Properties of Solutions PDF

Properties of Solutions 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 13 (c) 2014  Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is disperse...

Properties of Solutions 01006727 General Chemistry 01006708 Chemistry Brown Chemistry:The Central Science 13th Edition Chapter 13 (c) 2014  Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent. The ability of substances to form solutions depends on two key parameters Their natural tendency toward mixing. The intermolecular forces. 2  Natural Tendency toward Mixing Mixing of gases is a spontaneous process. Each gas acts as if it is alone to fill the container. Mixing causes more randomness in the position of the molecules, increasing a thermodynamic quantity called entropy. – The formation of solutions is favored by the increase in entropy that accompanies mixing. – However, interactions are also important – especially for liquid and solid solutions. 3  Intermolecular Forces of Attraction The ability of a solute and solvent to form attractive intermolecular interactions can favor the formation of a solution. 4 Energetics of Solution Formation 5  Attractions Involved When Forming a Solution Solute–solute interactions must be overcome to 8E disperse these particles when making a solution. Solvent–solvent interactions must be overcome to make room for the solute. Solvent–solute interactions occur as the particles mix. 6  Exothermic or Endothermic For a reaction to occur, Hreaction must be close to the sum of Hsolute and Hsolvent. Remember that the randomness from entropy is critical in the process too. Indeed it is a key driver of dissolution. & & = = 7  Factors That Affect Solubility Solubility is an endothermic process (∆H>0). - Why therefore do salts dissolve in water? Answer because of the reaction free energy (∆G < 0). – An increase in entropy drives the process. 60 - ↑ ∆G= ∆H – T∆S * remember we want a negative ∆G A number of factors affect solubility including: D - – Increased T, leads to increased Solubility – Increased P, leads to increased Solubility (for gas) - – Balance between solute-solute - & solute-solvent interactions – typically favors solid form. ⑧ = ⑦ + 8  Aqueous Solution vs. Chemical Reaction Just because a substance disappears when it comes in contact with a solvent, it does not mean the substance dissolved. It may have reacted, like nickel with hydrochloric acid. Ha P chemical reaction - 9  Opposing Processes The solution-making process and crystallization are opposing processes. - = When the rate of the opposing processes is equal, additional solute will not dissolve unless some crystallizes from solution. This is a saturated solution. - If we have not yet reached the amount that will result in crystallization, we have an - unsaturated solution. 10  Solubility Solubility is the maximum amount of solute - - 3) that can dissolve in a given amount of - solvent at a given temperature. - – Saturated solutions have that amount of - - solute dissolved. - – Unsaturated solutions have any amount of solute less than the maximum amount dissolved in solution. 11  Supersaturated Solutions In supersaturated solutions, the solvent holds - more solute than is normally possible at that temperature. – These solutions are unstable; crystallization can - - usually be stimulated by adding a “seed crystal” - or scratching the side of the flask. A 12  Solute–Solvent Interactions The stronger the solute– - - solvent interaction, the - greater the solubility of a - solute in that solvent. The gases in the table 66 only exhibit dispersion force. - The larger the gas, the more soluble it will be in water. 13 like dissolve likeOrganic Molecules in Water Polar organic molecules = dissolve in water better than nonpolar organic molecules. Hydrogen bonding increases solubility, since C–C and C– H bonds are not very polar. & 14  Liquid/Liquid Solubility & Liquids that mix in all proportions are miscible. - Liquids that do not mix in one another are immiscible. Because hexane is nonpolar and water is polar, they are immiscible. 15  Solubility and Biological Importance Molecules that are alike are likely to dissolve in each other Fat-soluble vitamins (like vitamin A) are nonpolar; they are readily stored in fatty tissue in the body. Water-soluble (polar) vitamins (like vitamin C) need to be included in the daily diet. 16  Pressure Effects The solubility of solids - and liquids are not - appreciably affected by pressure. 1 Gas solubility is affected by pressure. 17  Henry’s Law The solubility of a gas is proportional to the partial pressure of the E gas above the solution. 18  Henry’s Law If you double the partial pressure of a gas over a liquid at constant temperature. Which of these statements is then true? I (b) The Henry’s law constant is decreased by half. (a) The Henry’s law constant is doubled. (c) There are half as many gas molecules in the liquid. To(d) There are twice as many gas molecules in the liquid. (e) There is no change in the number of gas molecules in the liquid. ↑ Sa ki : Sg = kPg (d) There are twice as many gas molecules in the liquid. 19  Henry’s Law Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25 °C. The Henry’s law constant for CO2 in water at this temperature is 3.4x10-2 mol/L.atm. 33k 43102d) 3. 140 is Sg = kPg SCO2 = kPCO2 = (3.4x10-2 mol/L.atm)*4.0 atm = 0. 14 mol/L = 0.14 mol/L = 0.14 M Gu 20  Henry’s Law Calculate the concentration of CO2 in a soft drink after the bottle is opened and the solution equilibrates at 25 °C under a CO2 partial pressure of = MT 4x10mo) (3 0xc0Y at 3.0x10-4 atm. = = (3 Sg = kPg.. = 10410 SCO2 = kPCO2 = (3.4x10-2 mol/L.atm)* 3.0x10-4 atm mol/L = 10 MM = 40x10-6 mol/L = 40µM 21  Temperature Effects For most solids, as For all gases, as temperature increases, temperature increases, solubility increases. solubility decreases. – Not always true - some – Cold water has higher O2 increase greatly, some content than warm rivers. remain relatively constant, and others decrease. ↑ 1 - > - 22 -  Colligative Properties Colligative properties depend only on the quantity, not on the identity of the solute particles. Among colligative properties are: – Vapor-pressure lowering – Boiling-point elevation – Freezing-point depression – Osmotic pressure 23  Units of Concentration T Molarity = moles solute / L solution PPM = mass solute in mg / L solution %mass = mass solute / mass solution Molality = moles solute / Kg solvent 24  A solution is made by dissolving 13.5 g of glucose (C6H12O6) in 0.100 kg of Mass %, PPM -10- A 2.5g sample of groundwater was found to contain 5.4 µg of Zn2+. What is the concentration of Zn2+ in parts per million? ppm water. What is the mass percentage of – Density of water = 1 g/ml solute in this solution? %mass – - 1 ppm = 1 mg/L of glucose - glucose mes ppMinsEuthing 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑚𝑚𝑔𝑔 a % mass glucose = ∗ 100 = It 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑔𝑔𝑔𝑔𝑠𝑠 + 13.5𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑍𝑍𝑠𝑠2 5.4𝑥𝑥10−6 𝑔𝑔 ppm= ∗ 106 = ∗ 106 = 2.2 𝑝𝑝𝑝𝑝𝑝𝑝 13.5𝑔𝑔+100𝑔𝑔 𝑚𝑚𝑔𝑔𝑔𝑔𝑠𝑠 𝑖𝑖𝑠𝑠 𝐿𝐿 2.5𝑚𝑚𝑔𝑔 , = = 11.9% To ppm g go to = % mass water = 100-11.9% =88.1% Ester 9 % mg/L = 17. = 2. 2 % mass water = 100 - 11 9 = 88. 1 % = 2 2 ppm.. 25  M Sh Molarity & Molality A solution is made by dissolving 4.35 g What is the molality of a solution made by of glucose (C6H12O6, 180.2 g/mol) in dissolving 36.5 g of naphthalene 25.0 mL of water at 25 °C. Calculate (C10H8,128.1g/mol) in 425 g of toluene (C7H8, 92.1 g/mol)? the molality of glucose in the solution. = Water has a density of 1.10 g/mL. no glucose =ga mol glucose = 4.35 g = 0.0241 𝑝𝑝𝑚𝑚𝑚𝑚 0 - 0291 no molnaphthalen mol napthalene = glnd) 36.5 g 128.1 g/mol = 0.285 𝑝𝑝𝑚𝑚𝑚𝑚 180.2 𝑔𝑔/𝑚𝑚𝑔𝑔𝑔𝑔 0.285 𝑚𝑚𝑔𝑔𝑔𝑔 = 0 molality soln = 0.425 Kg toluene= 0.760 m. 285 mol *𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 1g/mL=1Kg/L M = molality soln = m you 0.0241 𝑚𝑚𝑔𝑔𝑔𝑔 0.025 𝐿𝐿 *1 Kg/L=0.964m m *𝑠𝑠𝑠𝑠 𝑡𝑡𝑡𝑠𝑠𝑠𝑠 𝑠𝑠𝑐𝑐𝑠𝑠𝑠𝑠 𝑝𝑝𝑚𝑚𝑚𝑚𝑐𝑐𝑚𝑚𝑠𝑠𝑡𝑡𝑚𝑚 = 𝑝𝑝𝑚𝑚𝑚𝑚𝑐𝑐𝑚𝑚𝑠𝑠𝑡𝑡𝑚𝑚 = 0. 964M m = 0 677. M = 0. 964M density of water is 1 g/h 26 mol frection , Xi = 41/ total Molarity & Molality ↳ mol/L An aqueous solution of hydrochloric A solution with a density of 0.876 acid (36.5 g/mol) contains 36% HCl by g/mL contains 5.0 g of toluene mass. Assume 100 g solution. (C7H8, 92.1g/mol) and 225 g of Water=18 g/mol. – (a) Calculate the mole fraction of HCl. benzene (78.1 g/mol).Calculate – (b) Calculate the molality of the solution. the molarity of the solution. moltolulene HCl = mo 0 m = 0. 59 = 36 𝑔𝑔 mole𝑠𝑠 HCl = = 0.99 mol 5.0 𝑔𝑔 36.5 𝑔𝑔/𝑝𝑝𝑚𝑚𝑚𝑚 mole𝑠𝑠 toluene = = 0.054 mol 92.1 𝑔𝑔/𝑝𝑝𝑚𝑚𝑚𝑚 mol of (3693 mol 20 = mole𝑠𝑠 H2O = 64 𝑔𝑔 18 𝑔𝑔/𝑝𝑝𝑚𝑚𝑚𝑚 = 3.6 mol 6. volume volume soln =solution a =263mL=0.263 L 225.0𝑔𝑔+ 5.0 𝑔𝑔 0.876 𝑔𝑔/𝑚𝑚𝑔𝑔 𝑚𝑚𝑔𝑔𝑔𝑔 𝐻𝐻𝐻𝐻𝑔𝑔 0.99 𝑚𝑚𝑔𝑔𝑔𝑔 0.99 0.054 mol m ma 22 XHCL= = = = 0.22 Molarity : = 0.263 L =0.21 M = 𝑀𝑀𝑔𝑔𝑔𝑔 𝐻𝐻2𝑂𝑂+𝐻𝐻𝐻𝐻𝑔𝑔 3.6 mol +0.99 mol 4.6 * H water = molality HCl = 0.99 mol = 15 m (63mL = 0 263L 0 22 0.064 𝐾𝐾𝑔𝑔 𝐻𝐻2𝑂𝑂.. = MH smol(9) = = 15m =27  Freezing-Point Depression - The phase diagram for a solution demonstrates that the freezing point is lowered while the boiling point is raised. – This is why sea-water doesn’t freeze (salt content). – Salt-water bonds must be broken before solid water can form. Effect is again entropic in origin 28  Boiling-Point Elevation Vapor pressures are lowered for solutions, it requires a higher temperature to reach atmospheric pressure. Hence, boiling point is raised. Effect is entropic in origin ↳ 29  Boiling-Point Elevation and Freezing-Point Depression The change in temperature (∆T) is directly proportional to molality (m) for ideal solutions – i (or activity factor) can be assumed to be 1 for a 1 M non-dissociating molecule The van’t Hoff factor is a correction for the fact that higher concentration solutions are not ideal. 30  i Freezing Point Depression 1 & = Ethylene glycol (EG) (C2H6O2, 62.7 g/mol) a nonvolatile nonelectrolyte is added to water. AT - = - (1)(1. 85 / 15. 32m) = = 89'C *9assume. 1 Kg solution Calculate the freezing point and boiling point and of a 25.0% by mass solution in water. Kb= oc-9 89 0.51oC/m, and Kf = 1.86 oC/m. Tf =. 0.250 𝐾𝐾𝑔𝑔 𝐸𝐸𝐸𝐸 1 ATp = -iRm molality = = = * 89'j 0.750𝐾𝐾𝑔𝑔 𝐻𝐻2𝑂𝑂 62.7 𝑔𝑔/𝑚𝑚𝑔𝑔𝑔𝑔 = 5.37 m g ∆Tf = -iKfm. AT = ikm ∆Tb = iKbm 1 Th = (1)(0 57c) (5 m) ∆Tf = -(1)(1.86 oC/m)(5.37 · 32 m = -10.0 oC ∆Tf= 0 oC – 10.0 oC = -10.0 oC. =E m 7 = 2 ∆Tb = (1)(0.51oC/m)(5.37 m) = 2.7 oC. ∆Tb= 100 oC + 2.7 oC = 102.7oC Tb = 100d + 2. 7d = 5. 32 M = 102. 7 31 A Tb Boiling Point Elevation = ikym 0717m)) i ↑ = 1 mole solute = (409(10. A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution = 2. 84x10"mol ∆Tb = iKbm 0.357 𝑔𝑔𝐻𝐻 was 0.357 °C higher than that of the pure Mole solute = = 0.0711 m mol 1 (5.02 °𝐻𝐻/𝑚𝑚) solvent. Kb for the solvent CCl4, Kb = 5.02 = - °C/m. Calculate the molar mass of the moler mass mol solute solute. (What is the unknown =0.0711 m ?( 40.0𝑥𝑥10−3 𝑘𝑘𝑔𝑔 mass10g ) moler ATs = iRpM mol solute = (0.0711 mol/kg soln) *(40.0𝑥𝑥10−3 𝑘𝑘𝑔𝑔 soln) 1 m = =2.84x10-3 mol solute [m) molar mass = 0.250 = g 88 s o lu t e 2.84x10−3 mol solute g/md = 88 g/mol = 0. 0711M M solute 9 = = 00. 32  The van’t Hoff Factor (i) The van’t Hoff factor #504-2 Kin + Sozgs takes into account ↓ 241 ET dissociation in 3 = = - solution. = Theoretically, we get 6 2 particles when NaCl E - - dissociates. So, i = 2. = – The exact amount that particles remain together is dependent on the concentration. 33  Van’t Hoff I factor A 1.00 m solution of acetic acid (CH3COOH) in benzene has a freezing point depression of 2.6 K. Calculate the value for i and suggest an explanation for its value. Kf benzene = 5.10 K/mol ST = - ikm ∆T = -iKfm 2.6 K = (i) (5.10 K/mol ) (1.00m) 70k) i (5 f 6k I = 2 -. -. i = 0.51 = Acetic acid is pooly soluble in benzene. It therefore dimerizes to give (CH3COOH)2. i.e. half the number of expected moles are present half number of expected moles are i = 0 -. 51 present. to soluble Acetic acid is poorly in benzene. I- dimerizes to give (CHS(OOH)2 34  Van’t Hoff I factor The freezing point depression of a Cif 100 % ionized , so the 0.10 m solution of HF(aq) solution is -0.201 °C. Calculate the percent dissociation of HF(aq). Kf water = 1.86 K/mol different in temperature HY would be - HF - + F i ∆Tf = -iKfm ↑ 0.201 K = (i) (1.86 K/mol ) (0.10mol) (1 -ikfm 86) (2)10. m , - -T. = i = 1.08 86K) 0 201K (i) (1 10 m 10 - ==... = = 0 372k. HF is only 8% ionized i= Ind coms solution = solute solvent If 100% ionized the difference in temperature would be 1.86 K i = 1. 08 M % ionized HE is only 8. 35  Vapor Pressure The boiling point of a liquid is the temperature at which the vapor pressure equal atmospheric pressure. For ideal solutions, the vapor pressure of a mixture containing a non-volatile solute (i.e sugar or salt) is equal to the mole-fraction- weighted sum of the components' vapor pressures – This results in the liquid solution boiling at higher temperatures – The same effect will lead to the solution freezing at a lower temperature 36 mole fruction , Xi Raoult’s Law = moli/total mol The vapor pressure of a at constant T volatile solvent is the product Vapour Pressure of the mole fraction of the solvent times the vapor pressure of the pure solvent. Psolution = PAoXA + PBoXB In ideal solutions, it is di assumed that each Mole fraction substance will follow Raoult’s Law. The partial pressures of 1 solvents are additive. = #lamB 5 37  Vapour Pressure a P = Glycerin (C3H8O3, 92.1 g/mol) is a nonvolatile - glycerin 26had nonelectrolyte. 1.26 g (92.1 g/mol) is added = - mol to water at 25 °C. Calculate the vapor pressure at 25 °C of a solution made by Psolution = Xsolvent.Posolvent adding the glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 °C is 23.8 torr, and its density is 1.00 g/mL. 20 1.26 𝑔𝑔 014. mole𝑠𝑠 gyc𝑠𝑠𝑚𝑚𝑠𝑠𝑠𝑠 = = 0.684 mol 92.1𝑔𝑔/𝑝𝑝𝑚𝑚𝑚𝑚 Hysin At Ya Pix 500 𝑔𝑔 Psolution = + mole𝑠𝑠 𝑤𝑤𝑐𝑐𝑡𝑡𝑠𝑠𝑚𝑚 = 18.0 𝑔𝑔/𝑝𝑝𝑚𝑚𝑚𝑚 = 27.8 mol 27.8 mol 27.8 mol O Xwater=27.8 mol +0.684 𝑚𝑚𝑔𝑔𝑔𝑔 = 28.484 𝑚𝑚𝑔𝑔𝑔𝑔 =0.976 Psolution = PBXB Pwater= 0.976*23.8torr = 23.2 torr ↓ Pglycerin = 23.8 torr - 23.2 torr = 0.6 torr = (23 8 tour) (0 99). = 23. 6 four 38. Gas Phase Observations: Positive Deviation from Raoult’s Law Most Compounds deviate from Raoult’s law including mixtures Vapour Pressure of acetone & ether Acetone & ether show PB positive deviation PA – Total vapor pressure Ideal line Observed line higher than expected Mole fraction – i.e. interact less strongly together than O they do in isolation. A B H3C CH3 H3C O CH3 34°C 56°C 39  Gas Phase Observations: Negative Deviation from Raoult’s Law Chloroform and Acetone show Vapour Pressure negative deviation from Raoult’s law PB – Total vapor pressure lower than expected PB Ideal line – i.e. interact more Observed line strongly together than Mole fraction they do in isolation. Oδ - δ+H Cl A B H3C CH3 Cl Cl 56 °C 61°C 40  Fractional Distillation Raoult’s law and the separation of petrochemical mixtures. – Liquid and vapor phases have different mole fractions which depend on their boiling point Consider a mixture of toluene:benzene (70:30) Vapor phase – Heat the mixture to till boiling point is reached– differences in 1 vapor pressure lead to: 1. enriched fraction of benzene 2 (~50% after 1st step) Liquid phase 3 2. enriched fraction of benzene (~70% after 2nd step) 3. enriched fraction of benzene (~85% after 3rd step) 4. Repeat the process using this BP= BP= 80.1oC 110.6oC fraction until pure benzene is obtained 41  Azeotropic Solution Azeotropes are solutions that at a particular concentration components exist in the same mole fraction in the liquid and vapor phase. – Once this composition is reached, the samples cannot be separated further by distillation – Need to find another physical method to separate azeotropic mixtures. https://en.wikipedia.org/wiki/Azeotrope 42  Osmosis Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking larger particles. The net movement of solvent molecules from solution of low to high concentration across a semipermeable membrane is osmosis. The applied pressure to stop it is osmotic pressure. i is the van’t Hoff Factor 43  Osmotic Pressure gi = The average osmotic pressure of blood is 7.7 atm at 25 °C. What molarity of glucose (C6H12O6, 180.2 π=iMRT g/mol) will be isotonic with blood? How many mg/ml? π 7.7 𝑚𝑚𝑎𝑎𝑚𝑚 M= = 𝐿𝐿.𝑎𝑎𝑎𝑎𝑎𝑎 =0.32 M – R=0.08206 L.atm/K.mol 𝑖𝑖𝑖𝑖𝑖𝑖 1 (0.0821 )(273𝐾𝐾+25). 𝐾𝐾.𝑎𝑎𝑚𝑚𝑚𝑚 - 𝑚𝑚𝑔𝑔𝑔𝑔 𝑔𝑔 # = iMRT g/ glucose= 0.32 𝐿𝐿 ∗ 180.2 𝑚𝑚𝑔𝑔𝑔𝑔 M = b Lates = 0.0018 𝑔𝑔/𝐿𝐿 + =254) 1.8 mg/L = 0.0018 mg/ml 0 37M 56 mg/mL 6 =. = I ↑ (0 mol) (180 29). 31. = 56l (b)(i)) = 44  Types of Solutions & Osmosis 1) Isotonic solutions: Same osmotic pressure; solvent passes the membrane at the same rate both ways. 2) Hypotonic solution: Lower osmotic pressure; solvent will leave this solution at a higher rate than it enters with. 3) Hypertonic solution: Higher osmotic pressure; solvent will enter this solution at a higher rate than it leaves with. 45  Osmosis and Blood Cells Red blood cells have semipermeable membranes. If stored in a hypertonic solution, they will shrivel as water leaves the cell; this is called crenation. If stored in a hypertonic solution, they will grow until they burst; this is called hemolysis. 46  Reverse Osmosis Reverse osmosis is the process of whereby pure water can be obtained from seawater or other contaminated sources. Pressure is applied to the sample to be purified, decreasing the volume of the vessel. According to Le Châtelier's principle water will diffuse across the membrane (lowering P) and leave an increasingly concentrated solution behind. 47  Colloids Suspensions of particles larger than individual ions or molecules, but too small to be settled out by gravity, are called colloids. 48  Tyndall Effect Colloidal suspensions can scatter rays of light. (Solutions do not.) This phenomenon is known as the Tyndall effect. 49