Announcements & Agenda (02/16/07)

Announcements& Agenda(02/16/07) • Please pick up exams if you haven’t. • Mon: Movie in VDW 102– You must attend & complete a worksheet! • Today • Solution Concentrations (7.4, 7.5) • Solution Properties & Some Other Mixtures (7.7) • Osmosis (7.7)

30 Good Practice Problems (Ch 7) 7.01, 7.03, 7.11, 7.13, 7.15, 7.17, 7.25, 7.27, 7.29, 7.31, 7.35, 7.37, 7.41, 7.45, 7.49, 7.53, 7.55, 7.61, 7.63, 7.65, 7.69, 7.71, 7.75, 7.77, 7.83, 7.87, 7.89, 7.93, 7.97, 7.99

Last Time: Nature of Solutions • Solution – homogeneous mixture of two or more substances • Solvent – the main medium, present in the largest quantity • Solute – material dissolved in the solvent • They don’t just have to be liquids:

Last Time: “Like Dissolves Like” • What does this mean? • Answer: compatibility of intermolecular forces (last week) • Typically, compounds with similar polarity will be soluble in each other. • Predicting trends based on properties…

Which of the following vitamins are fat-soluble and which are water soluble? Vitamins: organic molecules required for proper function but not made by the body. (Obtain by eating!) 2 categories: • fat-soluble (dissolve in fatty hydrocarbon-like tissues) • water-soluble Why we need at least some oil & fat in our diet!

Last Time: Solutes & Ionic Charge In water, • strong electrolytes produce ions and conduct an electric current. • weak electrolytes produce a few ions. • nonelectrolytes do not produce ions.

Solubility & Concentration (7.3-7.4) • the maximum amount of solute that dissolves in a specific amount of solvent. • expressed as grams of solute in 100 grams of solvent water. g of solute 100 g water Unsaturated vs. Saturated Solutions

Learning Check At 40C, the solubility of KBr is 80 g/100 g H2O. Identify the following solutions as either 1) saturated or 2) unsaturated. Explain. A. 60 g KBr added to 100 g of water at 40C. B. 200 g KBr added to 200 g of water at 40C. C. 25 g KBr added to 50 g of water at 40C. What happens to the “leftover” solute???

Effect of Temperature on Solubility • Depends on Temp! • Solids: usually increases as temperature inc. • Gases: usually decreases as temperature inc.

Solubility and Pressure Henry’s Law: • Gas solubility is directly related to gas pressure above the liquid • at higher pressures, more gas molecules dissolve in the liquid. Real life examples… soda, the bends, etc.

Percent Concentration (7.4) The amount of solute dissolved in a specific amount of solution. amount of solute amount of solution Comes in all sorts of fantastic flavors! • Mass Percent • Volume Percent • Mass/Volume Percent • Molarity

Mass Percent (% m/m) • concentration is the percent by mass of solute in a solution. mass percent = g of solute x 100 g of solute + g of solvent • amount in g of solute in 100 g of solution. mass percent = g of solute x 100 100 g of solution

Mass of Solution Add water to give 50.00 g solution 8.00 g KCl 50.00 g KCl solution

Calculating Mass Percent Calculation of mass percent (% m/m) requires the • grams of solute (g KCl) and • grams of solution (g KCl solution). g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute) x 100 = 16.0% (m/m) 50.00 g KCl solution

Volume Percent (% v/v) • percent volume (mL) of solute (liquid) to volume (mL) of solution. volume % (v/v) = mL of solute x 100 mL of solution • solute (mL) in 100 mL of solution. volume % (v/v) = mL of solute 100 mL of solution

Mass/Volume Percent (% m/v) • percent mass (g) of solute to volume (mL) of solution. mass/volume % (m/v) = g of solute x 100 mL of solution • solute (g) in 100 mL of solution. mass/volume % (m/v) = g of solute x 100 100 mL of solution

Percent Conversion Factors Two conversion factors can be written for each type of % value. TABLE 7.7 5% (m/v) glucose There are 5 g of glucose 5 g glucose and 100 mL solution in 100 mLof solution. 100 mL solution 5 g glucose

How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution? 1) 10.5 g NaOH 2) 75.0 g NaOH 3) 536 g NaOH

Solution 1) 10.5 g NaOH 75.0 g solution x 14.0 g NaOH = 10.5 g NaOH 100 g solution 14.0% (m/m) factor

How many milliliters of a 5.75% (v/v) ethanol solution can be prepared from 2.25 mL ethanol? 1) 2.56 mL 2) 12.9 mL 3) 39.1 mL

Solution 3) 39.1 mL 2.25 mL ethanol x 100 mL solution 5.75 mL ethanol 5.75% (v/v) inverted = 39.1 mL solution

Molarity and Dilution (7.5) Molarity (M) is • a chemist’s concentration term for solutions. • gives the moles of solute in 1 L solution. • moles of solute liter of solution

Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared • by weighing out 58.5 g NaCl (1.00 mole) and • adding water to make 1.00 liter of solution.

Dilution In a dilution • water is added. • volume increases. • concentration decreases.

Comparing Initial & Diluted Solutions In the initial and diluted solution, • the amount of solute is the same. • the concentrations and volumes are related by the following equations: For percent concentration: C1V1 = C2V2 initial diluted For molarity: M1V1 = M2V2 initial diluted

Summary: Properties of Solutions • contain small particles (ions or molecules). • are transparent. • do not separate*. • cannot be filtered. • do not scatter light.

Colloids • have medium size particles. • cannot be filtered. • can be separated by semipermeable membranes. • scatter light (Tyndall effect).

Examples of Colloids • Fog • Whipped cream • Milk • Cheese • Blood plasma • Pearls

Suspensions • have very large particles. • settle out. • can be filtered. • must be stirred to stay suspended. Examples include blood platelets, muddy water, and lotions.

Osmosis • water (solvent) flows from the lower solute concentration into the higher solute concentration. • the level of the solution with the higher concentration rises. • the concentrations of the two solutions become equal with time.

Osmosis Suppose a semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens? semi-permeable membrane 4% starch 10% starch H2O

Water Flow Equalizes • The 10% starch solution is diluted by the flow of water out of the 4% and its volume increases. • The 4% solution loses water and its volume decreases. • Eventually, the water flow between the two becomes equal. 7% starch 7% starch H2O

Osmotic Pressure • produced by the solute particles dissolved in a solution. • equal to the pressure that would prevent the flow of additional water into the more concentrated solution. • greater as the number of dissolved particles in the solution increases.

Osmotic Pressure of the Blood Red blood cells • have cell walls that are semipermeable membranes. • maintain an osmotic pressure that cannot change or damage occurs. • must maintain an equal flow of water between the red blood cell and its surrounding environment.

Isotonic Solutions • exerts the same osmotic pressure as red blood cells. • is known as a “physiological solution”. • of 5.0% glucose or 0.90% NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells. H2O

Hypotonic Solutions • has a lower osmotic pressure than red blood cells. • has a lower concentration than physiological solutions. • causes water to flow into red blood cells. • causes hemolysis: RBCs swell and may burst. H2O

Hypertonic Solutions • has a higher osmotic pressure than RBCs. • has a higher concentration than physiological solutions. • causes water to flow out of RBCs. • cause crenation: RBCs shrinks in size. H2O

Dialysis In dialysis, • solvent and small solute particles pass through an artificial membrane. • large particles are retained inside. • waste particles such as urea from blood are removed using hemodialysis (artificial kidney).

Announcements & Agenda (02/16/07)