Chem 1515 Section 2 Problem Set #4. Name Spring 1998

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1 Chem 1515 Section 2 Problem Set #4 Name Spring 1998 TA Name Lab Section # ALL work must be shown to receive full credit. Due Wednesday, February 4th PS4.1. Describe all the energy changes which must be considered in the solution process. Indicate the types of interactions which are important in the solution process and discuss what conditions favor and do not favor formation of a solution. There are two primary energy factors which must be considered in the solution process. The H solution and S solution. The first term, H solution depends on three steps; the energy required to separate the solute particles, the energy required to separate the solvent particles, and the energy released when new solute-solvent interactions form. The second term, S solution is related to the mixing process and always favors solution formation. When H solution is negative (exothermic) formation of a solution is favored. So if both terms favor formation of a solution is happens. If H solution is positive (endothermic) formation of a solution may or may not be favored depending on the relative magnitudes of H solution and S solution. Since it is a little difficult (at the moment) to get quantitative information on H solution and S solution we used a simple rule to help us predict solution formation.. like dissolves like. This rule is interpreted in terms of the intermolecular attractive forces between the pure solute and pure solvent particles. If they are similar we can expect a solution to form when the two components are mixed. If the forces are dissimilar we would predict no solution formation, i.e. a heterogeneous mixture. PS4.2. Which substance of each of the following pairs is likely to be more soluble in water? Explain each choice and, for the less soluble of each pair, suggest a better solvent. a) ethane(g) or acetic acid(l) Acetic acid will be more soluble than ethane because acetic acid can hydrogen bond with water. Ethane has only dispersion forces, and can not hydrogen bond with water. Another nonpolar solvent such as benzene, carbon tetrachloride or carbon disulfide would be a better solvent for ethane. b) chlorine(g) or hydrogen chloride(g) Hydrogen chloride will be more soluble than chlorine because hydrogen chloride is a polar covalent compound which when dissolved in water dissociates into ions. Chlorine has only dispersion forces, and can not hydrogen bond with water. Another nonpolar solvent such as benzene, carbon tetrachloride or carbon disulfide would be a better solvent for chlorine.

2 PS4.2. (Continued) c) hexane(g) or C 6 H 12 O 6 (s) Glucose will be more soluble than hexane because glucose can hydrogen bond with water. Hexane has only dispersion forces, and can not hydrogen bond with water. Another nonpolar solvent such as benzene, carbon tetrachloride or carbon disulfide would be a better solvent for hexane. d) ammonia(g) or phosphine(g) Ammonia will be more soluble than phosphine because ammonia can hydrogen bond with water. Phosphine is polar and can interact with water, but is not as soluble as ammonia. Another polar solvent such as dichloromethane, chloroform or acetone would be a better solvent. PS4.3. Describe the attractive forces present when NaCl(s), HCl(g) and acetic acid(l) dissolve in water. Use the space below to sketch diagrams depicting at the atomic level how each of the three substances interact with water molecules. This figure shows the ion-dipole interaction between the chloride ion and the water molecules. There are two more water molecules which could have been drawn. These are located in front and behind the chloride ion. Notice the orientation of the water molecules. The δ+ end of the dipole is oriented towards the negatively charged chloride ion. This figure shows the ion-dipole interaction between the sodium ion and the water molecules. There are two more water molecules which could have been drawn. These are located in front and behind the sodium ion. Notice the orientation of the water molecules. The δ end of the dipole is oriented towards the positively charged sodium ion. February 15, Spring 1998

3 PS4.3. (Continued) HCl(g) is a polar covalent molecule. When added to water hydrogen chloride completely ionizes to form H + and Cl ions. These species are hydrated by water. The H + exists as H 3 O + in solution. The H 3 O + ion hydrogen bonds to additional water molecules as shown. The chloride ion is oriented so it can interact (ion-dipole) with water molecules. February 15, Spring 1998

PS4.3. (Continued) Acetic acid, HC 2 H 3 O 2 is a weak acid and partially dissociates into acetate, C 2 H 3 O 2 -, and hydrogen, H +, ions. The interaction with water is via hydrogen bonding.

4 PS4.3. (Continued) Acetic acid, HC 2 H 3 O 2 is a weak acid and partially dissociates into acetate, C 2 H 3 O 2 -, and hydrogen, H +, ions. The interaction with water is via hydrogen bonding. The H 3 O + ion hydrogen bonds to additional water molecules as shown. HC 2 H 3 O 2 is a polar covalent molecule which partially ionizes when added to water. The water molecules hydrate the HC 2 H 3 O 2 molecule as shown hydrogen-bonding with OH group. Water will also hydrogen bond with the acetate, C 2 H 3 O 2 - ion as shown. Notice the hydrogen on the oxygen is absent in acetate compared to acetic acid. February 15, Spring 1998

5 PS4.4. Describe how the solubility of an ionic solid can depend on temperature. Describe how the solubility of a gas can depend on temperature. The solubility of ionic solids may increase or decrease with increasing temperature. It depends on whether the H solution is exothermic or endothermic. The solubility of gases always decreases with increasing temperature. PS4.5. A concentrated solution of acetic acid, CH 3 COOH, contains g CH 3 COOH dissolved in 85.6 g water. The density of the solution is g ml 1. Calculate; a) The weight percent CH 3 COOH in the solution, mass mass soln x g 122 g soln x 100 = 30.0 % b) the mol fraction of CH 3 COOH in the solution, g HC 2 H 3 O g = mol HC 2 H 3 O g H 2 O 18.0 g = 4.76 mol H 2 O χ = mol HC 2H 3 O 2 mol soln mol χ = 5.37 mol = c) molality of CH 3 COOH in the solution, mol HC 2 H 3 O 2 kg H 2 O mol HC 2 H 3 O kg = 7.13 molal d) the molarity of CH 3 COOH in the solution. 122 gm soln 1 ml g = 117 ml mol HC 2 H 3 O 2 L soln mol HC 2 H 3 O L = 5.2ar February 15, Spring 1998

6 PS4.6. An aqueous solution of sodium bromide is molal and has a density of g ml. Calculate the mol NaBr 1 kg H 2 O mol NaBr 103 gm = 667 g NaBr a) weight percent sodium bromide. 667 g NaBr 667 g NaBr g H 2 O b) mole fraction of sodium bromide. x 100 = 40.0% 1,000 g H 2 O 18.0g mol NaBr mole fraction NaBr = mol soln c) molarity of the solution. = = 55.6 mol H 2 O mol mol = mol NaBr Liter soln = molarity 1667 g soln 1 ml g mol = 1.18 Liters = 5.49 molar = 1.18 L February 15, Spring 1998

7 PS4.7. Describe how you would prepare the following aqueous solutions; a) ml of a M NaOH solution L mol 1 Liter = mol mol 40.0 g = 7.50 gm NaOH Add 7.50 gm NaOH to a 500 ml volumetric flask. Add about 400 mls of water and stir the mixture until all of the NaOH dissolves. Then add enough water until the final volume of solution is 500 mls. (Note: we do not have to worry about how much water is added, only the final volume.) b) 145 g of a 8.5 % (by weight) solution of CoCl 2 using CoCl 2 6H 2 O. 145 g soln 8.5 g CoCl g soln = 12.3 g of CoCl g CoCl 2 CoCl g CoCl 2 6H 2 O CoCl 2 = mol CoCl 2 6H 2 O mol CoCl 2 6H 2 O 238 g CoCl 2 = g CoCl 2 6H 2 O Of the g of CoCl 2 6H 2 O, g 12.3 g, g are water. We need 145 g of solution, 12.3 g is and g are water. But g of water are added from the CoCl 2 6H 2 O so only g of water need to mixed with g of CoCl 2 6H 2 O. c) 950. g (grams of solution) of a molal propylene glycol (C 3 H 8 O 2 ) solution mol C 3 H 8 O 2 1 kg H 2 O mol C 3 H 8 O gm = 57.0 gm C 3 H 8 O gm soln 57.0 g C 3 H 8 O g g = 51.2 gm C 3 H 8 O gm soln g C 3 H 8 O 2 = 899 g H 2 O Add 51.2 g C 3 H 8 O 2 to 899 g H 2 O and stir. February 15, Spring 1998

8 PS4.8. A concentrated solution of NaOH in water is prepared by mixing 30.0 g NaOH with 70.0 g of water and cooled back to 25 C. The molarity was found to be M. Calculate a) the molality of the solution 30.0 g NaOH 40 gm 0.75 mol NaOH kg H 2 O = 10.7 molal = 0.75 mol NaOH b) the density of the solution mass solution density = vol solution 1 liter of solution contains mol of NaOH. To determine the mass of solution in 1 liter of solution, then mol of NaOH 40.0 g 399 g solute 100 g solution 30.0 g solute density = = 399 g solute 1330 g solution 1000 ml solution = 1.33 g ml = 1330 g solution February 15, Spring 1998

Chapter 12. Physical Properties of Solutions. Chemistry, Raymond Chang 10th edition, 2010 McGraw-Hill

Chapter 12. Physical Properties of Solutions. Chemistry, Raymond Chang 10th edition, 2010 McGraw-Hill Chemistry, Raymond Chang 10th edition, 2010 McGraw-Hill Chapter 12 Physical Properties of Solutions Ahmad Aqel Ifseisi Assistant Professor of Analytical Chemistry College of Science, Department of Chemistry