Properties of Solutions Solution Terminology: Overview

Solutions & Colligative Properties of Solutions Solution Terminology: Overview • A solution is a homogeneous mixture. • Components of a solution: • solute, the component that is being dissolved into the solvent and the • solvent, the component that is doing dissolving. • Miscible vs. Ionization vs. Immiscible • Miscible means the solute dissolves into the solvent as “whole” unit; it does not ionize. • Immiscible means that the solute and solvent do NOT form a solution. • Solute & solvent are not compatible in polarities. • Ionization occurs when the solute dissolves in solvent forming ions. Solution Formation • DH of solution formation involves differences in attractive forces between the solute & solvent particles. • For a solution to form: 1. All solute (DHsolute or DH1) interactions (intermolecular attractions) must be broken. • NOTE: Process is endothermic. 2. Some, but not all solvent (DHsovlent or DH2) intermolecular interactions must be broken. • NOTE: Process is endothermic. 3. Solute and solvent (DHsoln. mix or DH3) interactions must be formed. • At least some of the energy to form a solution comes from making new solute–solvent interactions (attractions). • NOTE: This process is exothermic or slightly endothermic. • S DH1 + SDH2 = zero (~ even exchange of energy). • Entropy (DS) also plays a role in solution making. • A solution has greater entropy that just the solute and/or solvent particles alone. DHsoln = DHsolute + DHsolvent + DHsoln.mix Reference: www.chem.fsu.edu Solubility • Solubility is the ability of a solute to dissolve in a solvent to make a solution. • Factors that affect solubility: • Thermodynamics • Types of intermolecular attractive forces. • Environmental conditions • Temperature • Partial pressure (solute) • The ratio of volume of solvent to amount of solute. • Solubility is generally given in units of grams of solute that will dissolve per 100 grams of water (solvent). • Solubility curves are a graphical representation used to predict whether a solution is unsaturated, saturated or supersaturated for a given quantity of solvent (i.e,. usually 100 grams) at varies temperatures. Reference: sideplayer.com How to read a solubility curve Solubility Curves • On the line: • Saturated solution • Above the line: • Super saturated solution • Below the line: • Unsaturated solution Reference: opentextbc.ca Solubility Curves Problem: Is the solution saturated, super saturated or unsaturated? a. 150.0 g KNO3 in 100.0 g H2O at 50oC b. 90.0 g sugar in 50.0 g H2O at 0oC c. 40.0 g NaCl in 200.0 g H2O at 30oC Reference: opentextbc.ca Solute Solubility & Temperature Effects • A solute solubility in a solvent is DEPENDENT on the solvent’s temperature. • An INCREASE in temperature usually results in an INCREASE in solute solubility. • Example: • Sugar is more soluble in warm solvents. Solubility curves for gas solutes • The solubility of all gases, decrease as the solvent temperature increases. • A temperature increases, less solute (gas) remains in solution because the solute particles (gas) has a greater kinetic energy so it can escape (break intermolecular forces). • Examples: • Warm carbonated drinks fizzes more that cold carbonated drinks. • CO2 (g) is less soluble in warm solvents. Reference: openstax Gas Solute Solubility: The Pressure Connection • The greater the partial pressure of a solute (gas), the more soluble the gas is in the solvent. • The solubility of a gas (Cgas) is directly proportional to its partial pressure, (Pgas) above the solution. • Henry’s Law: Cgas = kHPgas • Cgas is the concentration of the gas most often in Molarity (M) • kH is called Henry’s Law Constant. • kH is unique to the solute (gas). • Henry’s Law Limitations: • Partial pressure of the gas solute is not very high. • Gas solute does not ionize in the solvent. • Gas solute doe not chemical interact with the solvent to form a new substances. • Temperature is not very low Reference: www.learner.org Problem: At 20°C, the concentration of dissolved oxygen in water is 1.38 × 10−3 mol/L with a partial pressure of 760 torr. Determine the solubility of oxygen when its partial pressure is 155 torr. Cgas = kHPgas Review of Concentration Units & More • Molarity (M) • The units of molarity (M) • Moles of solute per liter of solution (mol/L) • A mole is defined as (mass of sample (g)/molar mass of compound (g/mol)) • Molarity and ionic compound dissolution in water: K2SO4(aq) = 2K+(aq) + SO42−(aq) • A 1.0 M K2SO4 (aq) solution contains 1.0 mole of K2SO4 in each liter of solution. • So, if 1 L = 1.0 mole K2SO4, then 2 L = 2.0 moles K2SO4 • A 1.0 M K2SO4solution produces 2.0 M K+ and 1.0 M SO42− ions. • So, a 1.0 M K2SO4 produces a total of 3.0 M solution of ions; 2.0 M K+ & 1.0 M SO42− • Molarity & Molecular compound dissolution in water. • A 2.0 M solution of glucose means that there is two mole of glucose (solute) per liter of glucose solution. Other Concentration Units: The percentages The percent concentrations: • (Mass/Volume) x 100 = (w/v)% or (m/v)% • A 5.0 (w/v) % has 5.0 grams of solute in 100.0 mL of solution. • The specific units used for solute mass and solution volume may vary, depending on the solution, however the most common is grams/milliliter. • (Volume/Volume) x 100 = (v/v)% • A 70.0 (v/v)% has 70 mL of solute in 100.0 mL of solution. • 70.0 mL of solute + 30.0mL solvent = 100.0 mL of solution • (Mass/Mass) x 100 = (w/w)% or (m/m)% • A 6.5 (w/w)% has 6.5 grams of solute in 100.0 grams of solution. • 6.5 grams solute + 93.5 grams of solvent = 100.0 grams solvent Other Concentration Units: Molality & Mole Fraction Both molality & mole fractions are concentrations which ARE NOT affected by changes in temperature. • Molality (m):Moles of solute per kilogram of solvent • molality = moles of solute kilogram of solvent • Mole fraction (Ca): Mixture of A, B & C, the mole fraction of component A (Ca)is defined as: Ca = (mole of A) (mole A + mole B + mole C) • It is a unit less unit. • Summation of all mole fractions = 1 Going from one unit to another: Know the definition of the unit Strategy: 1. Write out the concentration as a ratio. Example: molality (m) is defined as m= mole solute kg solvent 2. Convert the solute part into the required unit. 3. Convert the solution part into the required unit. 4. Use the concentration ratio to calculate the new concentration unit. 5. Density is the link from going from mass to volume or volume to mass. 6. Remember that a solution is composed of a solvent & solute. Problem: Determine the molality, molarity, mole fraction of the solute, volume percent (v/v%) and mass percent (w/w %) for a 250.00 mL solution of HNO3 in water having a concentration of 1.50% (w/v). Useful information: Density of solution 0.9885 g/cm3 Density of concentrated HNO3 1.510 g/cm3 Colligative Properties: Physical Properties of Solutions • The addition of a solute to a pure solvent alters the physical properties of the solvent. • Example: addition of NaOH to water changes the pH from neutral to basic. • A solution’s colligative properties depends ONLY on the number of solute particles per solvent particles in solution and NOT on the solute type. • Colligative properties are about the physical properties of solutions (e.g., boiling points, melting points, vapor pressure). • Colligative Properties Raoulot's law Melting point depression Boiling point elevation Effect on the pure solvent Vapor pressure decreases temperature decreases temperature increases Concentration Units Associated with Colligative Properties • A solution’s colligative properties is depending only on the number of solute particles per solvent particles in solution. • An ideal solution is a solution whose physical properties depend only on the concentration of solute. • Concentration units that meet this criteria: • Molality • Mole fraction • The concentration unit molarity and the percentages are dependent the temperature of the solution. • Volume of a liquid is influence by temperature. • Density of liquid effected by temperature. Colligative Properties of Solutions • The colligative properties of solutions (e.g., vapor pressure, boiling point and freezing point) are dependent on the number of solute particles. • Ionic substances, such as NaCl, goes in solution it produces 2 moles of ions: Na+ & Cl• Molecular substances, such as glucose (C6H12O6) go into solution they dissolve as one non charged piece. • The van’t Hoff factor (i) is the ratio of moles of solute particles to moles of formula units dissolved. • Experimentally measured van’t Hoff factors are often lower than you might expect due to ion pairing in solution. • Ions become more widely separated the more dilute a solution, thus the residual interionic attractions become less. • Compound Theoretical Actual measured Value of i Value of i glucose 1 1 NaCl (Na+ & Cl-) 2 1.9 FeCl3 (Fe3+ & 3 Cl-) 4 3.4 Raoulot's Law is about Vapor Pressure of Solution • Raoulot's Law describes the relationship between the vapor pressures of the solution components and the concentrations of those components • The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. Psolution = Csolvent x Posolvent • The vapor pressure of a solution will always be lower than that of the pure solvent. The presence of nonvolatile solutes lowers the vapor pressure of a solution by impeding the evaporation of solvent molecules. Reference: openstax Problem: 1. Calculate the vapor pressure of an ideal solution which contains 49.1 g of glycerin, C3H5(OH)3, dissolved in 184.4 g of ethanol,C2H5OH. 2. Indicate the type of intermolecular forces present in the solution. Psolution = Csolvent x Posolvent Useful Information: Temperature is 40.0 oC Vapor pressure of ethanol is 0.178 atm at 40.0 oC. Boiling Point Elevation & Freezing Point Depression • Boiling point elevation: the boiling point of a solution will be greater compared to the boiling point of the pure solvent. • ∆TBP = kBP · m • Freezing point depression: the freezing point of a solution will less than the freezing point of the pure solvent. • ∆TFP = kFP · m Reference: elshareitforfree.blogspot.co.za Problem: • Determine the freezing point & boiling point of a solution of nitrobenzene dissolved in benzene. • The molal of the solution is 0.33 • Benzene boiling point is 80.3oC & its kBP is 2.53 oC/m • Benzene freezing point is 5.50 oC & its kFP is 5.12 oC/m Problem: 1. Determine the freezing point for seawater, given that 4.20 g NaCl is dissolved in 125.00 g of water at 25 oC. The kf for water is 1.86 oC/m. 2. Calculate the (w/v)% of seawater at 25 oC with a density of 1.030g/cm3 Osmosis and Osmotic Pressure Solution Chemistry Reference: openstak • Osmosis is an entropy diffusion-driven process involving the transfer of solvent molecules across a semipermeable membrane. • Osmosis results in the transfer of solvent molecules from a sample of low (or zero) solute concentration to a sample of higher solute concentration. • Osmosis is liquid property, but it is not a colligative property of matter. Osmotic Pressure: P = MRT • Osmotic pressure is a counterbalance pressure. • It is the minimum pressure required to prevent the flow of solvent molecules across a semipermeable membrane. • Applying a pressure greater than the osmotic pressure of a solution will reverse osmosis. • Solvent molecules from the solution are pushed into the pure solvent. Reference: openstax Problem: Determine the osmotic pressure of an aqueous solution when 1.64 g of calcium nitrate is dissolved in 275.0 mL of water at a temperature of 25.0 oC. R= 0.0821 L atm/K mol Problem: Determine the molecular mass of insulin given the following information: 7.00 g of insulin is dissolved in 1.00 liter of water at a pressure of 23.0 torr and a temperature of 25.0 oC. R= 0.0821 L atm/K mol Osmosis & Everyday Applications • Biological systems: • blood cell membranes Reference: openstax Hypotonic: Swelling/lysis Isotonic: Normal Hypertonic: Shrivel/crenation • Industrial process: • desalination Reference: chemistryhall.com