solutions (Continued) For a given solute in water, the energy changes are Esolute separation = 835 kJ, Esolvent separation = 98 kJ, and Esolvation = 805 kJ. Will this solute dissolve in water? Explain your answer.

12.2 Energy and the Formation of Solutions (Continued) For a given solute in water, the energy changes are Esolute separation = 835 kJ, Esolvent separation = 98 kJ, and Esolvation = 805 kJ. Will this solute dissolve in water? Explain your answer. Esolute separation + Esolvent separation = 933 kJ More energy is needed (933 kJ) than released (805 kJ), so the solute will likely not dissolve.

12.5 Getting Unlikes to Dissolve—Soaps and Detergents (Continued) Micelle Structure soap/detergent molecules form in water Minimize polar/nonpolar interactions between solvent and solute Nonpolar molecules dissolve into the center of the micelle.

12.6 Molarity (Continued) How many mL of a 1.500 M solution of NaCl do you need to obtain 100.0 g of NaCl?

When ions come in contact with one another in a solution 12.8 Reactions in Solution When ions come in contact with one another in a solution Diffusion—random ion movement about in solutions Solvent cage—ion surrounded by water molecules

12.8 Reactions in Solution (Continued) Stoichiometry Molarity is key for converting to and from moles. Once you know moles, you can understand stoichiometry.

12.8 Reactions in Solution (Continued)

12.9 Colligative Properties of Solutions (Continued) Dynamic equilibrium When the rate of a forward reaction is equal to the reverse process

12.9 Colligative Properties of Solutions (Continued) Vapor pressure Pressure exerted by the gas molecules of a liquid