Chemistry 121 Mines. Answer Key, Problem Set 13

Transcription

1 Chemistry 121 Mines Answer Key, Problem Set NT1; 2. NT2; 3. NT3; (Add: Do not just state the forces involved [too vague]; briefly indicate why you predict the solution would be formed); 5. NT4; 6. NT5; 7. NT6; 8. NT7; 9. NT Units of Concentration (and other physical quantities, like density, of solutions) 1. NT1. A solution of phosphoric acid was made by dissolving 10.0 g of H 3 PO 4 in ml of water. The resulting volume was 104 ml. Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of 1.00 g/ml. The following is (with slight modification) the solution from the published solutions guide (Thomas Hummel, Steven Zumdahl, and Susan Arena Zumdahl): (= M) (= 1.02 m) Molarity, Revisited (applied to mixtures of solutes or mixing of solutions) 2. NT2. Some K2Cr2O7, with a mass of g, is dissolved in enough water to make 500. ml of solution. (a) What is the molarity of the potassium dichromate? 1 mole of K2Cr2O7 has a mass of: 2(39.1) + 2(52.0) + 7(16.0) = g. So g is 2.335/294.2 ths of a mole or mol: g K Cr O g/mol K2Cr2O mol K Cr O PS13-1

2 Molarity = moles solute liter of solution mol K Cr O L solution M K Cr O = (b) What are the concentrations of the K + and Cr2O7 2- ions? [K + ] = 2 x [K2Cr2O7] = M (because there are two moles of K + ions per mole of K2Cr2O7) [Cr2O7 2- ] = [K2Cr2O7] = M (because there is one mole of Cr2O7 2- per mole of K2Cr2O7). (c) If 300. ml of M KNO3 were ADDED TO the 500. ml solution described in this problem, what would the final concentration of K + in the new solution be? (Note: assume that the final volume of the solution is 800. ml). Find TOTAL moles of K + in the new solution, and divide by the TOTAL volume of solution in L. Moles of K + from the 500. ml solution: mol K2Cr2O7 x 2 = mol K + Moles of K + from the 300. ml solution: (0.054 moles KNO3/L)(0.300 L)(1 mole K + /mol KNO3) = mol K + Total (new) volume = 800. ml = L So, mol K mol K + [K ]= M K L L Meaning of Solubility, Dynamic Equilibrium, Saturated vs. Unsaturated, Concentrated vs. Dilute 3. NT3. Consider the following snapshots of several different beakers, all at the same temperature, that have been sitting around for a while. Assume that if you waited awhile and looked again, the contents of the beakers would look essentially the same as they do now. Each contains water, which is not explicitly shown; each also contains a salt (see the key at the right), but not necessarily the SAME salt as in the others (i.e., one might have AgCl, one might have NaBr, etc.) (i) (ii) (iii) (iv) dissolved anion dissolved cation solid salt (undissolved) [NOTE: For all of the following, you should NOT consider the undissolved solid to be a part of the solution; it is a separate phase (a solid). Only ions that are dissolved are a part of a solution.] Answers: 1. (iii) and (iv); 2. (ii); 3. (iv); 4. (iv) NOTE: See Section 13.4 in Tro (especially Figure 13.9 on p. 581) for help with this problem! Explanations: 1. Which solution(s) are saturated? (iii) and (iv) These are the only ones that (you know) are saturated because there is some solid left undissolved (the solution is full or saturated with that solid). PS13-2

3 2. Which solution is the most concentrated? (ii) Concentration refers to HOW MANY PARTICLES are dissolved in a given volume of solution. The more particles there are in a given volume the higher the concentration. Since all volumes here are essentially the same, the one with the most particles DISSOLVED is the one with the highest concentration. 3. Which solution is the most dilute? (iv) Is the most dilute since it has the lowest concentration (fewest number of particles dissolved in a given volume). 4. Which beaker contains the salt that is the least soluble? (iv). The least soluble salt is the one that has the lowest concentration WHEN SATURATED. Beaker (iv) is the most dilute AND it is saturated, so it must contain the salt that is the least soluble. Factors that affect Solubility Determine whether each pair of compounds forms a homogeneous solution when combined. For those that form homogeneous solutions, indicate the type of forces that are involved. (Add: Do not just state the forces involved [too vague]; briefly indicate why you predict the solution would be formed) (a) CH3CH2CH2CH2CH3 and CH3CH2CH2CH2CH2CH3 Answer: Yes. We predict a solution to be formed since the self-self intermolecular forces between CH3CH2CH2CH2CH3 molecules are London forces, and that is also true for CH3CH2CH2CH2CH2CH3. When self-self IM forces are similar to self-other, there is no large, positive Hdissolution, and so dissolution is predicted to be favorable (due to the tendency for things to mix ). (b) CBr4 and H2O Answer: No. The self-self IM forces in CBr4 are only London forces, but the self-self IM forces in H2O are London + dipole-dipole + H-bonding. This means that when the water molecules are separated (to make room for the proposed solute, CBr4, some IM forces will have to be overcome which will not be compensated for by self-other IM forces with the nonpolar CBr4 upon dissolution. Thus Hdissolution will have a large and positive value, so much so that it will, in effect overcome the tendency for things to mix. (c) LiNO3 and H2O Answer: Yes (sort of). We predict a solution to be formed since the self-self intermolecular forces in these two substances are similar. LiNO3 is not molecular, but it is an ionic compound made up of a singly charged cation and singly charged anion, with the anion being particularly large (because it is polyatomic). As such, although the bonding is ionic, it is fairly weak for ionic and thus it is not all that much greater than the of the strong dipole-dipole and H-bonding forces between water molecules. And when an ionic compound dissolves (and dissociates!) in water, a new kind of intermolecular force appears: the ion-dipole force. This force is described in Chapter 11, but really belongs (in my opinion) in Chapter 13, since it is strictly a self-other kind of force. Basically, although it takes energy to separate the waters (due to London, dipoledipole, and H-bonding) and it takes energy to separate the ions in LiNO3 (due to ion-ion forces), a roughly similar amount of compensatory energy is released when the water hydrates the dissociated ions (due to ion-dipole forces). NOTE: This logic does not explain why so many ionic compounds are NOT soluble in water! It is clearly an overly simplistic argument. (d) CH3OH and CH3CH2CH2CH2CH3 PS13-3

4 Answer: No. The self-self IM forces in CH3CH2CH2CH2CH3 are only London forces, but the selfself IM forces in CH3OH are London + dipole-dipole + H-bonding. See answer to Part (b) above. 5. NT4. Which substance is most soluble in water? Give your reasoning. (i) C6H6 (ii) C2H5OH (iii) CH3Cl (iv) CO2 Comments/Explanation: To best predict solubility of a molecular substance in a molecular solvent, identify the intermolecular forces of attraction of the solute and solvent and then compare. The likelihood is that the solutes that have the greatest similarity of intermolecular forces with the solvent will be the most soluble in it. Why? At this point in your chemical education, the best explanation is that energy plays a partial role in dissolution. If the energy cost of separating the solute molecules from one another and the solvent molecules from one another are not at least reasonably compensated for by an energy release that comes from the solute molecules interacting with the solvent molecules, then the dissolution process tends to be less favorable. If the solute is polar or can hydrogen bond, the energy compensation will not likely come close to the energy cost unless the solvent is polar or can hydrogen bond as well (and the same logic applies if the solvent is polar). This is the basis for the simple phrase Like dissolves like. Polar solutes tend to dissolve in polar solvents, nonpolar solutes tend to dissolve in nonpolar solvents, and polar and nonpolar things do not dissolve in one another. And solutes that can participate in hydrogen bonding will be most soluble in solvents that can as well. So in this case, the solvent is water. Water is polar and its molecules can hydrogen bond to one another. C6H6 and CO2 are nonpolar molecules and so they will not be very soluble in water. CH3Cl is polar but cannot participate in hydrogen bonding. CH3OH is polar and can participate in hydrogen bonding, and so is predicted to be the most soluble in water (and it is). 6. NT5. Which solvent, water or carbon tetrachloride (CCl 4 ) would you choose to dissolve each of the following? (a) KrF2 (b) SF2 (c) SO2 (d) CO2 (e) MgF2 (f) CH2O (C is in the middle) (g) CH2=CH2 The following is essentially the solution from the published solutions guide (Thomas Hummel, Steven Zumdahl, and Susan Arena Zumdahl), although I ve tweaked it a bit by adding parenthetical comments. PS13-4

5 Polar (only one polar bond!), so more soluble in H2O Nonpolar (like all hydrocarbons), so more soluble in CCl4 7. NT6. Under what conditions of temperature and pressure is air (or most any other gas) most soluble in water? Give some reasoning. Answer: low temperature and high (partial) pressure (Note that these are the same two conditions under which a gas behaves most NON-ideally!) Reasoning/Comments: The solubility of gases depends on both temperature and pressure (unlike solids and liquids, which are only significantly dependent on temperature). It is most clear to see the effects of each if you start out with a system at dynamic equilibrium at a given T and P and then imagine how the opposing rates are affected by either a T change or a P change. Temperature effect: The solubility of gases decreases with increasing temperature, and so it increases with decreasing temperature. This makes sense if you think of the molecules of a gas already being separated from each other, but when in they are dissolved, they are surrounded by solvent molecules. That is, it takes energy for a dissolved gas molecule to escape the liquid, but it doesn t take energy for a gas molecule to be captured by the solvent. So an increase in temperature should increase the rate of escape more than the rate of capture, and so there will be less gas dissolved at dynamic equilibrium (which means a decreased solubility). Pressure effect. Consider gas dissolved in equilibrium with undissolved gas above it. The rates in and out of the solvent are the same. If you increase the partial pressure of the gas above the liquid, that means there will be MORE particles in a given volume, and so the rate of collisions into the solvent will become greater. More particles will end up in the (given) amount of solvent, and so the concentration will go up as dynamic equilibrium is reestablished (which means an increased solubility). PS13-5

6 Colligative Properties (Including strong electrolyte considerations) 8. NT7. Which aqueous solution (these are NOT pure substances!!!) has a) The smallest freezing point? b) The smallest boiling point? Explain. (i) 0.2 M Ca(NO3)2 (ii) 0.2 M CH3OH (iii) 0.2 M MgSO4 (iv) 0.2 M K3PO3 Answers: (a) Smallest freezing point: 0.2 M K3PO3 (highest [particles] = 0.8 M) (b) Smallest boiling point: 0.2 M CH3OH (lowest [particles] = 0.2 M) Explanation: The presence of particles dissolved in a solvent decreases the escaping tendency of solvent molecules, and thus inhibits both boiling and freezing. Thus the freezing point will decrease and the boiling point will increase with an increase in the concentration of PARTICLES dissolved (where particle means either an ion or a molecule). So, in a solution with the highest concentration of particles, the following will be true: the freezing point will be decreased the most (freezing point will be lowest) and the boiling point will be raised the most (boiling point will be greatest). It thus follows that the solution with the smallest concentration of particles will be the one that has the smallest decrease in freezing point (and thus the highest freezing point) as well as the smallest increase in boiling point (and thus the smallest boiling point). In order to determine the concentration of particles, one must first identify whether the solute is an electrolyte or a nonelectrolyte. If it is a nonelectrolyte, then the concentration of particles simply equals the concentration of solute dissolved. But if the solute is an electrolyte, one must then determine how many ions there are per formula unit AND whether or not the electrolyte is strong or weak (not a part of THIS particular problem, but it could be [if a weak acid is given, for example]). Then one can determine the total concentration of particles from the concentration of solute and the number of ions per formula unit. Ca(NO3)2 is ionic, so it is a strong electrolyte and it will separate into its ions in aqueous solution. In this case, that means 3 ions per formula unit (one Ca 2+ and two NO3 - ions). Thus the concentration of particles in a 0.2 M Ca(NO3)2 solution is actually 0.6 M particles. CH3OH is molecular (and not an electrolye) so a 0.2 M solution of CH3OH will have a concentration of particles of just 0.2 M. (lowest of all four). ***The OH here does not indicate that this is ionic! If it were indicating OH -, then the cation would have to be CH3 + and we have never seen this as a cation before [because it isn t]). -OH attached to a carbon is called an alcohol group; CH3OH is methanol, remember?!*** MgSO4 is ionic, but only has two ions per formula unit (Mg 2+ and SO4 2- ) so the concentration of particles in a 0.2 M MgSO4 solution is 0.4 M particles. K3PO3 is ionic and has four ions per formula unit (three K + ions and one PO3 3- ion), so a 0.2 M solution of K3PO3 will have a concentration of 0.8 M particles. (highest of all four). 9. NT8 (a) What is the mass percent concentration of the following solution: g of KBr is dissolved in 5.00 ml of water. (b) What is the molality of the same solution? (c) What would be the freezing point of the solution, assuming a k f of 1.86 C/m for water and a normal freezing point of 0.00 C? Answers: (a) 2.63 %; (b) m KBr; (c) C PS13-6

7 0.27 Answer Key, Problem Set 13 Reasoning: mass of solute (a) mass percent = x 100, so you need to find two things: (total) mass of solution mass of solute and mass of the solution. Mass of the solute here is given, but the volume of solvent (water, here) is given. So we need to use the density of water (assume it to be 1.00 g/ml) to calculate the mass of water from its volume: 5.00 ml x 1.00 g/ml = 5.00 g. Mass of solute: Given as g Mass of solution = mass of solute plus mass of solvent = g g g Mass percent x % g 5.00 g (b) Molality = moles of solute kg of solvent, so we need to find two things: the number of moles of KBr and the mass (in kg) of water. Moles KBr: Since the molar mass of KBr is = g/mol, g is a small fraction of a mole indeed! Specifically it is: g 19.0 g/mol mol KBr Mass of water: 5.00 g 5.00 g x 1 1 kg 000 g kg mol KBr So the molality kg H O 2 m(c) Tf = kf x mparticles = 1.86 C/m x (0.227 m) x 2 moles ions / mol KBr = C Thus, the freezing point is depressed by C. If the fp of (pure) water is 0.00 C, then the fp of this particular aqueous solution is C below 0.00 C: fp(sol n) = C = C = C PS13-7