Week 4 Properties of Solution PDF

A solution is a A colloid is a homogeneous heterogeneous mixture mixture and exists and exists as two or as a single phase. more phases, which The particles in a solution are may be visibly distinct. individual atoms, ion...

A solution is a A colloid is a homogeneous heterogeneous mixture mixture and exists and exists as two or as a single phase. more phases, which The particles in a solution are may be visibly distinct. individual atoms, ions, or small The particles in a colloid are typically molecules. macromolecules or aggregations of small molecules.  Mass % Number Number of Substance of Cell of Types Molecules Water ~70 1 5x1010 Ions 1 20 ? Sugars* 3 200 3x108 Amino acids* 0.4 100 5x107 Lipids* 2 50 3x107 Nucleotides* 0.4 200 1x107 Other small molecules 0.2 ~200 ? Macromolecules 23 ~5000 6x106 (proteins, nucleic acids, polysaccharides). *Includes precursors and metabolites. A solute dissolves in a solvent The solubility (S) of a solute is to form a solution. the maximum amount that Usually, the solvent is the most dissolves in a fixed quantity of abundant component. solvent at a given temperature. Substances that exhibit similar types of intermolecular force dissolve in each other. This is often expressed by saying “like dissolves in like.”  Figure 13.1 Types of intermolecular forces in solutions. Ion-dipole H bond Dipole-dipole (40-600) (10-40) (5-25) Ion-induced dipole Dipole-induced dipole Dispersion (3-15) (2-10) (0.05-40) When a solution forms, solute-solute attractions and solvent-solvent attractions are replaced by solute-solvent attractions. This can only occur if the forces within the solute and solvent are similar to the forces that replace them.  Figure 13.2 Hydration shells around an Na+ ion Ion-dipole forces orient water molecules around an ion. In the innermost shell here, six water molecules surround the cation octahedrally. Alcohols are organic compounds that have dual polarity. - The general formula for an alcohol is CH3(CH2)nOH. The –OH group of an alcohol is polar. - It interacts with water through H bonds and - with hexane through weak dipole-induced dipole forces. The hydrocarbon portion is nonpolar. - It interacts through weak dipole-induced dipole forces with water - and by dispersion forces with hexane.  Solubility in Solubility in Alcohol Model Water Hexane CH3OH (methanol) ∞ 1.2 CH3CH2OH (ethanol) ∞ ∞ CH3(CH2)2OH ∞ ∞ (propanol) (CH3(CH2)3OH (1- 1.1 ∞ butanol) (CH3(CH2)4OH (1- 0.30 ∞ pentanol) (CH3(CH2)5OH (1- 0.058 ∞ hexanol) *Expressed in mol alcohol/1000 g solvent at 20°C  Figure 13.3 Like dissolves like: solubility of methanol in water. H2O Methanol CH3OH Both H2O and methanol have H bonds between their molecules. A solution of water and methanol. PROBLEM: Predict which solvent will dissolve more of the given solute. (a) Sodium chloride in methanol (CH3OH) or in 1-propanol (CH3CH2CH2OH). (b) Ethylene glycol (HOCH2CH2OH) in hexane (CH3CH2CH2CH2CH2CH3) or in water. (c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH). PLAN: We examine the formulas of solute and solvent to determine the forces in and between solute and solvent. A solute is more soluble in a solvent whose intermolecular forces are similar to, and therefore can replace, its own. (a) Methanol. NaCl is ionic, so it dissolves in polar solvents through ion-dipole forces. Both methanol and 1-propanol have a polar –OH group, but the hydrocarbon portion of each alcohol interacts only weakly with the ions and 1-propanol has a longer hydrocarbon portion than methanol. (b) Water. Ethylene glycol molecules have two –OH groups, so they interact with each other through H bonding. H bonds formed with H2O can replace these H bonds between solute molecules better than the dipole-induced dipole forces that form with hexane. (c) Ethanol. Diethyl ether molecules interact through dipole-dipole and dispersion forces. They can form H bonds to H2O or to ethanol. However, ethanol can also interact with the ether effectively through dispersion forces because it has a hydrocarbon chain. Step 1: Solute particles separate from each other. This process is endothermic. Solute (aggregated) + heat → solute (separated) ΔHsolute>0 Step 2: Solvent particles separate from each other. This process is endothermic. Solvent (aggregated) + heat → solvent (separated) ΔHsolvent>0 Step 3: Solute and solvent particles mix and form a solution. This step is exothermic. Solute (separated) + solvent (separated) → solution + heat ΔHmix Hsolute + Hsolvent Hsoln < 0 Endothermic solution process Hmix < Hsolute + Hsolvent Hsoln > 0 Solvation is the process of surrounding a particle with solvent particles. In water, solvation is called hydration. ∆Hsolvation = ∆Hsolvent + ∆Hmix In water, ∆Hsoln = ∆Hsolute + ∆Hhydr The hydration of an ion is always exothermic because ion-dipole forces are very strong. + M (g) [or - X (g)] → + M (aq) [or - X (aq)] Hhydr of the ion (always < 0) The entropy (S) of a system is related to the number of ways a system can disperse its energy and therefore to the freedom of motion of the particles. Gases have the highest entropy of the three states of matter. A solution usually has higher entropy than the pure solute and pure solvent. An increase in entropy is favored in both physical and chemical processes. For NaCl, Hmix is much smaller than Hsolute; Hsoln is so much larger than the entropy increase due to mixing that NaCl does not dissolve. For octane, ∆Hsoln is very small, but the entropy increase due to mixing is large, so octane dissolves. A saturated solution contains the maximum amount of dissolved solute at a given temperature in the presence of undissolved solute. Undissolved solute is in equilibrium with dissolved solute. Solute (undissolved) solute (dissolved) An unsaturated solution contains less than the equilibrium concentration of dissolved solute. If more solute is added, it will dissolve. solute (undissolved) solute (dissolved) A supersaturated solution contains more than the equilibrium concentration of solute. It is unstable and any disturbance will cause excess solute to crystallize immediately. Temperature affects solubility. Most solids are more soluble at higher temperatures. Gases become less soluble as temperature increases. Pressure affects the solubility of gases – they become more soluble at higher pressure. As the pressure is increased, more gas particles collide with the liquid surface More gas particles dissolve until equilibrium is re-established. The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution. Sgas = kH X Pgas PROBLEM: The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 25oC. What is the solubility of CO2? The Henry’s law constant for CO2 dissolved in water is 3.3 x10−2 mol/L·atm at 25oC. PLAN: We know P for CO2 (4 atm) and the value of kH, so we substitute these into the Henry’s law equation. SOLUTION: SCO2= (3.3 x10 −2 mol/L·atm)(4 atm) = 0.1 mol/L PROBLEM: What is the molality of a solution prepared by dissolving 32.0g of CaCl2 in 271 g of water? PLAN: Molality is defined as moles of solute (CaCl2) divided by kg of solvent (H2O). We convert the mass of CaCl2 to moles using the molar mass, and then divide by the mass of H2O, being sure to convert from grams to kilograms. mass (g) of CaCl2 divide by M of (g/mol) amount (mol) of CaCl2 divide by kg of water molality (m) of CaCl2 solution SOLUTION: 1 mol CaCl2 32.0 g CaCl2 x = 0.288 mol CaCl2 110.98 g CaCl2 0.288 mol CaCl2 molality = = 1.06 m CaCl2 1 kg 271 g H2O x 103 g PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that contains 40.5 mg of Ca. (b) The label on a 0.750-L bottle of Italian chianti indicates “11.5% alcohol by volume.” How many liters of alcohol does the wine contain? (c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water? PLAN: (a) We convert mg to g of Ca2+, find the mass ratio of Ca2+ to pill and multiply by 106. (b) We know the volume % of the alcohol and the total volume, so we can find the volume of alcohol. (c) We convert g of solute and solvent to moles in order to calculate the mole fractions. SOLUTION: 1g 40.5 mg Ca2+ x 103 mg (a) x 106 = 1.16x104 ppm Ca2+ 3.50 g 11.5 L alcohol (b) 0.750 L chianti x = 0.0862 L alcohol 100. L chianti (c) 1 mole moles isopropyl alcohol = 142 g x = 2.36 mol C3H7OH 60.09 g SOLUTION: 1 mole moles water = 58.0 g x = 3.22 mol H2O 18.02 g 2.36 mol C3H7OH  C3H7OH = = 0.423 2.36 mol C3H7OH + 3.22 mol H2O 3.22 mol H2O  HO= = 0.577 2 2.36 mol C3H7OH + 3.22 mol H2O To convert a term based on amount (mol) to one based on mass, you need the molar mass. To convert a term based on mass to one based on volume, you need the solution density Molality involves quantity of solvent, whereas the other concentration terms involve quantity of solution. PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution as a hair bleach. An aqueous solution H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its (a) Molality (b) Mole fraction of H2O2 (c) Molarity PLAN: (a) To find the mass of solvent we assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution for the mass of peroxide. (b) Convert g of solute and solvent to moles before finding X. (c) Use the density to find the volume of the solution. SOLUTION: (a) From mass % to molality: g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O 1 mol H2O2 30.0 g H2O2 x = 0.882 mol H2O2 34.02 g H2O2 0.882 mol H2O2 molality = = 12.6 m H2O2 70.0 g x 1 kg 103 g SOLUTION: (b) From mass % to mole fraction: 1 mol H2O 70.0 g H2O x = 3.88 mol H2O 18.02 g H2O 0.882 mol H2O2 X = H2O2 3.88 mol H O + 0.882 mol H O = 0.185 2 2 2 SOLUTION: (c) From mass % and density to molarity: 1 mL volume (mL) of solution = 100.0 g x = 90.1 mL 1.11 g mol H2O2 0.882 mol H2O2 molarity = = = 9.79 M H2O2 L soln 1 L soln 90.1 mL x 3 10 mL