Chapter 16 Solutions

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 l For gases in a liquid, as the temperature goes up-the solubility goes down l For gases in a liquid, as the pressure goes up-the solubility goes up

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 =

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?