Unit 8- Solutions Aqueous Boiling point Colligative property Concentrated Dilute Molarity Parts per million (ppm) Percent by volume Percent mass Saturated Solubility Solute Solution Solvent Supersaturated Unsaturated Vapor Vapor pressure
Mixtures Homogeneous Heterogeneous Solutions Can be in all phases of matter (s, l or g) Solids- metals mixed form an alloy. Brass= Zn + Cu Gases- ex: air Liquids- usually a solid or liquid dissolved in a liquid Heterogeneous Temporary mixtures are suspensions Ex: muddy water Stable mixtures are colloids Ex: milk Particles have charge, they repel each other so they can’t clump and settle out
Characteristics and parts of solutions Homogeneous mixture Are clear and don’t disperse light Can have color Will not settle out Will pass through a filter Parts Solute- substance that is being dissolved present in a smaller amount Solvent- substance that dissolves solute Present in greater amount Water- most common solvent Water solutions = aqueous (aq)
Separating mixtures Physically- decanting, centrifuge, filtration, evaporation By solubility- chromatography By boiling point- distillation
Solubility How much solute will dissolve in a certain amount of solvent at a certain temperature Materials with high solubility are soluble Materials with low solubility are insoluble “like dissolves like” Liquids that are soluble in each other are miscible If they aren’t they are immiscible
Factors that affect solubility “like dissolves like”; degree of solubility (how much dissolves) Soap* - long nonpolar chain with polar end Grease= nonpolar, water= polar (soap connects the two) Ionic substances- dissociate- ions separate Takes E, entropy (randomness) increases as ions dissociate and then decreases as water molecules assemble around the ions Solute type Non polar solvent Polar solvent Nonpolar Soluble Insoluble Polar Ionic soluble http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html
Soap P O- CH3 CH2 7
Soap P O- CH3 CH2 Hydrophobic non-polar end 8
Soap P O- CH3 CH2 Hydrophilic polar end 9
P O- CH3 CH2 _ 10
A drop of grease in water Grease is non-polar Water is polar Soap lets you dissolve the non-polar in the polar. 11
Hydrophobic ends dissolve in grease 12
Hydrophilic ends dissolve in water 13
Water molecules can surround and dissolve grease. 14
Factors con’t Temperature Pressure As temp of water increases, solids become more soluble (they separate), gases become less soluble (they come together) Pressure Has little or no effect on solubility in solids and liquids Does effect the solubility of gases in liquids, as pressure increases, solubility of a gas in the liquid increases **greater surface area = increased dissolving process
Factor review Factor Affect on Solid Solute Affect on Gaseous Solute Particle Size Stirring Amount of dissolved solute Temperature Reducing particle size by crushing increases the rate by increasing surface area. Not applicable Increases the rate by exposing fresh solvent to solute and increasing kinetic energy. Decreases the rate by increasing kinetic energy, thereby reducing solubility. As the amount of dissolved solute increases, the rate decreases. Reducing particle size by crushing increases the rate by increasing surface area. As the temperature increases, the rate increases. As the temperature increases, the rate decreases.
Looking at solubility in the Reference Tables Table F - Shows ionic compounds Predicts whether a precipitate will form when 2 ionic compounds are mixed A Rx will take place if 1 or both of the products are listed as insoluble
Looking at solubility in the Reference Tables Table G- Shows the amount of a substance that can be dissolved in 100g of water at different temperatures Lines show the max amount of solute at a given temp
Solids- lines that increase as temp increases 3 lines- NH3, HCl, SO2- show decrease in solubility as temp increases- they are GASES
If given pt is below the line the solution is unsaturated TEMPERATURE MASS of SOLUTE SATURATED UNSATURATED SUPERSATURATED If given pt is below the line the solution is unsaturated If given pt is above the line the solution is supersaturated If given pt is on the line the solution is saturated- solubility equilibrium
The morphology diagram tells us a great deal about what kinds of snow crystals form under what conditions. For example, we see that thin plates and stars grow around -2 C (28 F), while columns and slender needles appear near -5 C (23 F). Plates and stars again form near -15 C (5 F), and a combination of plates and columns are made around -30 C (-22 F). Furthermore, we see from the diagram that snow crystals tend to form simpler shapes when the humidity (supersaturation) is low, while more complex shapes at higher humidities. The most extreme shapes -- long needles around -5C and large, thin plates around -15C -- form when the humidity is especially high. Why snow crystal shapes change so much with temperature remains something of a scientific mystery. The growth depends on exactly how water vapor molecules are incorporated into the growing ice crystal, and the physics behind this is complex and not well understood. It is the subject of current research in my lab and elsewhere. http://www.its.caltech.edu/~atomic/snowcrystals/primer/morphologydiagram.jpg
Saturation- occurs at a pt of solubility equilibrium
Saturation Test 1- Add a crystal Saturated Solution The crystal falls to the bottom. Unsaturated Solution The crystal dissolves. Supersaturated Solution Excess solute precipitates.
Interpreting Solubility Curves Which compound which is the least soluble at 10°C? How many grams of potassium nitrate needed to saturate 100 mL of water at 52°C? One hundred mL of a potassium chloride solution is saturated at 10°C. How many additional grams are needed to saturate the solution at 50°C? At what temperature do saturated solutions of sodium chloride and potassium chloride contain the same mass of solute per 100 mL of water? How many more grams of sulfur dioxide can be dissolved in 100 mL of water at 20°C than at 90°C? KClO3 90 g 12 g 37°C 7 g
Determining concentration Concentration- ratio of solute to solvent- (g/L) Dilute- solution with little solute Concentrated- solution with a lot of solute Molarity- M- moles/liter of soln M = moles of solute ÷ liters of solution *many times grams must be converted to moles before molarity is calculated For dilute solutions- ppm- parts per million ppm = Mass (Solute) Mass (Solution) × 1,000,000 ppm ______________ Percent mass = Mass (Solute) Mass (Solution) × 100% ______________ Percent volume = Volume (Solute) Volume (Solution) × 100% ______________
Concentration Problems What is the concentration of salt in a 20.0 mL solution containing 18.3 g of dissolved NaCl? Procedure: Divide the mass of the solute by the volume of the solvent or the solution. Concentration = 18.3 g 20.0 mL = 0.92 g/mL __________
Percent by Mass Problems What is the percent by mass of ethanol in a solution containing 2.3 g of ethanol (C2H5OH) dissolved in 10.0 g of water? Step 1: Find the mass of the solution 10.0 g + 2.3 g = 12.3 g Step 2: Divide the mass of the solute by the mass of the solution and multiply by 100 % Percent mass = 2.3 g 12.3 g × 100% = 19 % __________
Percent by Volume Problems What is the percent by volume of glycerine in a solution containing 18.2 mL of glycerine (C3H6O3) dissolved in 85.0 mL of water? Step 1: Find the volume of the solution 18.2 mL + 85.0 mL = 103.2 mL Step 2: Divide the volume of the solute by the volume of the solution and multiply by 100 % Percent volume = 18.2 mL 103.2 mL × 100% = 17.6 % ______________
Parts per Million Problems About 0.0035 g of hydrogen sulfide are dissolved in 10.0 g of water. Express this in parts per million. Step 1: Find the mass of the solution 10.0 g + 0.0035 g = 10.0035 g Step 2: Divide the mass of the solute by the mass of the solution and multiply by 1,000,000 ppm. ppm = 0.0035 g 10.0035 g × 1,000,000 ppm = 350 ppm ______________
Molarity Sample Problems Find the molarity of 100. mL of a solution that contains 0.25 moles of dissolved solute. Step 1: Convert all volumes to liters Step 2: Substitute values into the definitional equation 100. mL × 0.001 L ____________ 1 mL = 0.100 L M = 0.25 mol ____________ 0.100 L = 2.5 M
Sample Problem 2 Find the molarity of 250. mL of a solution that contains 4.0 g of dissolved sodium hydroxide (NaOH). Step 1: Find the GFM Na = 23 × 1 = 23 O = 16 × 1 = 16 H = 1 × 1 = 1 40 Step 2: Convert all volumes to liters Step 3: Substitute values into the correct equation 250. mL × 0.001 L ____________ 1 mL = 0.250 L
Sample Problem 3 Step 1: Convert all volumes to liters How many moles of solute are dissolved in 30 mL of a 2 M solution? Step 1: Convert all volumes to liters Step 2: Substitute values into the correct equation 30 mL × 0.001 L ____________ 1 mL = 0.03 L 0.06 mol
Sample Problem 4 How many grams of silver nitrate (AgNO3) are needed to prepare 200. mL of a 0.10 M solution? Step 1: Find the GFM Ag = 108 × 1 = 108 N = 14 × 1 = 14 O = 16 × 3 = 48 170 Step 2: Convert all volumes to liters Step 3: Substitute values into the correct equation 200. mL × 0.001 L ____________ 1 mL = 0.200 L = 3.4 g
How to prepare a solution Add desired amount of solute to volumetric flask Add some DI water and mix until dissolved Fill flask to the line on the neck, close and mix What mass of sodium bicarbonate is required to prepare 2 L of a .250M sodium bicarbonate solution? Knowns: concentration= .250M vol of soln= 2L - unknown: mass of Na2CO3 = ____g - Find moles of using molarity equation then convert moles to grams
Preparing Solutions sample problems: How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution? How many grams of CaCl2 are needed to make 625 mL of a 2.0 M solution?
Physical Properties of Solutions Conductivity- Depends on solution having charged particles that can move Electrolyte- substances that conduct electricity when melted or dissolved in water (ionic substances) Sports drinks etc. Strong electrolytes- completely dissociate Weak electrolytes- only some ions dissociate (weak conductivity) Nonelectrolyte- no ions; doesn’t conduct (molecular substances)
Colligative Properties Dependent on # of dissolved particles A physical effect of solute on solvent Degree of effect determined by conc. of dissolved particles Freezing pt depression, boiling pt elevation, lowering vapor pressure
Freezing Point depression Crystallization process in water is disrupted by solute particles Salt on sidewalks and roads Making ice-cream 1 mole of any particles will have the same effect on freezing pt of water 1 mole of particles lowers freezing pt 1.86oC of 1000g of water
Vapor Pressure The pressure that a vapor exerts Table H shows vapor pressure of liquids at different temperatures
BP elevation and VP lowering Boiling pt- when VP = atmospheric pressure As temp increases, VP increases until it reaches atmospheric pressure; then it starts to boil Table H will show at what temp water will boil at different pressures As solute is added to a solution, vapor pressure decreases; therefore it takes longer for the VP to reach the atmospheric pressure and the water boils at a higher temperature