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202-NYB-05 Lecture 4 notes

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202-NYB-05 Lecture 4 notes 202-NYB-05

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These notes finish going over osmotic pressure, and give a brief intro to equilibrium.
Chemistry of Solutions
Nadia Schoonhoven
Class Notes
osmotic pressure, Equilibrium
25 ?




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This 9 page Class Notes was uploaded by CatLover44 on Sunday September 25, 2016. The Class Notes belongs to 202-NYB-05 at Dawson Community College taught by Nadia Schoonhoven in Fall 2016. Since its upload, it has received 4 views. For similar materials see Chemistry of Solutions in Chemistry at Dawson Community College.

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Date Created: 09/25/16
Chemistry of Solutions Course Number: 202-NYB-05 Lecture no. 4 Date: Tuesday, September 6, 2016 Professor: Nadia Schoonhoven Topics Covered: osmotic pressure, colligative properties of electrolyte solutions, the equilibrium condition Osmotic Pressure • Grapes provide good examples for understanding the basics of osmotic pressure. When washing grapes, the concentration of the solute inside a grape is greater than the concentration of the water surrounding it. o The water surrounding the grape wants to dilute the solutes inside the grape. o Only water can pass through the grape’s membrane (semipermeable), so since it seeks to enter the grape, if the grapes are left in water long enough, the solute concentration inside the grape increases to the point that the grape explodes. • Osmosis is defined as the net flow of solvent that passes through a semipermeable membrane. o All molecules in a solution bump into a semipermeable membrane, but only the solvent can go through it. The side of the solution containing more water molecules has a higher rate of escaping water molecules. The solvent wants to move to the side of the membrane that has more solute (less solvent). • So far, the colligative properties we know are boiling-point elevation, and freezing-point depression. A third one is osmotic pressure. o Osmotic pressure is the pressure that must be applied to stop the solvent from passing a semipermeable membrane (to make the solution reach equilibrium). o We calculate osmotic pressure with the following formula: § ∏ = MRT, where ∏ is osmotic pressure, M is molarity (moles of solute per litre of solution), R is the ideal gas constant (0.08206 L atm / mol K) and T is temperature in Kelvins. § The units for R must match with the units of the other values. For example, if you're using R = 0.08206 L atm / mol K, temperature must be in Kelvins and osmotic pressure must be in atmospheres. § This equation is very similar to the ideal gas law equation, PV = NRT • Note: osmosis is the reasoning behind why we salt meats to preserve them; distilling alcohol is a process that involves boiling points; and salting roads to prevent them from becoming icy in the winter, and the use of anti-freeze in the cooling systems of cars, involves freezing-point depression. • When an osmotic system reaches equilibrium, no osmosis is actually occurring (no net flow of solvent across semipermeable membrane). • The concentration gradient (low solvent concentration to solvent high concentration) is countered when pressure is exerted by the higher collusion of liquid on one side. This is the osmotic pressure, • Osmometry is useful because it lets us precisely determine solute concentration. Exercise: to determine the molar mass of a protein, 1.00 x 10 g of it are dissolved in 1.00 mL of solution. The osmotic pressure of this solution was 1.12 torr at 25°C. Calculate the molar mass of this protein. ∏ = MRT ∏ = 1.12 torr, V = 0.001 L, T = 298.15 K. ∏ = MRT becomes M = ∏ / RT 1.12 torr x 1 atm / 760 torr = 0.0014737 atm, because 1 atm = 760 torr. M = (0.0014737 atm) / (0.08206 L atm/ mol K) (298.15 K) = 0.0000602341 mol/L, or 6.02 x 10 -5 mol/L. M x V = n, so (0.0000602341 mol/L) x (0.001 L) = 6.02 x 10 mol.8 Mass/ molar mass = moles, and mass / moles = molar mass, so: MM = (1.00 x 10 g) / (6.02 x 10 mol) = 16.6019 g/mol = 1.66 x 104 g/mol (3 sig figs). • Dialysis is an application of osmotic pressure. In dialysis, the membrane lets solvent and other small molecules pass through it. This occurs naturally in plant and animal cells. Blood is cleaned by using a solution that has the same osmotic pressure (isotonic). o If a solution with an osmotic pressure that is greater than that of blood (hypertonic) this can cause blood cells to shrivel up. 2 § The concentration of solute is greater outside the blood cells than it is inside the cell, so the blood inside the cell seeks to leave it so it can lower the concentration of solution outside of it. o If a solution with an osmotic pressure less than blood is injected into the bloodstream (hypotonic), this can cause blood cells to explode. § The concentration of solute is lower inside the cell than it is on the outside of the cell, so the solvent around the cell wants to enter to cell to increase its concentration. § Remember: concentration is directly related to osmotic pressure! Exercise: what concentration of sodium chloride in water is needed to produce an aqueous solution that's isotonic with blood, whose osmotic pressure is 7.70 atm at 25°C. ∏ = MRT becomes M = ∏ / RT ∏ = 7.70 atm, T = 298.15 K, R = 0.08206 L atm / mol K M = (7.70 atm) / (0.08206 L atm/ mol K) (298.15 K) = 0.31472 mol/L NaCl dissolves into two ions, Na and Cl , so the concentration of sodium chloride is: 0.31472 M / 2 = 0.157 M. Note: each particle affects the concentration, osmotic pressure, boiling point and freezing point of a solution. • Reverse osmosis occurs if the external pressure of a solution is larger than the osmotic pressure, causing solvent to leave a solution instead of enter it. o Reverse osmosis is useful because it is used in the process of desalination of sea water; applying a pressure that's greater than the sea water’s osmotic pressure produces fresh water. Exercise: in an ideal solution, what happens when a non-volatile solute is added to a solvent? Solution: vapour pressure decreases, freezing point decreases, boiling point increases, and osmotic pressure increases. • Electrolyte solutions demonstrate colligative properties, which only involve the numbers of particles in an ideal solution. 3 o Consider a solution of 0.100 mol/kg glucose (molecular dissolution) with a freezing-point depression of 0.186°C, and another solution of NaCl (ionic dissolution) of the same molality. o The freezing-point of the NaCl solution is not double that of the glucose solution, even though NaCl dissolves into two ions in aqueous solutions, and glucose doesn't. This occurs because of ion pairing: some ions stay paired up during the process of dissolution. • The van’t Hoff factor, i, relates the number of particles released when a substance dissolves. Study the table below: Exercise: which solution should have the highest boiling point: 0.20 m ethylene glycol (HOCH CH OH, non-electrolyte); 0.12 m NaI; 0.10 m CaCl ; or 0.12 m Na SO ? 2 2 2 2 4 Solution: 0.12 m Na SO ,2usi4g the van't Hoff factor. 0.20 m ethylene glycol: 1 x 0.20 = 0.2 (molecular dissolution) 0.12 m NaI: 2 x 0.12 = 0.24 (ionic dissolution, releases 2 ions) 0.10 m CaCl : 2 x 0.10 = 0.30 (ionic dissolution, releases 3 ions) + 2- 0.12 m Na SO2: an4 3 x 0.12 = 0.36 (ionic dissolution, releases 3 ions: 2Na and SO 4 ). 4 Exercise: complete the table, and rank the solutions from lowest to highest vapour pressure, boiling point, freezing point, and osmotic pressure. Solute Solution Species in van't Hoff Total concentration Solution Factor concentration of particles + - KCl 0.25 M K , Cl 2.00 0.500 M (0.25 M x 2) C 6 O12 6 0.40 M C , H , O 2- 1.00 0.40 M (0.40 M x 1) + 2- (NH )4 2 4 0.20 M 2NH ,4SO 4 3.00 0.60 M (0.20 M x 3) Recall: osmotic pressure is directly related to concentration. o The solution with the lowest vapour pressure is (NH ) SO .4 2ncen4ration is directly related to vapour pressure, and (NH ) SO4 2s t4e greatest concentration, while C H O has th6 12 6 lowest. Therefore, from lowest to highest vapour pressure, (NH ) SO < K4 2< C4H O . 6 12 6 o The presence of a nonvolatile solute increases the boiling point of a solution. C H O has6 12 6 the least amount of particles in solution so it's boiling point is the lowest, while (NH ) 4 2 4 releases three particles when dissolved so it has the highest boiling point. In order of lowest to highest boiling point, C 6 O12 K6l < (NH ) SO .4 2 4 o From lowest to highest, the freezing points of the solutions are (NH ) SO <4 2l <4C H O ., 6 12 6 the opposite of the ascending order of boiling points. o Osmotic pressure is dependent on concentration, so from lowest to highest, the osmotic pressures of the solutions are C H O6<12Cl6< (NH ) SO . 4 2 4 Exercise: calculate the freezing point and the boiling point of a 0.050 m FeCL aqueous3solution using the observed and expected van't Hoff factors. The molal freezing and boiling point constants for water are 1.86°C kg/mol and 0.51°C kg/mol. Expected i = 4, observed i = 3.4 ΔT = i Km 1. FP, expected: ΔT = (0°C- ((4) (1.86°C kg/mol) (0.050 mol/kg)) = 0°C – 0.37°C = – 0.37°C 2. FP, observed: 5 ΔT = (0°C – ((3.4) (1.86°C kg/mol) (0.050 mol/kg)) = 0°C – 0.32°C = -0.32°C 3. BP, expected: ΔT = (100°C + ((4) (0.51°C kg/mol) (0.050 mol/kg)) = 100°C + 0.102°C = 101.10°C 4. BP, observed: ΔT = (100°C + ((3.4) (0.51°C kg/mol) (0.050 mol/kg)) = 100°C + 0.0867°C = 100.87°C • Study the following diagram because understanding it can help you solve problems involving the van't Hoff factor: Exercises 1. A 1.48 M solution of H C3H6O5(c7tric acid) in water has density 1.10 g/mL. Calculate the mass percent, molality, mole fraction, and normality of citric acid. Citric acid has three acidic protons and its molar mass is 192.124 g/mol. (i = 4). 6 Mass percent = mass of solute / mass of solution x 100% Molality = moles of solute/ kg of solvent Mole fraction of citric acid = moles of citric acid / (moles of citric acid + moles of water) Normality = number of equivalents. 1.48 M = 1.48 mol. solute per litre of solution, so moles of citric acid = 1.48 mol. 192.124 g/mol x 1.48 mol. = 284.3 g of citric acid. Assuming 1 L, 1000 mL x 1.10 g/mL = 1100 g of solution. 1100 g solution – 284.3 g citric acid = 815.7 g water = 0.8157 kg solvent. 815.7 g x 1 mol. / 18.02 g = 45.27 mol. water. Solution: - Mass percent = (284.3)/(1100) x 100% = 25.9% - Molality = 1.48 mol./ 0.8157 kg = 1.81 m - Mole fraction of citric acid = (1.48)/(1.48 + 45.27) = 0.0317 - Normality = 1.48 M x 3 acidic ions = 4.44 N 2. A 25.00 mL solution containing 4.562 mg of an ionic compound with formula MCL3 exhibits an osmotic pressure of 83.1 +\- 0.6 mm Hg at 22°C. What's the most likely identity of M: Al, Cr, Mn, Fe, or Co? Use osmotic pressure to find the moles of solute and the total number of moles of dissolved ions. ∏ = MRT => ∏ = (n/v)RT => n = (∏V)/(RT) = ((83.1 mm Hg/760 mm Hg/atm) x (0.02500 L/ (0.08206 L atm/ mol K)(295.15 K)) = 1.1286 x 10-4 mol. ions. Calculate the molar mass based on the 4 ions that MCl release3. 7 1.1286 x 10-4 mol. / 4 = 2.823 x 10-5 mol. MCl . 3 Mass / Molar Mass = moles => mass / moles = molar mass, so: 0.004562 g / 2.823 x 10-5 mol. = 161.6 g/mol = Molar Mass of Compound. For the Molar Mass of M: 161.6 g/mol – 3(35.45g/mol) = 55.3 g/mol. % error in osmotic pressure: 100% x (0.6/88.1) = 0.72% 55.3 g/mol +/- (0.0072 x 55.3 g/mol) = 55.3 g/mol +/- 0.4 g/mol, so M can either be Fe or Mn due to the percent error. 3. The electrolyte in car batteries (12 V lead storage batteries) is a 3.75 M sulfuric acid solution with a density of 1.230 g/mL. Calculate the mass percent and molality of the sulfuric acid, and find its freezing point. A problem with the same values was solved in an earlier lecture, so they will be used in this solution. m = 3.75 mol. / 0.8623 kg = 4.35 mol/kg H S2 . 4 Using the van't Hoff Factor with the estimated value of 3, because H S2 pro4uces 3 ions in solution: ΔT = 0°C – (3 x 4.35 mol/kg x 1.86°C kg/mol) = 0°C - 24.3°C = -24.3°C. 3. Cigars are best stored in a humidor at 18°C and 55% relative humidity. This means the pressure of water vapour should be 55% of the vapour pressure of pure water at the same temperature. The proper humidity can be maintained by placing an aqueous solution of glycerol, C H3(5H) (M3 = 92.09 g/mol) inside the humidor. Calculate the concentration of glycerol (in mass%) required to lower the vapour pr sure of water to be desired value. Assume ideal gas behaviour, with glycerol as a nonvolatile component. Determine the desired vapour pressure of water in the humidor. 8 P = 55%, so P = Xsolute P°water => Xsolute = (0.55 atm)/(1 atm) = 0.55. Convert the mole fraction of glycerol to mass %: Assume the total number of moles in solution is 1 (ideal), so 1 mole – 0.55 mole = 0.45 moles of solvent. Mass of solute = 0.55 mol x 92.09 g/mol = 50.6495 g Mass of solvent = 0.45 mol x 18.02 g/mol = 8.109 g Mass % = (8.11 g) / (50.65 g + 8.11 g) x 100% = 14% (2 sig figs) Chemical Equilibrium (Short Review) • Any reversible reaction can reach equilibrium. • We know a system reached equilibrium if the concentration of its reactants and products stopped changing, and if the rate of the forward reaction equals to the rate of the reverse reaction. • There are some instances where chemical reactions go so slowly that their reactants and products don't seem to be changing, but they really are. 9


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