Presentation on theme: "UNIT 9 SOLUTIONS. Overview Types of Mixtures Suspension Colloid Solution Concentration Mass percent Mole fraction Molality Molarity Dilution."— Presentation transcript:
UNIT 9 SOLUTIONS
Overview Types of Mixtures Suspension Colloid Solution Concentration Mass percent Mole fraction Molality Molarity Dilution Solution Stoichiometry Solubility “Like dissolves like” Factors affecting Henry’s Law Solubility product (intro) Common ion effect (intro) Saturation Electrolytes Colligative Properties Vapor Pressure Lowering Freezing Point Depression Boiling Point Elevation Osmotic Pressure
Vocabulary Solute Solute – substance being dissolved Solvent Solvent – the dissolving medium
Types of Mixtures Colloid Colloid – heterogeneous mixture with intermediate sized solute particles that do not settle out of the mixture Suspension Suspension – heterogeneous mixture with large solute particles that can settle out of the mixture Solution Solution – homogeneous mixture of two or more components
Colloids Heterogeneous mixture with intermediate sized solute particles that do not settle out of the mixture When filtered, particles do not separate out Particles do not separate when left to stand Colloid particles make up the dispersed phase Example When large soil particles settle out of muddy water, the water is still cloudy Emulsion and foam are used to classify colloids Emulsifying agents help keep colloid particles dispersed
Examples of Colloids Class of ColloidPhasesExample SolSolid dispersed in liquidPaints, mud GelSolid network extending throughout liquid Gelatin, jell-o Liquid emulsionLiquid dispersed in liquidMilk, mayonnaise FoamGas dispersed in liquidShaving cream, whipped cream Solid aerosolSolid dispersed in gasSmoke, auto exhaust Liquid aerosolLiquid dispersed in gasFog, mist, clouds, aerosol spray Solid emulsionLiquid dispersed in solidCheese, butter Solid foamGas dispersed in solidMarshmallow, styrofoam Solid solSolid dispersed in solidRuby glass
Tyndall Effect Colloids scatter light, making a beam visible. Solutions do not scatter light.
Suspension Heterogeneous mixture with large solute particles that can settle out of the mixture Particles are so large that they settle out unless the mixture is constantly stirred Particles can be separated from the mixture through filtration Example Muddy water
Solution In a solution… The solute can’t be filtered out The solute always stays mixed Particles are always in motion A solution will have different properties than the solvent Solvation Solvation : interactions between solute and solvent Hydration Hydration : when the solvent is water
Other Solution Examples SoluteSolventExample Gas Air GasLiquidSoda Liquid Alcohol in water SolidLiquidSalt in water GasSolidHydrogen in palladium LiquidSolidMercury in silver Solid Silver in gold
Comparison ColloidSuspensionSolution Heterogeneous Homogeneous Particle size = nm, dispersed; can be aggregates or small molecules Particle size = over 1000 nm, suspended; can be large particles or aggregates Particle size = nm, can be atoms, ions, molecules Do not separate on standing Particles settle outDo not separate on standing Cannot be separated by filtration Can be separated by filtration Cannot be separated by filtration Scatter light (Tyndall effect) May scatter light but are not transparent Do not scatter light
Concentration Concentrated solution – strong solution Dilute solution – “Watered-down” solution Low concentrations Pollutants often found in air and water are typically found at very low concentrations. Two common units are used to express these trace amounts. Parts per million (ppm) = Parts per million (ppm) = (volume solute/volume solution) × 10 6 Parts per billion (ppb) = Parts per billion (ppb) = (volume solute/volume solution) × 10 9 Example. Example. One cm 3 of SO 2 in one m 3 of air would be expressed as 1 ppm or 1000 ppb.
Solution Concentration Mass percent: Mass percent: the ratio of mass units of solute to mass units of solution, expressed as a percent Mole Fraction (X): Mole Fraction (X): the ratio of moles of solute to total moles of solution
Solution Concentration Molality (m): Molality (m): moles of solute per kilogram of solvent
Assumptions Assume that solutions with water as the solvent have the density of pure water (1 mL = 1 gram) 1 ml of water = 1 gram of water 1000 ml of water = 1 liter = 1000 grams
Solution Concentration Molarity (M): Molarity (M): the ratio of moles of solute to liters of solution Recognizes that compounds have different formula weights A 1 M solution of glucose contains the same number of molecules as 1 M ethanol. [ ] - special symbol which means molar (mol/L ) M = moles solutemol liters of solution L =
Molarity Examples Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH. M NaOH =10 mol NaOH / 2.0 L =5.0 M What’s the molarity of a solution that has g HCl in 2.0 liters? First, you need the FM of HCl. Mass HCl = g/mol Next, find the number of moles. moles HCl = g HCl / g/mol = 0.50 mol Finally, divide by the volume. M HCl = 0.50 mol / 2.0 L = 0.25 M
Solution Preparation Solutions are typically prepared by: 1. Dissolving the proper amount of solute and diluting to volume. 2. Dilution of a concentrated solution.
Dissolving Solute in Proper Volume How do you prepare ml of a M solution of sodium chloride. First, you need to know how many moles of NaCl are in ml of a 0.5 M solution. M = mol/liters and mL = liters 0.5 M = (X mol)/( liters) X = mols NaCl Next, we need to know how many grams of NaCl to weigh out mol NaCl ÷ mol = grams Finally, you’re ready to make the solution. Weigh out exactly grams of dry, pure NaCl and transfer it to a volumetric flask. Fill the flask about 1/3 of the way with pure water and gently swirl until the salt dissolves. Now, dilute exactly to the mark, cap and mix.
Diluting Solutions to Proper Concentration Once you have a solution, it can be diluted by adding more solvent. This is also important for materials only available as solutions M 1 V 1 = M 2 V 2 1 = initial2 = final Any volume or concentration unit can be used as long as you use the same units on both sides of the equation.
Diluting Solutions to Proper Concentration What is the concentration of a solution produced by diluting ml of 1.5 M NaOH to liters? M 1 V 1 = M 2 V 2 M 1 = 1.5 MM 2 = ??? V 1 = mlV 2 = 2000 ml (1.5)(100.0) = (M 2 )(2000) M 2 = M
Solution Stoichiometry Extension of earlier stoichiometry problems. Extension of earlier stoichiometry problems. First step is to determine the number of moles based on solution concentration and volume. Final step is to convert back to volume or concentration as required by the problem. You still need a balanced equation and must use the coefficients for working the problem.
Stoichiometry Example Determine the volume of M HCl that must be added to completely react with 250 ml of 2.50 M NaOH HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (l) We have 250 ml of a 2.50 M solution 2.50 mol/L = (mol NaOH )(0.250 L) mol NaOH = mol NaOH = × 1 mol HCl /1 mol NaOH = mol HCl
Stoichiometry Example (cont…) Now we can determine what volume of our M HCl solution is required. M = mol HCl /liter HCl = (0.625 mol)/(X liters) = 6.26 L of HCl
Solubility of Solutions “Like” dissolves “like” Nonpolar solutes dissolve best in nonpolar solvents Examples Fats in benzene, steroids in hexane, waxes in toluene Polar and ionic solutes dissolve best in polar solvents Examples Inorganic salts in water, sugars in small alcohols Miscible Miscible – pairs of liquids that dissolve in one another Immiscible Immiscible – pairs of liquids that do not dissolve in one another
Factors Affecting Dissolution Factors that increase the rate of dissolving a solute Increasing surface area of solute More contact surfaces for solvent to dissolve solute Agitating (stirring/shaking) solution Increase surfaces of solute exposed to solvent Heating the solvent Solvent molecules move faster Average kinetic energy increases At higher temperatures, collisions between solvent molecules and solute are more frequent
Henry’s Law Solubility of a gas is directly proportional to the partial pressure of that gas on the surface of the liquid In carbonated beverages, solubility of CO2 is increased by increasing the pressure Carbonation escapes when the pressure is reduced to atmospheric pressure so CO2 escapes Effervescence – the rapid escape of gas from a liquid in which it is dissolved S g = kP g S g = solubility of gas K = Henry’s law constant (table p 536) P g = partial pressure of gas over solution
Effects on Solubility of Gases To increase the solubility of gases Increase pressure Gas is forced into solution under pressure Soda goes “flat” when exposed to atmospheric pressure Decrease temperature Increased temperature means increased kinetic energy With increased kinetic energy, more gas molecules can escape from the liquid Soda goes “flat” faster in warm temperatures than cold
Solubility Product Salts are considered “soluble” if more than 1 gram can be dissolved in 100 mL of water. Salts that are “slightly soluble” and “insoluble” still dissociate to some extent Solubility product (K sp ) Solubility product (K sp ) is the extent to which a salt dissociates in solution Greater the value of K sp, the more soluble the salt
Solubility Product For the reaction:A a B b (s) ↔ aA(aq) + bB(aq) Solubility product:K sp = [A] a [B] b [ ] = concentration Example: CaF 2 (s) ↔ Ca +2 (aq) + 2F - (aq) K sp = [Ca +2 ][F - ] 2
Ksp Values for Salts at 25°C
Common Ion Effect AgCl(s) is added to water The Ag + ions and the Cl - ions completely dissociate NaCl(aq) is mixed with AgCl(aq) The Na + ions do not have an effect on the Ag + or the Cl - ions There are already Cl - ions in solution The addition of a “common ion” can cause some of the salt to precipitate out of solution The salt with the smaller K sp value will precipitate first
Saturation of Solutions Supersaturated solution Supersaturated solution – contains more solute than can dissolve in the solvent Saturated solution Saturated solution – a solution that cannot dissolve any more solute under the given conditions Unsaturated solution Unsaturated solution – a solution that is able to dissolve additional solute
Electrolytes Electrolyte Electrolyte – solution that conducts electricity ionic compounds in polar solvents dissociate (break apart) in solution to make ions may be strong (100% dissociation) or weak (less than 100%) Strong Electrolyte – all or almost all of compound dissociates Example: strong acids (H2SO4, HNO3, HClO4, HCl, HBr, HI) Weak Electrolyte – Small amount of compound dissociates Example – weak acids (HF, acetic acid) Nonelectrolyte Nonelectrolyte – solution that does not conduct electricity solute is dispersed but does not dissociate Example: sugar (dissolves but does not dissociate), organic acids (contain carboxyl groups)
Concentration of Ions Strong electrolytes – ions completely dissociate Concentration of ions Relative concentration of ions depends on chemical formula Example 1: In a solution of 0.1 M NaCl Concentration of Na + = 0.1 M Concentraion of Cl - = 0.1 M Example 2: In a solution of 0.1 M Na2SO4 Concentration of Na + = 0.1 M × 2 = 0.2 M (account for 2 sodium ions) Concentraion of SO4 -2 = 0.1 M
Colligative Properties Properties that depend on the concentration of solute particles but not on their identity Vapor pressure lowering Freezing point depression Boiling point elevation Osmotic Pressure
Vapor Pressure Lowering When a solute is added to a solution, the vapor pressure of the solution decreases (Raoult’s Law) Increased number of solute particles means fewer water molecules Fewer molecules exist to escape from the liquid Molecules have lower tendency to leave the liquid phase
Vapor Pressure Lowering Raoult’s Law: P = XP° P = vapor pressure of the solution P° = vapor pressure of the pure solvent X = mole fraction of the solvent
Vapor Pressure Lowering (Example) Calculate the vapor pressure of the solution when 235 grams of sugar (C12H22O11) are added to 650 mL of water at 40°C. The vapor pressure of water at 40°C is 55 mm Hg. 235 grams C12H22O11 = mol C12H22O mL H2O= 650 g H2O= mol H2O X= /( ) = P = XP° P = (0.981)(55mmHg) = 54.0 mmHg
Freezing Point Depression When solute is added to a solution, freezing point decreases Particles in first beaker can easily shift into solid phase from liquid phase at normal freezing temperature Particles in second beaker are blocked by solute particles and cannot as easily move into the solid phase from the liquid phase so a colder temperature is needed to shift into the solid phase
Freezing Point Depression ∆t f = iK f m ∆t f = change in freezing temperature i = van’t Hoff factor (number of particles into which the added solute dissociates) K f = molal freezing-point depression constant m = molality
Boiling Point Elevation When a solute is added to a solution, the boiling point of the substance increases Fewer water molecules exist due to addition of solute particles Fewer particles leave liquid phase Higher temperature needed to increase kinetic energy and allow liquid particles to escape
Boiling Point Elevation ∆t b = iK b m ∆t b = change in boiling temperature i = van’t Hoff factor (number of particles into which the added solute dissociates) K b = molal boiling-point elevation constant m = molality
Osmotic Pressure External pressure that must be applied to stop osmosis Osmosis – movement of solvent through a semi-permeable membrane (allows passage of some particles but not others) from the side of lower solute concentration to the side of higher solute concentration Semi-permeable membrane - allows passage of some particles but not others
Osmotic Pressure Semi-permeable membrane allows passage of water molecules but not solute particles Solute particles prevent water molecules from striking surface of semi- permeable membrane On pure water side, water molecules free to hit surface
Osmotic Pressure (cont…) Rate of water molecules leaving pure water side is greater than rate of water molecules leaving solution side Height of solution rises until pressure exerted by the height of solution is large enough to forces water molecules back through the membrane at equal rate to which they enter the solution side
Osmotic Pressure П = RTi = MRTi λ = osmotic pressure (atm) n = moles of solute R = gas constant T = temperature (K) V = volume of solution i = van’t Hoff factor (number of particles into which the added solute dissociates) M = molarity of solution
Van’t Hoff Factor Salt lowers the freezing point of water nearly twice as much as sugar does. Why is this? Nonelectrolyte solutions – particles do not dissociate into ions Strong electrolyte solutions – particles completely dissociate into ions Sugar is a nonelectrolyte so sugar dissolves to produce only one particle in solution. Salt is a strong electrolyte so it dissociates completely to produce two ions in solution (Na + and Cl - ). Salt has twice as many particles to lower the freezing point as sugar. This is accounted for with the van’t Hoff factor (i) What is the van’t Hoff factor for Ba(NO3)2? _________
Forces of Attraction A 0.1 m solution of KCl lowers the freezing point more than a 0.1 m solution of MgSO4. Why is that? Forces of attraction (charges of ions) interfere with movements of aqueous ions Only in very dilute solutions is the forces of attraction between ions negligible Ions of higher charge attract each other more strongly and cluster together more acting as a single unit more than multiple particles MgSO4 has ions of +2 and -2. Ions are more attracted to each other more than in KCl (ions are +1 and -1). MgSO4 particles cluster together more than KCl particles so they have less effect on colligative properties.