Physical Properties of Solutions

Transcription

1 1 Physical Properties of Solutions Chapter 13 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 2 A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount 2 3 A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 3

2 4 Three types of interactions in the solution process: solvent-solvent interaction solute-solute interaction solvent-solute interaction Molecular view of the formation of solution H soln = H 1 + H 2 + H like dissolves like Two substances with similar intermolecular forces are likely to be soluble in each other. non-polar molecules are soluble in non-polar solvents CCl 4 in C 6 H 6 polar molecules are soluble in polar solvents C 2 H 5 OH in H 2 O ionic compounds are more soluble in polar solvents NaCl in H 2 O or NH 3 (l) 5 6 Concentration Units The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. Percent by Mass mass of solute % by mass = x 1% mass of solute + mass of solvent Mole Fraction (X) X A = mass of solute = x 1% mass of solution moles of A sum of moles of all components 6

3 7 Molarity (M) Concentration Units Continued M = liters of solution Molality (m) m = mass of solvent (kg) 7 8 What is the molality of a 5.86 M ethanol (C 2 H 5 OH) solution whose density is.927 g/ml? m = mass of solvent (kg) Assume 1 L of solution: 5.86 moles ethanol = 27 g ethanol 927 g of solution (1 ml x.927 g/ml) M = liters of solution mass of solvent = mass of solution mass of solute = 927 g 27 g = 657 g =.657 kg m = mass of solvent (kg) = 5.86 moles C 2H 5 OH.657 kg solvent = 8.92 m 8 9 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = liters of solution What mass of KI is required to make 5. ml of a 2.8 M KI solution? M KI M KI volume of KI solution moles KI grams KI 1 L 2.8 mol KI 166 g KI 5. ml x x x = 232 g KI 1 ml 1 L soln 1 mol KI 9

4 1 Preparing a Solution of Known Concentration 1 11 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) = Moles of solute after dilution (f) M i V i = M f V f How would you prepare 6. ml of.2 M HNO 3 from a stock solution of 4. M HNO 3? M i V i = M f V f M i = 4. M M f =.2 M V f =.6 L V i =? L V i = M fv f M i =.2 M x.6 L 4. M =.3 L = 3. ml Dilute 3. ml of acid with water to a total volume of 6. ml. 12

5 13 Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point the point at which the reaction is complete Indicator substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color Titrations can be used in the analysis of Acid-base reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 Redox reactions 5Fe 2+ + MnO H + Mn Fe H 2 O What volume of a 1.42 M NaOH solution is required to titrate 25. ml of a 4.5 M H 2 SO 4 solution? WRITE THE CHEMICAL EQUATION! H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 M rxn M volume acid moles red moles base volume base acid coef. base 4.5 mol H 2 SO 4 2 mol NaOH 1 ml soln 25. ml x x x = 158 ml 1 ml soln 1 mol H 2 SO mol NaOH 15

6 ml of.1327 M KMnO 4 solution is needed to oxidize 25. ml of an acidic FeSO 4 solution. What is the molarity of the iron solution? WRITE THE CHEMICAL EQUATION! 5Fe 2+ + MnO H + Mn Fe H 2 O M rxn V volume red moles red moles oxid M oxid red coef. oxid ml =.1642 L 25. ml =.25 L.1327 mol KMnO 4 5 mol Fe L x x x =.4358 M 1 L 1 mol KMnO 4.25 L Fe Temperature and Solubility Solid solubility and temperature solubility increases with increasing temperature solubility decreases with increasing temperature Temperature and Solubility O 2 gas solubility and temperature solubility usually decreases with increasing temperature 18

7 19 Pressure and Solubility of Gases The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry s law). c = kp c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution k is a constant for each gas (mol/l atm) that depends only on temperature low P high P low c high c 19 2 Colligative Properties Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 1 Raoult s law If the solution contains only one solute: X 1 = 1 X 2 P 1 - P 1 = P = X 2 P 1 P 1 = vapor pressure of pure solvent X 1 = mole fraction of the solvent X 2 = mole fraction of the solute 2 21 Ideal Solution P A = X A P A P B = X B P B P T = P A + P B P T = X A P A +X B P B 21

8 22 Boiling-Point Elevation T b = T b T b T b is the boiling point of the pure solvent T b is the boiling point of the solution T b > T b T b > T b = K b m m is the molality of the solution K b is the molal boiling-point elevation constant ( C/m) for a given solvent Freezing-Point Depression T f = T f T f T f T f is the freezing point of the pure solvent is the freezing point of the solution T f > T f T f > T f = K f m m is the molality of the solution K f is the molal freezing-point depression constant ( C/m) for a given solvent

9 25 What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 322 g of water? The molar mass of ethylene glycol is 62.1 g. m = T f = K f m = mass of solvent (kg) K f water = 1.86 o C/m 478 g x 1 mol 62.1 g 3.22 kg solvent = 2.41 m T f = K f m = 1.86 o C/m x 2.41 m = 4.48 o C T f = T f T f = T f T f T f =. o C 4.48 o C = o C Osmotic Pressure (π) Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. Osmotic pressure (π) is the pressure required to stop osmosis. dilute more concentrated Osmotic Pressure (π) solvent solution time High P Low P π = MRT M is the molarity of the solution R is the gas constant T is the temperature (in K) 27

10 28 A cell in an: isotonic solution hypotonic solution hypertonic solution Colligative Properties Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. Vapor-Pressure Lowering P 1 = X 1 P 1 Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure (π) T b = K b m T f = K f m π = MRT 29 3 Colligative Properties of Electrolyte Solutions.1 m NaCl solution.1 m Na + ions &.1 m Cl - ions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles..1 m NaCl solution.2 m ions in solution van t Hoff factor (i) = actual number of particles in soln after dissociation number of formula units initially dissolved in soln nonelectrolytes NaCl CaCl 2 i should be

11 31 Colligative Properties of Electrolyte Solutions Boiling-Point Elevation T b = i K b m Freezing-Point Depression Osmotic Pressure (π) T f = i K f m π = imrt CH-13 HW Questions and Problems Pages , 4.56, 4.58, 4.6, 4.64, 4.66, 4.68, 4.78, 4.8. Pages , 13.1, 13.14, 13.16, 13.18, 13.2 (molarity only), 13.22,