Aqueous Solutions and Solubility Ch 3

Chapter 3 Aqueous Solutions and Solubility Problem 2.65. For each solution in this problem we are asked to calculate the molar concentration, molarity = moles per liter of solution, of a known mass of solute present in a known volume of solution. To get the molarity, we need to convert the given mass to moles and divide by the solution volume (in liters) to find the number of moles per liter: mass solute moles solute molarity These steps can be combined: 1 1 mol solute volume solution, L g solute In each case, we will need the molar mass of the solute and will calculate this first. molarity solute, mol·L–1 = (mass solute, g) (a) The molar mass of NaCl is: 22.99 g (1 mol Na) = 22.99 g 1 mol Na 35.45 g = 35.45 g (1 mol Cl) 1 mol Cl molar mass NaCl = 58.44 g = 58.4 g (accurate enough for the rest of the data) The molarity of NaCl in 0.120 L of solution containing 4.5 g NaCl is: molarity NaCl = (4.5 g NaCl) 1 mol NaCl 58.4 g 1 0.120 L = 0.64 M (b) The molar mass of NH4Cl is: 14.01 g (1 mol N) = 14.01 g 1 mol N 1.008 g = 4.03 g (4 mol H) 1 mol H 35.45 g = 35.45 g (1 mol Cl) 1 mol Cl molar mass NH4Cl = 53.49 g = 53.5 g (accurate enough for the rest of the data) The molarity of NH4Cl in 0.25 L of solution containing 1.3 g NH4Cl is: 1 1 mol NH 4 Cl = 0.097 M 53.5 g 0.25 L Or we get about 0.10 M, if the mass is only known to a precision of about 1 part in 13. molarity NH4Cl = (1.3 g NH4Cl) (c) The molar mass of AgNO3 is: 107.9 g = 107.9 g (1 mol Ag) 1 mol Ag 14.01 g = 14.01 g (1 mol N) 1 mol N 16.00 g = 48.00 g 1 mol O molar mass AgNO3 = 169.9 g = 170 g (accurate enough for the rest of the data) (3 mol O) The molarity of AgNO3 in 1.3 L of solution containing 1.85 g AgNO3 is: 1 1 mol AgNO3 = 0.0084 M 1.3 L 170 g Or we get about 0.008 M, if the volume is only known to a precision of about 1 part in 13. molarity AgNO3 = (1.85 g AgNO3) Problem 2.66. (a) We are asked to calculate the molar concentration, molarity = moles per liter of solution, of a known mass of glucose present in a known volume of solution. To get the molarity, we need to convert the given mass to moles and divide by the solution volume (in liters) to find the number of moles per liter: mass glucose moles glucose molarity These steps can be combined: 1 mol glucose 1 molarity glucose, mol·L–1 = (mass glucose, g) g glucose volume solution, L We will need the molar mass of glucose, C6H12O6, and will calculate this first. 12.01 g (6 mol C) = 72.06 g 1 mol C 1.008 g = 12.10 g (12 mol H) 1 mol H 16.00 g = 96.00 g (6 mol O) 1 mol O molar mass glucose = 180.16 g molarity glucose = (5.405 g glucose) 1 mol glucose 180.16 g 1 1.000 L = 0.03000 M (b) There are many ways to solve this part of the problem and any one that you can explain and justify to get the correct answer is OK. The way presented here may be useful in other situations involving moles/millimoles (mmol) and liters/milliliters (mL); these are especially common in biochemistry. Since 1 L = 1000 mL, a solution that contains x moles of a solute in one liter of solution will contain x millimoles of a solute in one milliliter of solution (one-thousandth of the solute in one-thousandth of the solution). For example, the solution in part (a) contains 0.03000 mmol of solute in each 1.000 mL of solution. The “rule” is this: the numeric value of a solution concentration in mmol·mL–1 equals the concentration in mol·L–1. In the problem we have here, we wish to convert a desired number of millimoles to the equivalent volume (in mL) of solution of a known concentration: 1.000 mL 0.950 mmol glucose = (0.950 mmol glucose) = 31.67 mL 0.03000 mmol You may have chosen a different route to calculate this volume, and, as we said at the beginning, that’s OK. Whatever way you choose is perfectly acceptable — as long as you understand the concepts and arrive at the correct answer. Problem 2.67. In order to prepare a 1.00 M solution of any solute, you have to dissolve one molar mass of the solute in one liter of solution (or an equivalent fraction of the molar mass and the volume). For this problem, we need to know the molar mass of the solutes each student used. The molar mass of CuSO4 is: 63.55 g (1 mol Cu) = 63.55 g 1 mol Cu 32.07 g = 32.07 g (1 mol S) 1 mol S 16.00 g = 64.00 g (4 mol O) 1 mol O molar mass CuSO4 = 159.62 g The molar mass of CuSO4·5H2O adds the mass of the five moles of water, 90.08 g, to the mass of the CuSO4 for a total of 249.70 g·mol–1. Both students used the correct procedure for making a solution of known concentration, but the male student neglected to account for the water of hydration and weighed too little of the solute: so prepared a solution of too low a concentration: 1 mol 159.60 g CuSO4·5H2O = (159.60 g CuSO4·5H2O) 249.70 g CuSO4 5H2 O = 0.639 mol He prepared a solution of too low a concentration: 0.639 mol·L–1 = 0.639 M. The female student used a solid of the correct composition, CuSO4, and prepared a solution of the correct composition: 1.000 mol·L–1 = 1.000 M. Problem 2.68. To make a solution with a fairly exact concentration of a solid solute, we need to weigh the solute carefully, add the solid to a volumetric flask, dissolve the solid in some water in the flask, and then make up the solution to the exact volume of the flask. Volumetric flasks are commonly available in a variety of sizes: 10 , 25 , 50 , 100 , 250 , 500 , and 1000 mL. We need about 170 mL of 0.10 M NaOH. In order to waste as little as possible of the solid reagent, we should use the smallest volumetric flask, 250-mL, that will give us enough of the solution. This is a problem for which we know the concentration of the desired solution, 0.10 M, and its volume, 0.250 L, so we need to convert the volume to the equivalent number of moles of solute and then moles to mass using the molar mass of NaOH (40.0 g·mol–1) to get the mass of NaOH that must be dissolved. The calculation may be done in a single step by combining the two conversions: 40.0 g 0.10 mol NaOH 0.250 L soln = (0.250 L solution) = 1.00 g NaOH 1 mol NaOH 1 L solution You may not have known the standard sizes of volumetric flasks and have chosen a different volume of solution to prepare. That’s OK. If you substitute your volume into this equation and get the result you calculated, you solved the problem correctly. Problem 2.69. We are first asked to determine the mass of sodium chloride, NaCl, required to make 250. mL of a 0.90% (mass to volume %) sodium chloride (normal saline) solution and then to find the molarity of this solution. The mass to volume % concentration is not discussed in the textbook, but its meaning is clear: the mass of solute is a given percentage (parts in 100) of the volume of the solution. The only ambiguities are the units to be used for mass and volume. In context, it makes sense to use grams and milliliters, because they will give a reasonable result (where “reasonable” means physically significant or possible). A 0.90% solution will have 0.90 g of solute per 100. mL of solution. In this case, volume of solution is given, 250. mL, and we wish to find the mass of NaCl solute: 0.90 g NaCl 250. mL solution = (250. mL solution) = 2.25 g NaCl 100. mL solution To find the molarity of this solution, we need to convert the mass of NaCl in a given volume (in liters) to moles of NaCl, using the molar mass of NaCl (58.4 g·mol–1), and then divide by the volume to get molarity (mol·L–1). We can either work with the mass of NaCl in 100. mL of solution (0.90 g) or the mass in 250. mL (2.25 g). Let’s use the former: 0.90 g NaCl = (0.90 g NaCl) 1 mol 58.4 g NaCl 1 0.100 L = 0.154 M Check to see that using 2.25 grams of NaCl in 250. mL of aqueous solution to calculate the molarity gives the same answer. Problem 2.70. We have a mass of solute, 25 g of urea, (NH2)2CO, in 2.5 L of solution (urine) and are asked to calculate the molarity of the solution. We need to convert the mass of urea to moles, using the molar mass of urea (60 g·mol–1), and then divide by the volume to get molarity (mol·L–1): 25 g urea = (25 g urea) 1 mol 60 g urea 1 2.5 L = 0.17 M Problem 2.71. The label on a sports drink tells us that 240 mL of the solution contains 30 mg of potassium. KH2PO4 is the only solution ingredient listed on the label that can provide this potassium. We are asked to determine the mass of KH2PO4 required to provide 30 mg of potassium [as K+(aq)] in the solution and to find the molarity of KH2PO4 in the solution. Note that each mole of KH2PO4 that dissolves in the solution provides a mole of K+(aq) in the solution. There are several ways to approach this problem, but they all require knowing the molar masses of K+ (39.1 g·mol–1) and of KH2PO4 (136 g·mol–1), because we need to convert masses to moles and moles to masses in this problem. In one approach, we follow these pathways: mass K+ moles K+ = moles KH2PO4 moles KH2PO4 mass KH2PO4 moles KH2PO4 molarity KH2PO4 Another approach takes these pathways: mass K+ mass KH2PO4 (using relative masses in one mole KH2PO4) mass K+ moles K+ molarity K+ = molarity KH2PO4 Note that the actual number of conversions in each approach is the same, so neither is more efficient or preferred, although the second approach can be condensed into two steps. We will go through both to show that the results are the same. In the first approach, we have (after converting milligrams to grams and milliliters to liters): 0.030 g K+ = (0.030 g K+) 1 mol 39.1 g K+ = 7.7 10–4 mol K+ = 7.7 10–4 mol KH2PO4 7.7 10–4 mol KH2PO4 = (7.7 10–4 mol KH2PO4) 136 g KH2 PO 4 1 mol = 0.10 g 7.7 10–4 mol KH2PO4 = (7.7 10–4 mol KH2PO4) 1 0.240 L 10–3 M KH2PO4 KH2PO4 = 3.2 In the second approach we have: 0.030 g K+ = (0.030 g K+) 136 (g KH 2 PO4 ) (mol KH2 PO4 ) + –1 39.1 (g K ) (mol KH2 PO 4 ) 0.030 g K+ = (0.030 g K+) 1 mol 39.1 g K+ 1 0.240 L = 3.2 –1 = 0.10 g KH2PO4 10–3 M K+ = 3.2 10–3 M KH2PO4 Thus, the results are the same, no matter which approach we use. If you used yet a different approach, got the same results, and can explain the concepts behind what you did, that’s fine. Problem 2.72. We need to remember that the relationship of numbers of atoms of the elements to a given number of molecules of a compound is the same as the relationship of numbers of moles of atoms of the elements to the given number of moles of the compound. For example, in two molecules of ammonium acetate, NH4C2H3O2, there are two atoms of nitrogen, N, and in two moles of NH4C2H3O2 there are two moles of nitrogen atoms, N. Also, in two moles of NH4C2H3O2, we have: 2 mol NH4C2H3O2 = (2 mol NH4C2H3O2) 2 mol C 1 mol NH 4 C2 H 3O2 = 4 mol C 2 mol NH4C2H3O2 = (2 mol NH4C2H3O2) 7 mol H 1 mol NH 4 C2 H 3O2 = 14 mol H 2 mol NH4C2H3O2 = (2 mol NH4C2H3O2) 2 mol O 1 mol NH 4 C2 H 3O2 = 4 mol O Summing the number of moles of all atoms (N, C, H, and O) in 2 mol NH4C2H3O2, we have 24 (= 2 + 4 + 14 + 4) mol of atoms. We are also asked for the total number of moles of ions in two moles of NH4C2H3O2. Although no discrete molecules of NH4C2H3O2 are present in a crystal of the ionic compound, the chemical formula indicates that the ratio of cations to anions is 1:1, so one formula unit (equivalent to a molecule) contains one ammonium cation, NH4+, and one acetate anion, C2H3O2–. Two formula units (molecules) of NH4C2H3O2 contain four ions (two cations and two anions), so two moles of NH4C2H3O2 contain four moles of ions: 2 mol NH4C2H3O2 = (2 mol NH4C2H3O2) 2 mol ions 1 mol NH 4 C2 H 3O2 = 4 mol ions There is more than one way to solve a problem like this and, if you solved it differently, but got the correct answers and can explain the concepts you used, that’s OK. Problem 2.73. We are asked to determine the mass of calcium phosphate that can be made by mixing 125. mL of 0.100 M calcium chloride with 125. mL of 0.100 M sodium phosphate. To solve this problem, we will first need the formulas for calcium chloride (CaCl2), sodium phosphate (Na3PO4), and calcium phosphate (Ca3(PO4)2). The formulas are based on balancing out the known charges on the cations and anions, which we can get from Table 2.2, if we don’t remember them. Combined with the solution concentration data, the first two formulas provide us enough information to determine the molarity of the reactant ions, Ca2+(aq) and PO43–(aq), in the solutions and, hence, the number of moles of each of these ions present in the mixed solution. We see that there is one mole of Ca2+(aq) in solution for each mole of CaCl2 that is dissolved. Thus, the concentration and number of moles of calcium cation is: 0.100 M CaCl2 = 0.100 mol CaCl2 1 L solution 2+ 1 mol Ca 1 mol CaCl 2 = 0.100 M Ca2+(aq) 0.100 mol Ca 2+ 0.100 M Ca (aq) = (0.125 L solution) = 1.25 1 L solution 2+ 10–2 mol Ca2+(aq) The same chain of reasoning gives the concentration and number of moles of phosphate anion: 0.100 mol Na 2 PO4 0.100 M Na3PO4 = 1 L solution 3– 1 mol PO4 1 mol Na 2 PO4 = 0.100 M PO43–(aq) 0.100 mol PO3– 4 0.100 M PO4 (aq) = (0.125 L solution) = 1.25 1 L solution 3– 10–2 mol PO43– (aq) Use the formula for the product of reaction, a precipitate of calcium phosphate (Ca3(PO4)2(s)), to write a balanced reaction for its formation from the ions: 3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s) The equation tells us that the mole ratio of Ca2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5 (= 3 2 ) mol of Ca2+(aq) is required to react completely with 1 mol of PO43–(aq). Since our mixed solution contains an equal number of moles of each ion, there is not enough Ca2+(aq) to react with all the PO43–(aq), so Ca2+(aq) is the limiting reactant in the mixture. We use the ratio of moles of calcium ions in a mole of calcium phosphate to convert number of moles of calcium ion that react to moles of solid calcium phosphate precipitated: 1 mol Ca 3 (PO 4 )2 1.25 10–2 mol Ca2+ = (1.25 10–2 mol Ca2+) 3 mol Ca 2+ = 4.17 10–3 mol Ca3(PO4)2(s) The molar mass of Ca3(PO4)2 is 310.2 g, so the mass of Ca3(PO4)2(s) formed is: 4.17 10–3 mol Ca3(PO4)2 = (4.17 10–3 mol Ca3(PO4)2) 310.2 g Ca 3 (PO4 )2 1 mol Ca 3 (PO4 )2 = 1.29 g Ca3(PO4)2(s) Problem 2.74. The coefficients in the balanced ionic equation give the relative number of moles of each reactant and product; 2Fe3+(aq) + SO32–(aq) + 3H2O(l) 2Fe2+(aq) + SO42–(aq) + 2H3O+(aq) These ratios can also be expressed in millimoles (1 mol = 1000 mmol). See the solution for Problem 2.66. The volume of 0.100 M SO32–(aq) needed to react exactly and completely with 24.0 mL of 0.200 M Fe3+(aq) is: 24.0 mL Fe3+ = (24.0 mL 2– 0.200 mmol Fe 3+ 1 mmol SO2– 1.000 mL SO3 3 Fe3+) 2– 1.000 mL Fe 3+ 2 mmol Fe3+ 0.100 mmol SO 3 = 24.0 mL SO32– solution Problem 2.75. (a) We are asked to determine the number of moles of each of the four ions, Na+, SO42–, Ba2+, and Cl–, in a mixture prepared by mixing 50.0 mL of a 0.45 M Na2SO4 solution with 50.0 mL of a 0.36 M BaCl2 solution. The number of moles of an ion in the mixture is the same as the number moles of that ion in its original solution. We need to convert the volumes (in liters) to moles of ionic compound solute, using the solution molarity, and then to moles of individual ions, using the ratio of moles of ions to moles of solute: 0.45 mol Na 2 SO 4 0.050 L Na2SO4 = (0.050 L Na2SO4) 1L + 2 mol Na 1 mol Na 2 SO4 = 0.045 mol Na+ 0.050 L Na2SO4 = (0.050 L Na2SO4) 0.45 mol Na 2 SO 4 1L 2– 1 mol SO4 1 mol Na 2 SO4 = 0.023 mol SO42– 0.050 L BaCl2 = (0.050 L BaCl2) = 0.018 mol Ba2+ 0.36 mol BaCl2 1L 2+ 1 mol Ba 1 mol BaCl 2 – 0.36 mol BaCl2 0.050 L BaCl2 = (0.050 L BaCl2) 1L 2 mol Cl 1 mol BaCl 2 = 0.036 mol Cl– (b) We are asked to determine how many moles of Ba2+(aq) are required to react with all the SO42–(aq) in the mixture, if the SO42–(aq) reacts with Ba2+(aq) to give BaSO4(s). We use the 1:1 reactant ratio to find the number of moles of Ba2+(aq) required: 2– 0.023 mol SO4 (aq) = (0.023 mol SO4 2– 2+ 1 mol Ba ) 2– 1 mol SO 4 = 0.023 mol Ba2+(aq) (c) We are asked to determine how many moles of SO42–(aq) are required to react with all the Ba2+(aq) in the mixture, if the SO42–(aq) reacts with Ba2+(aq) to give BaSO4(s). We use the 1:1 reactant ratio to find the number of moles of SO42–(aq) required: 1 mol SO2– 4 0.018 mol Ba (aq) = (0.018 mol Ba ) 1 mol Ba 2+ 2+ 2+ = 0.018 mol SO42–(aq) (d) To determine whether Ba2+(aq) or SO42–(aq) is the limiting reactant in this mixture, we compare the number of moles of each ion in the solution, from part (a), to the number of moles of that ion, from parts (b) and (c), required to react completely with the other ion. If the solution does not contain the requisite number of moles of one of the ions, that ion is the limiting reactant, since it will be used up before all the other ion has reacted. In this case, we see that 0.023 mol Ba2+(aq) are required to react with all the SO42–(aq), but the solution contains only 0.018 mol Ba2+(aq). Therefore, Ba2+(aq) is the limiting reactant in this mixture and it will react with 0.018 mol SO42–(aq), leaving about 0.005 mol SO42–(aq) behind in solution. Problem 2.76. The objective of this problem is to use our solubility rules, including the few exceptions we have noted, to predict the products of mixing four pairs of aqueous solutions and to determine the limiting reagent for each reaction and the mass of the precipitate (assuming that all precipitation reactions go to completion). To find the limiting reagent in each case we will also need to write a net ionic equation for the reaction that occurs (a) 125 mL of 0.15 M BaBr2 are mixed with 125 mL of 0.15 M Na3PO4. The mixed solution contains a multiply-charged cation, Ba2+(aq) and a multiply-charged anion, PO43–(aq), which react to form an insoluble precipitate: 3Ba2+(aq) + 2PO43–(aq) Ba3(PO4)2(s) The number of moles of each of these ions in the mixture is: 0.15 mol BaBr2 0.125 L BaBr2 = (0.125 L BaBr2(aq)) 1 L BaBr2 (aq) 2+ 1 mol Ba 1 mol BaBr2 = 0.019 mol Ba2+(aq) 0.125 L Na3PO4 = (0.125 L Na3PO4(aq)) = 0.019 mol PO43–(aq) 0.15 mol Na 3 PO 4 1 L Na 3 PO 4 ( aq) 3– 1 mol PO4 1 mol Na 3PO4 The equation tells us that the mole ratio of Ba2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5 (= 3 2 ) mol of Ba2+(aq) is required to react completely with 1 mol of PO43–(aq). Since our mixed solution contains an equal number of moles of each ion, there is not enough Ba2+(aq) to react with all the PO43–(aq), so Ba2+(aq) is the limiting reactant in the mixture. We use the ratio of moles of barium ions in a mole of barium phosphate to convert number of moles of barium ion that react to moles of solid barium phosphate precipitated: 1 mol Ba 3 (PO 4 )2 = 6.3 10–3 mol Ba3(PO4)2(s) 0.019 mol Ba2+ = (0.019 mol Ba2+) 2+ 3 mol Ba The molar mass of Ba3(PO4)2 is 602 g, so the mass of Ba3(PO4)2(s) formed is: 6.3 10–3 mol Ba3(PO4)2 = (6.3 10–3 mol Ba3(PO4)2) 602 g Ba 3 (PO4 )2 1 mol Ba 3 (PO4 )2 = 3.8 g Ba3(PO4)2(s) (b) 85 mL of 0.40 M NH4Cl are mixed with 65 mL of 0.50 M KNO3. The mixed solution contains only singly-charged cations and anions which form soluble ionic compounds, so there is NO APPARENT REACTION in this case. (c) 85 mL of 0.40 M (NH4 )2S are mixed with 65 mL of 0.50 M ZnCl2. The mixed solution contains a multiply-charged cation, Zn2+(aq) and a multiply-charged anion, S2–(aq), which react to form an insoluble precipitate: Zn2+(aq) + S2–(aq) ZnS(s) The number of moles of each of these ions in the mixture is: 0.065 L ZnCl2 = (0.065 L ZnCl2(aq)) 0.50 mol ZnCl2 1 L ZnCl2 (aq ) 2+ 1 mol Zn 1 mol ZnCl 2 = 0.033 mol Zn2+(aq) 0.085 L (NH4 )2S = (0.085 L (NH4 )2S(aq)) 0.40 mol (NH4 )2 S 1 L (NH4 )2 S(aq) 2– 1 mol S 1 mol (NH 4 )2 S = 0.034 mol S2–(aq) These ions react in a 1:1 ratio, so the ion whose concentration is lower, the Zn2+(aq), will be completely used up in the reaction. Thus, Zn2+(aq) is the limiting reactant and the mass of ZnS(s), molar mass = 97.4 g, that will be formed is: 97.4 g ZnS 1 mol ZnS = 3.2 g ZnS(s) 0.033 mol Zn2+ = (0.033 mol Zn2+) 2+ 1 mol ZnS 1 mol Zn (d) 15.0 mL of 0.20 M AgNO3 are mixed with 15.0 mL of 0.40 M NaBr. The mixed solution contains only singly-charged cations and anions, but these include Ag+(aq) and Br–(aq), which form a precipitate, AgBr(s), that is an exception to our rules: Ag+(aq) + Br–(aq) AgBr(s) The number of moles of each of these ions in the mixture is: 0.20 mol AgNO3 0.0150 L AgNO3 = (0.0150 L AgNO3(aq)) 1 L AgNO 3 (aq ) + 1 mol Ag 1 mol AgNO3 = 0.0030 mol Ag+(aq) 1 mol Br – 0.40 mol NaBr 0.0150 L NaBr = (0.0150 L NaBr(aq)) 1 L NaBr(aq) 1 mol NaBr – = 0.0060 mol Br (aq) These ions react in a 1:1 ratio, so the ion whose concentration is lower, the Ag+(aq), will be completely used up in the reaction. Thus, Ag+(aq) is the limiting reactant and the mass of AgBr(s), molar mass = 188 g, that will be formed is: 0.0030 mol Ag+ = (0.030 mol Ag+) 1 mol AgBr 1 mol Ag+ 188 g AgBr 1 mol AgBr = 5.6 g AgBr(s) Problem 2.77. (a) What is the white precipitate formed when 50. mL of an aqueous 0.1 M SrCl2 solution are mixed with 50. mL of an aqueous 0.1 M Na3PO4 solution? The mixed solution contains a multiply-charged cation, Sr2+(aq) and a multiply-charged anion, PO43–(aq), which react to form an insoluble precipitate, Sr3(PO4)2(s). The complete ionic equation representing the reaction that occurs in the mixed solution is: 3Sr2+(aq) + 6Cl–(aq) + 6Na+(aq) + 2PO43–(aq) Sr3(PO4)2(s) + 6Cl–(aq) + 6Na+(aq) (b) The complete ionic reaction shows that all the chloride ion from its original solution is still present in the mixture when the precipitation is complete. The number of moles of chloride and its mass may be calculated from the number of moles in its original solution volume: 0.050 L SrCl2(aq) = (0.050 L SrCl2(aq)) 0.1 mol SrCl 2 1 L SrCl2 (aq ) – 2 mol Cl 1 mol SrCl 2 = 0.01 mol Cl–(aq) 0.01 mol Cl– = (0.01 mol Cl–) 35.45 g 1 mol Cl – = 0.35 g Cl– The concentrations of the solutions are only given with one significant digit, so we really should not report this result with two figures, but we will need the greater precision for the later calculation to show the mixed solution’s electrical neutrality. (c) The number of moles of each of the other ions added to the mixed solution is: 0.1 mol SrCl 2 0.050 L SrCl2(aq) = (0.050 L SrCl2(aq)) 1 L SrCl2 (aq ) 2+ 1 mol Sr 1 mol SrCl 2 = 0.005 mol Sr2+(aq) 0.1 mol Na 3 PO4 0.050 L Na3PO4(aq) = (0.050 L Na3PO4(aq)) 1 L Na 3 PO 4 (aq) = 0.015 mol Na+(aq) + 3 mol Na 1 mol Na 3PO4 0.1 mol Na 3 PO4 0.050 L Na3PO4(aq) = (0.050 L Na3PO4(aq)) 1 L Na 3 PO 4 (aq) 3– 1 mol PO4 1 mol Na 3PO4 = 0.005 mol PO43–(aq) The equation in part (a) tells us that all of the Na+(aq) added to the mixture is still present after the precipitation reaction is complete. Now we need to find out whether any Sr2+(aq) or PO43– (aq) ions remain unreacted. Without doing any calculations, we can predict which ion will still be present after the precipitation is complete. Note that we found in part (b) that 0.01 mol of Cl– (aq) is present in the mixture and now we have 0.015 mol of Na+(aq) also left in the mixture. But the solution must be electrically neutral, so there must be more negative ions present. The only possibility is that some PO43–(aq) ions remain unreacted. Thus, it must be the case that Sr2+(aq) is the limiting reactant. Let’s see if that is the case and find out how much PO43–(aq) remains unreacted. The equation in part (a) tells us that the mole ratio of Sr2+(aq) to PO43–(aq) ions that react is 3:2. Thus, 1.5 (= 3 2 ) mol of Sr2+(aq) is required to react completely with 1 mol of PO43–(aq). Since our mixed solution contains an equal number of moles of each ion, there is not enough Sr2+(aq) to react with all the PO43–(aq), so Sr2+(aq) is the limiting reactant in the mixture, as we reasoned in the previous paragraph. We use the ratio of moles of strontium ions in a mole of strontium phosphate to convert number of moles of strontium ion that react to moles of solid strontium phosphate precipitated and then to find the number of moles of PO43–(aq) left unreacted: 1 mol Sr3 (PO4 )2 = 0.0017 mol Sr3(PO4)2(s) 0.005 mol Sr2+ = (0.005 mol Sr2+) 3 mol Sr 2+ Since each mole of Sr3(PO4)2 contains two moles of PO43–, we have 0.0034 mol of PO43– removed from solution as Sr3(PO4)2(s). The amount of PO43–(aq) remaining in solution is 0.0016 mol [= (0.005 mol) – (0.0034 mol)]. (d) The total number of moles of negative charge on the remaining PO43–(aq) anions, from part (c), is 0.005 mol [= (3 charges·mol–1)(0.0016 mol)]. This amount of negative charge added to that from Cl–(aq), 0.01 mol, is equivalent to the positive charge from the Na+(aq), 0.015 mol, so the solution is electrically neutral, as we said in the reasoning in part (c). Problem 2.78. We are asked whether gases are very soluble in water. There is no simple answer to this question, because the answer depends upon the interactions of the gas with water. Non-polar gas molecules like nitrogen and oxygen have only limited solubility because they are so volatile and interact only weakly by dispersion forces with the water. At the other end of the solubility scale are gases like ammonia and hydrogen chloride that hydrogen bond strongly with and/or react with water to form quite concentrated solutions. Intermediate solubilities are shown by gases like carbon dioxide that have polar bonds (dipole-dipole attractions with water) and react somewhat with water (as CO2(aq) does to form small amounts of H2CO3(aq)). In all cases, the solubility of gases is temperature dependent and decreases as temperature increases. Problem 2.79. We are asked to predict whether the noble gases (He, Ne, Ar, Kr, and Xe) have a low solubility in water (less than 1 g·L–1) or a high solubility in water (greater than 10 g·L–1). Recall from Section 2.11 that only when water reacts with a gas will the gas have a very high solubility. Since the noble gases are all rather unreactive (that is why they are called noble -- aloof and not mingling well with the more “common” elements), they show no tendency to react with or bond with water. The electron distribution around the noble gas nuclei is spherically symmetric, which makes them non-polar and further suggests that they are insoluble in water. Problem 2.80. (a) We are asked to determine how many moles of nitrogen gas, N2(g), dissolve in 10.0 L of water at a temperature of 25 C and gas pressure of 101 kPa (one atmosphere). The data in Table 2.6 give the solubility of N2(g) under these conditions as 0.018 g·kg–1 (grams of gas per kilogram of water). The density of water at 25 C and one atmosphere pressure is 1.00 kg·L–1 [Chapter 1, Figure 1.29(a)], so 10.0 L of water is 10.0 kg of water and the mass of N2 (g) that dissolves in this much water is 0.18 g [= (0.018 g·kg–1)(10.0 kg)]. Thus, the number of moles of nitrogen dissolved in 10.0 L of water is: 0.18 g N2 = (0.18 g N2) 1 mol N2 28 g N2 = 6.4 10–3 mol N2 (a) We are asked to determine how many moles of oxygen gas, O2(g), dissolve in 0.100 L of water at a temperature of 25 C and gas pressure of 101 kPa (one atmosphere). The data in Table 2.6 give the solubility of O2(g) under these conditions as 0.039 g·kg–1. As in part (a), 0.100 L of water is 0.100 kg of water and the mass of O2 (g) that dissolves in this much water is 0.0039 g [= (0.039 g·kg–1)(0.100 kg)]. Thus, the number of moles of oxygen dissolved in 0.100 L of water is: 0.0039 g O2 = (0.0039 g O2) 1 mol O2 32 g O2 = 1.2 10–4 mol O2 Problem 2.81. The solubility of H2(g) in water is temperature dependent. The solubility at 25 C is 7.68 10–4 mol L–1 and, at 0 C, the solubility is 9.61 10–4 mol L–1. Note that the solubility of this nonpolar molecule is quite low at both temperatures. There is evidently very little attraction of water molecules for hydrogen molecules. There are only dispersion interactions and with only two electrons, it is difficult to induce dipoles in hydrogen molecules. Since there is very little attraction between H2 and H2O, perhaps the explanation for the temperature dependence of the solubility lies in the motion of H2 at different temperatures. At 25 C, H2 are moving faster than 0 C. Thus, more molecules will have enough energy to escape into the gas phase at 25 C than at 0 C. This means that the solubility of H2 should be greater at 0 C than at 25 C, as observed. Problem 2.82. The solution that is formed when hydrogen bromide gas, HBr(g), dissolves in water to form an acidic solution is called hydrobromic acid, just as the analogous solution of HCl(g) is called hydrochloric acid, Problem 2.83. A saturated solution of HBr(g) in water (hydrobromic acid) is approximately 8.9 M in HBr(aq) and the density of the solution is about 1.5 kg·L–1. We are asked to express the solubility of HBr(g) in units of g·kg–1. The density of the solution tells us that a liter of the solution has a mass of 1.5 kg and we know, from the molarity, that this liter contains 8.9 mol HBr. The molar mass of HBr is 80.9 g, so the mass of HBr dissolved in the liter of solution is: 80.9 g HBr = 7.2 102 g HBr = 0.7 kg HBr 8.9 mol HBr = (8.9 mol HBr) 1 mol HBr The mass of water in the solution is 0.8 kg [= (1.5 kg solution) – (0.7 kg solute)], so the concentration of HBr(aq) (in units of g·kg–1)is: 7.2 10 2 g HBr = 9 102 g·kg–1 0.8 kg water Note that the mass of HBr(g) that dissolves in a kilogram of water is greater than the mass of HCl(g) that dissolves in water, 695 g·kg–1, but the number of moles of HBr(g) that dissolve, about 11, is only about half as much as the moles of HCl(g) that dissolve, about 19. On a molar basis, HCl(g) is much more soluble. Problem 2.84. (a) We are asked to illustrate each of the potential interactions that might explain the high solubility of HCl(g) in diethyl ether. For simplicity, diethyl ether, C2H5O C2H5, is written as R– O–R in these sketches showing (i) dipole-dipole, (ii) hydrogen bonding, and (iii) ionization of HCl in ether solution: (b) An experiment that might be done to eliminate or confirm one or more of these possibilities is testing the electrical conductivity of the solution, If it conducts, then we know that at least some HCl ionization, interaction (iii), must occur. This does not rule out contributions to the solubility from the other interactions. If the solution does not conduct, ionization is ruled out as a contributor to the solubility. The other interactions are hard to distinguish experimentally. NOTE: Stephen Branz, San Jose State University, has provided an interesting discussion of the possible reasons for the high solubility of HCl(g) in diethyl ether. If ionization reactions are primarily responsible for the solubility of HCl(g), we would expect the dielectric strength of the solvent to greatly affect the solubility. Solubility data for HCl(g) in water, MeOH, EtOH, and Et2O are from the Merck Index, 9th ed., and all other data are from standard reference tables. Solvent Solubility g·(100 mL)–1 at 30 °C Dielectric Constant water 67.3 78.4 methanol 43.0 33.6 ethanol 38.1 25.0 diethyl ether 19.5 4.3 benzene “soluble” * 2.3 pentane n/a # 1.8 chloroform n/a # 4.8 * no quantitative data available; several sources report “soluble.” # presumably very low Although there is indeed a reduction in the solubility of HCl going from water to the alcohols and ether, the differences are (qualitatively) not reflective of the differences in dielectric constants. Another way to look at the data is to calculate the solubility relative to the number of oxygen atoms in a 100 mL sample of the solvent. Solvent g HCl / mol “O” (in 100 mL solvent) water methanol ethanol diethyl ether 12.1 17.4 22.2 20.3 Assuming that there is a specific interaction between HCl and the oxygen of the solvent (and this might be anywhere from formation of the oxonium ion, R2OH+, to a strong hydrogen bond), it is justifiable to consider the solubility relative to the number of oxygen atoms in the solvent. By this measure, of the four oxygenated solvents, water is the worst solvent for HCl! A (perhaps naive) qualitative explanation is that of the four solvents in the table, only water has two H’s and two lone pairs for hydrogen bonding (or oxonium ion formation). Water must sacrifice H bonds (among water molecules) in order to interact with/ionize HCl. The other three solvents each have an “excess of lone pairs” and can more effectively solvate the HCl. The foregoing discussion is based on correlations with no physical data (such as conductivity, for which we have found no data) to back it up, but the correlation is quite good. Other support for a possible specific interaction (or interactions) between HCl and ether molecules should be noted: (a) Although we don’t have quantitative data for benzene, strong hydrogen bonding to the pi system is quite well documented and undoubtedly accounts for the high solubility of HCl in benzene. (b) The boiling points of ether and pentane (35-36 °C) are nearly identical, reflective of their very similar sizes and shapes. If the dipole moment of ether were of great importance in interactions with itself or other molecules, we would predict that its boiling point would be higher than that of pentane. (c) There are many well-known organic reactions that support the formation of low concentrations of ionic species in non-polar solvents (for example, electrophilic aromatic substitution). (d) Some non-polar solvents can be surprisingly good solvents for ionic species (for example, chloroform dissolves very high concentrations of simple organic ammonium chlorides even though those same salts are quite insoluble in ether -- these solvents have very similar dielectric constants). Problem 2.85. We would expect hydrogen chloride gas, HCl(g), to be less soluble in hexane than in water. Hydrogen chloride can react with water to form hydronium cations and chloride anions, equation (2.16). This reaction with the solvent increases the solubility of HCl(g) in water. There is no parallel reaction of HCl with hexane, CH3CH2CH2CH2CH2CH3, a non-polar solvent, so its hexane solubility is much lower. Problem 2.86. (a) The data in Table 2.6 are for the solubility of the gases at 101 kPa (one atmosphere) pressure. Assume that the amount of a gas that dissolves in water is directly proportional to its pressure over the solution; the lower the pressure, the less gas dissolved. (This is Henry’s Law for the solubility of non-polar gases in liquids.) In air, the nitrogen pressure is 80% of one atmosphere, so only 80% as much nitrogen will dissolve: (0.80)[0.018 g·(kg water)–1] = 0.014 g·(kg water)–1. Oxygen pressure is 20% of one atmosphere: (0.20)[0.039 g·(kg water)–1] = 0.008 g·(kg water)–1. (b) In one kilogram of water saturated with gases from the air, there are 0.014 g of nitrogen and 0.008 g of oxygen, or 0.022 g total of these gases. The mass percent of oxygen is: 0.008 g O2 100% = 36% 0.22 g air (c) In a mole of air (Avogadro’s number of nitrogen and oxygen molecules), 80% of the molecules are nitrogen and 20% are oxygen. The mass of nitrogen in a mole of air is (0.80 mol)(28 g·mol–1) = 22.4 g. The mass of oxygen is (0.20 mol)(32 g·mol–1) = 6.4 g. The total mass of a mole of air is 28.4 g and the mass percent of oxygen in air is: 6.4 g O2 100% = 22% 28.8 g air The mass percent of oxygen in the gases dissolved in water is greater than the mass percent of oxygen in the (major) gases in the atmosphere. Problem 2.87. An acid is a compound whose aqueous solution has a pH below 7. Problem 2.88. A base is a compound whose aqueous solution has a pH above 7. Problem 2.89. An acid solution that has a pH of 1 contains a higher concentration (100 times higher) of hydronium ions than an acid solution that has a pH of 3. From the data given, this is the only conclusion we can draw. We have no way to characterize acid A relative to acid B, because we do not know the concentrations of acid A and acid B in the solutions. Problem 2.90. Aqueous solutions with pH < 7, that is [H3O+(aq)] > 1.0 10-7 M and [OH–(aq)] < 1.0 10-7 M, are acidic. Aqueous solutions with pH > 7, that is [H3O+(aq)] < 1.0 10-7 M and [OH–(aq)] > 1.0 10-7 M, are basic. Aqueous solutions with pH = 7, that is [H3O+(aq)] = 1.0 10-7 M and [OH–(aq)] = 1.0 10-7 M, are neither acidic nor basic. Each of these solutions is characterized as acidic, basic, or neither on the basis of the definitions just stated. (a) pH < 7 acidic + -7 (b) [H3O (aq)] = 1.0 10 M neither – -7 (c) [OH (aq)] > 1.0 10 M basic (d) pH > 7 basic (e) [H3O+(aq)] > 1.0 10-7 M acidic (f) [OH–(aq)] < 1.0 10-7 M acidic + -7 (g) [H3O (aq)] < 1.0 10 M basic – -7 (h) [OH (aq)] = 1.0 10 M neither Problem 2.91. We calculate the pH of each of these solutions using the definition of pH [equation (2.18)]: pH = –log10[H3O+(aq)]. This is very easy to do on any scientific calculator, because they all have a “log” key that returns the logarithm to the base 10, log10, of whatever value is displayed. That is how these results were calculated. The number of significant digits following the decimal point in a pH value is determined by the number in the pre-exponential factor in the concentration value. In all the cases here, the pre-exponential has two digits and the pH has two digits after the decimal point. (a) [H3O+(aq)] = 1.0 10-2 M pH = 2.00 + -10 (b) [H3O (aq)] = 1.0 10 M pH = 10.00 + -4 (c) [H3O (aq)] = 5.0 10 M pH = 3.30 + -8 (d) [H3O (aq)] = 5.0 10 M pH = 7.30 Problem 2.92. If the reaction of HCl(g) and water goes to completion to form H3O+(aq) and Cl–(aq), the molar concentration of HCl(aq) in the solution is the same as the molar concentration of H3O+(aq). In this problem, we are asked to find the [HCl(aq)], which is equal to [H3O+(aq)], that will result in a solution of a given pH. Use Figure 2.24 to find the [H3O+(aq)] that corresponds to a given pH and that will also be the appropriate [HCl(aq)]. (a) For a pH = 4, Figure 2.24 shows that [H3O+(aq)] = 10–4 M, so we need [HCl(aq)] = 10–4 M. (b) For a pH = 2, Figure 2.24 shows that [H3O+(aq)] = 10–2 M, so we need [HCl(aq)] = 10–2 M. Problem 2.93. By analogy with HCl(g) being named hydrogen chloride and HCl(aq) named hydrochloric acid, we can name other gaseous hydrides that dissolve to form acidic solutions: HI(g) is hydrogen iodide. H2S(g) is hydrogen sulfide. HI(aq) is hydroiodic acid. H2S(aq) is hydrosulfuric acid.