Chapter 16 Solutions Killarney High School

Section 16.1 Properties of Solutions l OBJECTIVES: – Identify the factors that determine the rate at which a solute dissolves.

Section 16.1 Properties of Solutions l OBJECTIVES: – Calculate the solubility of a gas in a liquid under various pressure conditions.

Solution formation l Nature of the solute and the solvent – Whether a substance will dissolve – How much will dissolve l Factors determining rate of solution... – stirred or shaken (agitation) – particles are made smaller – temperature is increased l Why?

Making solutions l In order to dissolve, the solvent molecules must come in contact with the solute. l Stirring moves fresh solvent next to the solute. l The solvent touches the surface of the solute. l Smaller pieces increase the amount of surface area of the solute. Solution Formation Flash

Temperature and Solutions l Higher temperature makes the molecules of the solvent move around faster and contact the solute harder and more often. – Speeds up dissolving. l Usually increases the amount that will dissolve (exception is gases)

How Much? l Solubility- The maximum amount of substance that will dissolve at a specific temperature (g solute/100 g solvent) l Saturated solution- Contains the maximum amount of solute dissolved l Unsaturated solution- Can still dissolve more solute l Supersaturated- solution that is holding more than it theoretically can; seed crystal will make it come out

Liquids l Miscible means that two liquids can dissolve in each other – water and antifreeze, water and ethanol l Partially miscible- slightly – water and ether l Immiscible means they can’t – oil and vinegar

Solubility? l For solids in liquids, as the temperature goes up-the solubility usually goes up (Fig. 16.5, p.373) l For gases in a liquid, as the temperature goes up-the solubility goes down (Fig. 16.6, p.374) l For gases in a liquid, as the pressure goes up-the solubility goes up (Fig. 16.6, p.373)

Gases in liquids... l Henry’s Law - says the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid – think of a bottle of soda pop, removing the lid releases pres. l Equation: S 1 S 2 P 1 P 2 =

Cloud seeding l Ever heard of seeding the clouds to make them produce rain? l Clouds- mass of air supersaturated with water vapor l Silver Iodide (AgI) crystals are dusted into the cloud l The AgI attracts the water, forming droplets to attract others

Section 16.2 Concentration of Solutions l OBJECTIVES: – Solve problems involving the molarity of a solution.

Section 16.2 Concentration of Solutions l OBJECTIVES: – Describe how to prepare dilute solutions from more concentrated solutions of known molarity.

Section 16.2 Concentration of Solutions l OBJECTIVES: – Explain what is meant by percent by volume [ % (v/v) ], and percent by mass [ % (m/v) ] solutions.

Concentration is... l a measure of the amount of solute dissolved in a given quantity of solvent l A concentrated solution has a large amount of solute l A dilute solution has a small amount of solute – thus, only qualitative descriptions l But, there are ways to express solution concentration quantitatively

Molarity - most important l The number of moles of solute in 1 Liter of the solution. l M = moles/Liter; such as 6.0 molar l What is the molarity of a solution with 2.0 moles of NaCl in 250 mL of solution? l Sample 16-2, page 378

Making solutions l Pour in a small amount of solvent l Then add the solute (to dissolve it) l Carefully fill to final volume. – Fig , page 509 l Also: M x L = moles l How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution?

Making solutions l 10.3 g of NaCl are dissolved in a small amount of water, then diluted to 250 mL. What is the concentration? l How many grams of sugar are needed to make 125 mL of a 0.50 M C 6 H 12 O 6 solution?

Dilution Adding water to a solution

Dilution l The number of moles of solute doesn’t change if you add more solvent! l The # moles before = the # moles after l M 1 x V 1 = M 2 x V 2 l M 1 and V 1 are the starting concentration and volume. l M 2 and V 2 are the final concentration and volume. l Stock solutions are pre-made to known Molarity

Practice l 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? l You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make? l Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. How do you make the required solution?

Percent solutions... l Percent means parts per 100, so l Percent by volume: = Volume of solute x 100% Volume of solution l indicated %(v/v) l What is the percent solution if 25 mL of CH 3 OH is diluted to 150 mL with water?

Percent solutions l Percent by mass: = Mass of solute(g) x 100% Volume of solution(mL) l Indicated %(m/v) l More commonly used l 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution? l How many grams of salt are there in 52 mL of a 6.3 % solution?

Section 16.3 Colligative Properties of Solutions l OBJECTIVES: – Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution.

Section 16.3 Colligative Properties of Solutions l OBJECTIVES: – Explain on a particle basis why a solution has an elevated boiling point, and a depressed freezing point compared with the pure solvent.

Colligative Properties Depend only on the number of dissolved particles Not on what kind of particle

Vapor Pressure decreased l The bonds between molecules keep molecules from escaping. l In a solution, some of the solvent is busy keeping the solute dissolved. l Lowers the vapor pressure l Electrolytes form ions when they are dissolved = more pieces. NaCl  Na + + Cl - (= 2 pieces) l More pieces = bigger effect

Boiling Point Elevation l The vapor pressure determines the boiling point. l Lower vapor pressure = higher boiling point. l Salt water boils above 100ºC l The number of dissolved particles determines how much, as well as the solvent itself.

Freezing Point Depression l Solids form when molecules make an orderly pattern. l The solute molecules break up the orderly pattern. l Makes the freezing point lower. l Salt water freezes below 0ºC l How much depends on the number of solute particles dissolved. Simulation

Section 16.4 Calculations Involving Colligative Properties l OBJECTIVES: – Calculate the molality and mole fraction of a solution.

Section 16.4 Calculations Involving Colligative Properties l OBJECTIVES: – Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound.

Molality l a new unit for concentration l m = Moles of solute kilogram of solvent l m = Moles of solute 1000 g of solvent l What is the molality of a solution with 9.3 mole of NaCl in 450 g of water?

Why molality? l The size of the change in boiling point is determined by the molality.  T b = K b x m x n  T b is the change in the boiling point l K b is a constant determined by the solvent (Table 16.2, page 387). l m is the molality of the solution. l n is the number of pieces it falls into when it dissolves.

What about Freezing? l The size of the change in freezing point is also determined by molality.  T f = -K f x m x n  T f is the change in freezing point l K f is a constant determined by the solvent (Table 16.3, page 388). l m is the molality of the solution. l n is the number of pieces it falls into when it dissolves.

Problems l What is the boiling point of a solution made by dissolving 1.20 moles of NaCl in 750 g of water? l What is the freezing point? l What is the boiling point of a solution made by dissolving 1.20 moles of CaCl 2 in 750 g of water? l What is the freezing point?

Mole fraction l This is another way to express concentration l It is the ratio of moles of solute to total number of moles of solute + solvent (Fig , p.386) n a n a + n b X = Sample 18-8, page 521

Molar Mass l We can use changes in boiling and freezing to calculate the molar mass of a substance l Find: 1) molality 2) moles, and then 3) molar mass l Sample 16-10, page 388

A solution of 7.50g of a nonvolatile compound in g of water boils at degrees celsius at 1 atm. What is the molar mass of the solute? 1. Use the BP elevation to calculate the molality of the solution. Tb = Kb x m x n thus m = tb KbKb Kb = C x kg of water Mol of solute 0.78 C C / m m = = 1.5 m

2. Now calculate the moles of solute I the solution 1.5 mol solute Kg of water Kg0.034 mol solute 3. Use the # of moles of solute and its mass to get the molar mass Molar mass = mass of solute Moles of solute Molar mass = 7.50 g mol Molar mass = g/mol

Key Equations l Note the key equations to solve problems in this chapter.