Solutions for Chapter 2 End-of-Chapter Problemsquantum.bu.edu/.../text/answers/Chapter02.pdfDecember...

December 2004 ACS Chemistry Chapter 2 suggested solutions 1

Solutions for Chapter 2 End-of-Chapter Problems

Problem 2.1. Student answers will vary. This is an example. Use a pipette to collect samples from various places throughout the mixture. Analyze each sample for properties like density, boiling point, freezing point, or specific gravity. Compare the results to each other. If the solution is homogeneous all the results should be the same (within the experimental uncertainty).

Problem 2.2. Some examples of homogeneous solutions you might find at the supermarket or a pharmacy might be: soft drinks, saline solutions for contact lenses, rubbing alcohol, liquid bleach, and vinegar. Some examples of heterogeneous solutions (suspensions) might be: milk, Milk of Magnesia, Peptobismol, calamine lotion, and paint.

Problem 2.3. (a) An energy diagram for the change of a liquid (water) going to a gas is shown in Figure 1.35 and this problem is here to help recall the earlier introduction. The energy diagram for this general case with the energy change labeled is:

(b) For gaseous solvent molecules solvating a solute to form a liquid solution, we can think about the reaction occurring two steps, first condensing the gas to a liquid and then dissolving the solute in the liquid. The first step is exothermic, because condensing a gas to a liquid always releases energy. The second step, dissolution of the solute, may be either exothermic or endothermic, although not endothermic enough to make the overall two-step process endothermic, because the energy released in condensation of solvents is quite substantial. Thus, the overall process will be exothermic, but the solvation step may by endothermic or exothermic. If solvation by the liquid solvent is endothermic, we have:

Aqueous Solutions and Solubility Chapter 2

2 ACS Chemistry Chapter 2 suggested solutions

If solvation by the liquid solvent is exothermic, we have:

Problem 2.4. (a) The solution feels cool to the touch because the process is drawing needed thermal energy from your hand, providing the energy needed for the overall solution process. This is a sketch of the energy diagram showing this endothermic solution process:

(b) The solution feels warm to the touch because the process is adding thermal energy to your hand, because the overall solution process releases energy. This is a sketch of the energy diagram showing this exothermic solution process:

Chapter 2 Aqueous Solutions and Solubility

ACS Chemistry Chapter 2 suggested solutions 3

Problem 2.5. The energy diagram for the overall solution process for the exothermic dissolution of calcium chloride, CaCl2, in water is:

Problem 2.6. Whether you can predict that a substance will be soluble in water by looking at its line formula depends to some extent on how the formula is actually written. For example, the line formula C2H6O, does not tell you how the atoms are connected, so the best you can do is predict that the molecule will probably be polar (because of the oxygen bonded to other less electronegative atoms) and, since it is a small molecule is likely to be reasonably soluble in water. However, if the line formula is written as C2H5OH, you can predict that this low molecular mass alcohol should be very soluble (actually it is miscible mixes in any ratio) in water. If the line formula is written as CH3OCH3, you can predict that this low molecular mass ether should be somewhat soluble in water, but probably not as soluble as the alcohol (which can act as donor of an H to form H bonds as well as an acceptor of H from water to form other H bonds. A structural formula will always show the connectivity of the atoms and will, therefore, always provide the kind of information we just discussed for simple line formulas written to help us understand the connectivity. The regions of more positive and more negative charge in the molecule will be relatively easy to locate and an estimate of the polar and non-polar contributions to the solubility (or insolubility) will help you predict whether the substance will dissolve in water or not.

Problem 2.7. The like-dissolves-like expression reflects the fact that attractions between solute molecules and some attractions between solvent molecules must be replaced by solute-solvent attractions when a solution forms. If the new attractions are similar to those replaced, we expect that a solution will be more easily formed. A polar liquid, such as water, is generally the best solvent for polar compounds, especially those with H bonding sites. Non-polar liquids, such as hexane are better solvents for non-polar compounds like, for example, wax, whose molecules are mostly attracted by dispersion forces.



Problem 2.8. (a) Two representations of the testosterone structure are:

The dipoles are easier to show on the structure on the left (with fewer atoms shown explicitly):

(b) The acetylsalicylic acid (aspirin) structure, with polar bonds circled and direction of dipoles shown, is:

(c) The methyl salicylate (oil of wintergreen) structure, with polar bonds circled and direction of dipoles shown, is:



Problem 2.9. (a) The dashed lines in this diagram represent the network of hydrogen bonds that can form between ethanol and water.

Note that there are two H bonds between H atoms in water molecules and nonbonding electron pairs on the oxygen of ethanol, and another H bond between the OH groups hydrogen and a nonbonding electron pair on the oxygen in water. The water molecules have other nonbonding pairs on oxygen and covalently bonded hydrogen, all of which are capable of extending the network of hydrogen bonds. In Chapter 1, we found that there are, on the average, fewer than four H bonds per water molecule in liquid water. The three H bonds formed by ethanol fits right into this structure and the rather small ethyl group, CH2CH3, probably does not perturb the structure very much, so ethanol molecules fit well into the liquid water structure and thus account for their miscibility with water. (b) The alcohol group in pentanol, CH3CH2CH2CH2CH2OH, can also H bond with water, as shown for ethanol in part (a). However, the solubility of pentanol is expected to be a good deal lower than the solubility of ethanol because the nonpolar hydrocarbon part of the molecule is larger and has a much larger effect on the structure of the liquid water. The solubility of pentanol in water is about 0.3 M (about 27 gL1).

Problem 2.10. (a) Cyclohexane, C6H12, is a nonpolar liquid while methanol, CH3OH, is a polar liquid. The interactions among cyclohexane molecules in the liquid are induced dipole attractions (dispersion forces). For methanol, the largest interactions among the molecules in the liquid are H bonds; methanol acts much like water as a solvent. In a mixture of methanol and cyclohexane, the molecules of the two liquids lose some of their freedom of movement to make way for each other. This type of reorganization of unlike molecules is unfavorable and limits their mutual solubility, so they are not miscible. They are, however, rather soluble in one another: 100 mL of methanol dissolves 57 g of cyclohexane. Interestingly, ethanol (with its slightly larger alkyl group) and cyclohexane are miscible with one another, so we have to be careful not to push the like dissolves like (or unlikes do not dissolve) simplistic idea to extremes. (b) Naphthalene, C10H8, is a nonpolar solid and water is a polar, hydrogen-bonding liquid. The reorganization of the H-bonding structure of liquid water in order to accommodate naphthalene molecules is unfavorable. In this case water molecules would lose some of their freedom of movement while making way for naphthalene molecules. (c) Both naphthalene, C10H8(s), and benzene, C6H6(l), are nonpolar molecules. The reorganization involved in mixing two nonpolar compounds, both of which interact mainly by



induced dipole attractions (dispersion forces), favors the mixed state, so the solid naphthalene dissolves in the liquid benzene. (d) Molecules of water can form hydrogen bonds with 1-propanol, CH3CH2CH2OH, like those shown for methanol in the solution to Problem 2.9(a). The nonpolar part of the molecule, like those of methanol and ethanol, apparently does not disturb the liquid water structure enough to make the interactions unfavorable. Thus mixing of the two liquids, water and 1-propanol, is not impaired and they mix in all proportions (are miscible).

Problem 2.11. Since gasoline, C8H18, is a nonpolar molecule, we predict that it will be pretty insoluble in water. Its interactions with water molecules will be unfavorable as the freedom of movement of the water molecules is impaired.

Problem 2.12. (a) The Lewis structures for 1-hexanol, CH3(CH2)5OH, and 1,6-hexanediol, HO(CH2)6OH, with their regions for hydrogen bonding identified, are:

1-hexanol

1,6hexanediol, HO(CH2)6OH

(b) 1,6-hexanediol is predicted to be more soluble in water than 1-hexanol, because compounds with multiple polar groups present more opportunity for hydrogen bonding with water molecules.

Problem 2.13. Recall (from Section 2.2) that, for alcohols, the solubilizing and hydrogen bonding effects of the polar hydroxy (OH) group becomes less and less important as the hydrocarbon portion of the alcohol increases in size. To a first approximation, the solubility is related directly to the ratio of carbon atoms to polar groups. In part (a), the polar group is an alcohol and in part (b), the polar group is an amine, NH2, which can also hydrogen bond with water. The ratios and rank orders of solubilities are shown (#1 is the most soluble) here: carbon/polar grp ratio solubility rank (a) CH3-CH2-CH2-CH2-OH 4/1 3 CH3-CH2-CH2-CH2-CH2-OH 5/1 4 HO-CH2-CH2-CH2-CH2-OH 4/2 = 2/1 2 HO- CH2-CH2-CH2-OH 3/2 = 1.5/1 1



carbon/polar grp ratio solubility rank

(b) CH3CH2CH2CH2NH2 4/1 1 (CH3)2CHCH2CH2NH2 5/1 2 (CH3)3CCH2CH2NH2 6/1 3 (CH3 CH2)3CCH2NH2 8/1 4

Problem 2.14. [There are many books (especially beginning biochemistry texts) and web resources that have the structures of these sugars. We are not interested here in the exact stereochemistry of these molecules (for which most students are unprepared at this stage), but in the connectivity that shows the several alcohol groups in each one. The structures here are simply modeled after that for glucose in Figure 2.6 without showing the nonbonding electrons.]

The structure of fructose might be found in either the pyranose (six-membered ring) or furanose (five-membered ring) form:

C CC

OCC

CH2OHHOHO

HHO

H HOH

H

H

CC C

CO CH2OH

H

OHHH

HO

HOJ2CHO

In both structures, there are five alcohol (OH) groups that can H bond with water, as well as a ring oxygen atom with nonbonding electrons that can also donate electron pairs to H bonds with water. Looking, as in the solution to Problem 2.13, at the ratio of number of carbons to number of alcohol groups, we have 6/5 = 1.2/1. This ratio is comparable to that for methanol (1/1) and ethanol (2/1), both of which are miscible with water. The high solubility of fructose is expected.

The structures of lactose and sucrose are a bit more complicated, because they are disaccharides combinations of two simpler sugars, two glucose molecules in lactose and glucose and fructose in sucrose:

C CC

OCC

CH2OHH

HHO H OH

H

OH

HC C

COC

C

CH2OHHHO

HHO H OH

H

O

H

C CC

OCC

CH2OHHHO

HHO H OH

H

O

H

C

C CC

O

OH

CH2OHCH2OH

HOH

HH

lactose sucrose

Lactose and sucrose both have eight alcohol (OH) groups that can H bond with water, as well as the two ring oxygen atoms and the oxygen atom bonding the rings together which have nonbonding electrons that can also donate electron pairs to H bonds with water. (One alcohol group from each of the simple sugars has been lost in forming the disaccharides via a reaction we can write as R1OH + R2OH R1OR2 + H2O.) Looking again at the ratio of number of carbons to number of alcohol groups, we have 12/8 = 1.5/1. Again, this ratio is comparable to that for the simplest alcohols and helps to explain the high solubility of these sugars.



Problem 2.15. The more polar groups present in a molecule the more soluble it will be in water. Conversely, the fewer the polar groups and the larger the nonpolar portion of a molecule the less soluble it will be in water and the more soluble it will be in nonpolar solvents.

CC

CC

CC

CC

CC

OC

C

CC

C

CCH

H

HH H

H

H H HH

HH

C CH

HH HHHH

H H

H

H

HH

H

H

H

C

CO

CC

OOH

H

OC

HC

OH

OH

H

H

H

vitamin A vitamin C

Vitamin A has a single polar alcohol group and a substantial nonpolar part, so it will be insoluble in water, but relatively soluble in nonpolar solvents like the fats in our body. Vitamin A is a fat-soluble vitamin. Vitamin C has three polar alcohol groups that can H bond with water as well as two other oxygen atoms with nonbonding electrons that can also donate electron pairs to H bonds with water. Using our ratio of number of carbon atoms to number of alcohol groups (as in the solutions to Problems 2.13 and 2.14), we have 5/4 = 1.25/1, which indicates that vitamin C should be quite soluble in water. Vitamin C is a water-soluble vitamin.

Problem 2.16. (a) Vitamins that are soluble in the fatty tissues in our bodies will tend to stay in the body for a substantial period of time. Vitamins that are soluble in water will dissolve in fluids like our blood plasma and can be transported to the kidneys, removed from the blood stream, and excreted. The fat-soluble vitamins D, E, and K can be stored in your body, although they are slowly being used and lost and need to be part of your diet (which is part of the definition of what constitutes a vitamin), but not in large amounts. Water soluble, vitamin B is not stored and needs to be included in your daily diet at a larger dosage, in order to maintain an appropriate level in your body. (b) Vitamin A can be stored in your body, while vitamin C should be included in your daily diet in relatively large amounts. (c) Since only fat-soluble vitamins (vitamins A, D, E, and K) can be stored in your body, true hypervitaminosis has been observed only for these vitamins. This does not mean that you are entirely safe ingesting excessive doses of water-soluble vitamins. Vitamin C is a case in point. One fad is to take massive doses of vitamin C, in order to prevent colds. The problems with this regimen occur when the dosage is lowered. Your body becomes used to the high level and, if it is reduced to the normal level, you begin to show the signs of scurvy, a vitamin C deficiency disease.

Problem 2.17. In order for the light bulb to glow when testing a solution for electrical conductivity, as in Investigate This 2.10, ions must be present in the solution.



Problem 2.18. In the representations of the Fe3+(aq) and NO3

(aq) ions in the Web Companion, Chapter 2, Section 2.3, page 1, movies, both ions have six water molecules around them in their hydration layer. The positively charged Fe3+(aq) ions attract the negative (oxygen) ends of the water molecules, so the water molecules are oriented with their oxygen atoms directed toward the ion. The negatively charged NO3

(aq) ions attract the positive (hydrogen) ends of the water molecules, so the water molecules are oriented with their hydrogen atoms directed toward the ion.

Problem 2.19. Solution A, ethanoyl chloride (acetyl chloride), CH3C(O)Cl, dissolved in water, must contain ions, because it conducts and electric current. Solution B, 2-chloroethanol, ClCH2CH2OH, dissolved in water, does not conduct an electric current, so does not contain ions. The identity of the ions in solution A cannot be determined from these data.

Problem 2.20. (a) Barium chloride, BaCl2(s), dissolves in water to give barium cations, Ba2+(aq), and chloride anions, Cl(aq). (b) Potassium chloride, KCl(s), dissolves in water to give potassium cations, K+(aq), and chloride anions, Cl(aq). (c) Sodium triphosphate, Na3PO4(s), dissolves in water to give sodium cations, Na+(aq), and phosphate anions, PO4

3(aq). (e) Ammonium chloride, NH4Cl(s), dissolves in water to give ammonium cations, NH4+(aq), and chloride anions, Cl(aq). (f) Sodium sulfide, Na2S(s), dissolves in water to give sodium cations, Na+(aq), and sulfide anions, S2(aq). (g) Magnesium sulfate, MgSO4(s), dissolves in water to give magnesium cations, Mg2+(aq), and sulfate anions, SO4

2(aq).

Problem 2.21. (a) If current flow through a conducting solution were simply caused by a flow of electrons through the solution, we would not expect any chemical changes to occur in the solution. However, in Investigate This 2.10, you probably observed bubbles of gas formed at the electrodes when electric current passed through the conducting solutions. Formation of the gases is a sign that chemistry occurs as a result of current passing through the solution. Some mechanism for chemistry to occur, coupled with current flow, is necessary to explain the observations. Movement of ions in the solution provides a mechanism that explains both current flow and the observation that chemical reactions occur at the electrodes, presumably when the ions pick up electrons from or deliver them to the electrodes. See Chapter 10, Section 10.1 for a discussion of the chemistry that occurs at electrodes. (b) The Web Companion, Chapter 2, Section 2.3, page 2, represents the motion of cations and anions in a solution, so you can see how they are always in motion. You have to indicate which ion moves which way when responding to this page, so the assumption is made that the ions move and are the source of the electric current in the solution. If your friend is not convinced by the argument in part (a), s/he may not be convinced by an illustration of the mechanism that is



in the textbook. The movement of colored ions in a conducting medium, as in electrophoresis (introduced in Chapter 8, Section 9.5), might be more convincing

Problem 2.22. Neither solid NaCl nor solid HgCl2 will conduct electricity because any ions present are not free to move and transport the charge. An aqueous solution of sodium chloride is a good conductor because sodium cations, Na+(aq), and chloride anions, Cl(aq), surrounded by polar water molecules, are free to move about in the solution. Although HgCl2 is soluble in water, the aqueous solution does not conduct electricity. Evidently no ions are present when this compound goes into solution. The molecule itself stays together in solution as HgCl2(aq). This is not the usual circumstance for what appears to be a salt, but mercury(II) chloride molecules stay bonded together even in aqueous solution.

Problem 2.23. This is a representation of a positively charged ion surrounded by polar molecules like our simple ellipsoids with positive and negative ends in Chapter 1, Figures 1.15 and 1.16.

If you are the positive ion, you will feel a good deal of attraction from the partial negative charges on the surrounding solvation sphere of polar molecules. Other positive or negative ions elsewhere in the solution are farther away and have little influence on your behavior. They, too, are surrounded by solvating polar solvent molecules and are little influenced by your charge.

Problem 2.24. (a) MgBr2(s) dissolves in water to produce Mg2+(aq) and Br(aq) ions, so the solution will conduct an electric current (by movement of the ions). (b) CH3OH(l) dissolves in water and forms H bonds with the water molecules, but produces no ions, so the solution will not conduct an electric current (there are no ions to carry the current). (c) NaOH(s) dissolves in water to produce Na+(aq) and OH(aq) ions, so the solution will conduct an electric current (by movement of the ions). (d) CH3OCH3(g) dissolves in water and forms some H bonds with the water molecules, but produces no ions, so the solution will not conduct an electric current (there are no ions to carry the current). (e) KNO3(s) dissolves in water to produce K+(aq) and NO3(aq) ions, so the solution will conduct an electric current (by movement of the ions). (h) CH3CH2CH2CH3(g) dissolves to a very limited extent in water and produces no ions, so the solution will not conduct an electric current (there are no ions to carry the current).



Problem 2.25. (a) Alkali metals (column I of the periodic table) lose their single valence electron to form positive ions (cations) with a 1+ charge. Familiar examples are Na+ and K+ and members of a family ( column of the table) have similar properties. (b) Oxygen family elements (column VI of the periodic table) gain two electrons in their valence shell to form negative ions (anions) with a 2 charge. The oxide, O2, and sulfide, S2, anions are familiar examples and we expect other members of the family to have similar properties. (c)Alkaline earth metals (column II of the periodic table) lose their two valence electrons to form positive ions (cations) with a 2+ charge. Familiar examples are Mg2+ and Ca2+ and we expect members of the family to have similar properties. (d) Halogens (elements in column VII of the periodic table) gain an electron in their valence shell to form negative ions (anions) with a 1 charge. The chloride, Cl, and bromide, Br, anions are familiar examples and we expect other members of the family to have similar properties.

Problem 2.26. (a) The magnesium cation, Mg2+, and bromide anion, Br, combine to form the electrically neutral ionic compound magnesium bromide, MgBr2. (b) The calcium cation, Ca2+, and nitrate anion, NO3, combine to form the electrically neutral ionic compound calcium nitrate, Ca(NO3)2. (c) The magnesium cation, Mg2+, and sulfate anion, SO42, combine to form the electrically neutral ionic compound magnesium sulfate, MgSO4. (d) The potassium cation, K+, and oxide anion, O2, combine to form the electrically neutral ionic compound potassium oxide K2O.

Problem 2.27. (a) Na2SO4 is sodium sulfate, an ionic compound of the Na+ cation and SO42 anion. (b) MgCl2 is magnesium chloride, an ionic compound of the Mg2+ cation and Cl anion. (c) (NH4)2CO3 is ammonium carbonate, an ionic compound of the NH4+ cation and CO32 anion. (d) Al2S3 is aluminum sulfide, an ionic compound of the Al3+ cation and S2 anion.

Problem 2.28. (a) The electrically neutral formula for barium nitrate, a combination of the Ba2+ cation and NO anion, is Ba(NO3)2. (b) The electrically neutral formula for ammonium phosphate, a combination of the NH4+ cation and PO4

3 anion, is (NH4)3PO4. (c) The electrically neutral formula for calcium oxide, a combination of the Ca2+ cation and O2 anion, is CaO. (d) The electrically neutral formula for potassium sulfate, a combination of the K+ cation and SO4

2 anion, is K2SO4.



Problem 2.29. (a) MgS is magnesium sulfide, an ionic compound of the Mg2+ cation and S2 anion. (b) Na3PO4 is sodium phosphate (or sometimes trisodium phosphate), an ionic compound of the Na+ cation and PO4

3 anion. (c) NH4NO3 is ammonium nitrate, an ionic compound of the NH4+ cation and NO3 anion. (d) LiOH is lithium hydroxide, an ionic compound of the Li+ cation and OH anion.

Problem 2.30. (a) The electrically neutral formula for calcium iodide, a combination of the Ca2+ cation and I anion, is CaI2. (b) The electrically neutral formula for sodium fluoride, a combination of the Na+ cation and F anion, is NaF. (c) The electrically neutral formula for potassium carbonate, a combination of the K+ cation and CO3

2 anion, is K2CO3. (d) The electrically neutral formula for barium hydroxide, a combination of the Ba2+ cation and OH anion, is Ba(OH)2.

Problem 2.31. The completed grid for ionic compound formation by the cation/anion pairs is:

CO32 PO4

3 F

Mg2+ MgCO3 magnesium carbonate

Mg3(PO4)2 magnesium phosphate

MgF2 magnesium

fluoride

NH4+ (NH4)2CO3

ammonium carbonate

(NH4)3PO4 ammonium phosphate

NH4F ammonium

fluoride

Al3+ Al2(CO3)3 aluminum carbonate

AlPO4 aluminum phosphate

AlF3 aluminum fluoride

Na+ Na2CO3 sodium

carbonate

Na3PO4 sodium

phosphate

NaF sodium fluoride

Problem 2.32. (a) Milk of Magnesia is a suspension of the sparingly soluble ionic compound Mg(OH)2, magnesium hydroxide. (b) Epsom salt is MgSO4, magnesium sulfate. (c) Plaster of Paris is CaSO4, calcium sulfate. (d) Caustic soda (often called lye) is NaOH, sodium hydroxide. (e) Soda ash is Na2CO3, sodium carbonate.



Problem 2.33. The reaction equations for formation of the common cations or anions of these elements are: (a) potassium K(g) K+(g) + e (b) calcium Ca(g) Ca2+(g) + 2e (d) sulfur S(g) + 2e S2(g) (e) bromine Br(g) + e Br-(g)

Problem 2.34. By definition, ionization always involves the separation of a negative charge (an electron) and a positive charge (the cation that remains after the electron has departed). If energy is released when opposite charges come together (energy has a negative value for the process), then the reverse process requires the input of energy (energy has a positive value for the process). Mathematically, the energy of coulombic attraction, E (Q1Q2)/d, equation (2.2), expresses the energy of attraction of opposite charges when Q1 and Q2 have opposite signs. This attraction must be overcome (a reversal of the mathematical sign from negative to positive) in order to separate the opposite charges.

Problem 2.35. We are asked whether lattice energies, such as those in Table 2.3, are consistent with electrical attraction energies characterized by equation (2.2), E (Q1Q2)/d. If they are consistent, we would expect the lattice energies to be roughly proportional to the product, |Q1Q2| for a series of ionic compounds. The distance separating the ions, d in equation (2.2), might also affect the lattice energies, but simple cations are about the same size, as are simple anions, so the distances of separation are probably not very different and we will focus on the effects of charge. We are interested only in relative values, so let us assign values of Q as the charges we write on the ions. For our comparisons, two of the ionic compounds from Table 2.3 are chosen here (but any other of the 1:1 cation-to-anion compounds could be chosen):

Compound Cation Q1 Anion Q2

Q1Q2 Elattice, kJmol1

NaBr Na+ 1+ Br 1- 1 751

MgS Mg+2 2+ S2 2- 4 3406

We see here that the lattice energies are roughly proportional to Q1Q2 . An increase by a factor of four in the product of the charges is accompanied by an increase of about 4.5 in the lattice energy. Thus, the lattice energy data are consistent with coulombic attraction energy. Other factors, such as the distance of separation of the ions and the geometric arrangement of the ions with respect to one another, also affect lattice energies, but charge is most important.

Problem 2.36. (a) Based on coulombic electrical attraction, CaBr2(s) would have greater forces of attraction and repulsion than KBr(s), if the distance separating the charges is the same in both crystals. The double charge on Ca2+ will result in larger coulombic forces, because they are directly related to the size of the charges interacting.



(b) The data in Table 2.3 supports our answer in part (a). The lattice energy for CaBr2, 2176 kJmol1, is about three times the lattice energy of KBr, 689 kJmol1.

Problem 2.37. The problem statement gives us Elattice = 2176 kJmol1 and Eion form = 966 kJmol1 for calcium bromide, CaBr2(s). Combining these values in an energy diagram, we have:

We equate the energies for the two pathways from separated atoms to the ionic crystal to find Extal form: Extal form = Eion form + ( Elattice) = (966 kJmol1) + (2176 kJmol1) Extal form = 1210 kJmol1 (as shown on the energy diagram)

Problem 2.38. This is the table of lattice energies (in kJmol1) we are to use for this problem.

F Cl Br

Li+ 1046 861 818

Na+ 929 787 751

K+ 826 717 689

(a) As we go across any row of this table, the size of the anion increases, because the size of the ions in a group (column) of the periodic table increases as we go down the group and that is what we are doing going from F to Br. We note that the lattice energy decreases across each row, indicating that the lattice energy decreases as the size of the anion increases (and the cation remains constant). This makes sense, because the distance between the cations and anions in the crystal increases and we see from equation 2.2 that the energy of attraction between unlike charges decreases as the distance between them increases. (b) Analysis of the data for any column of this table is exactly like that in part (a) with the roles of the cation and anion reversed. As we go down a column the size of the cation increases while the size of the anion remains constant. The lattice energy decreases as the size of the cation increases, because the distance between the cations and anions in the crystal increases and the energy of attraction between unlike charges decreases as the distance between them increases.



(c) The lattice energy of a salt decreases as the size of its ions increases. The lattice energies are largest when the ions are the smallest. (d) For CsI, the cation, Cs+, is larger than any of the alkali metal cations in the table and the anion, I, is larger than any of the anions in the table. The lattice energy for CsI should be even lower than the lowest value in the table (that for KBr) and certainly considerably smaller than that of NaCl. The lattice energy for CsI is 604 kJmol1.

Problem 2.39. This is the energy diagram for the formation of one mole of ionic crystals of MgCl2 that we are to use for this problem.

(a) The lattice energy, Elattice, for MgBr2 is the energy required to convert one mole of the compound from its solid crystal to separated ions in the gas phase. On the diagram, we see that 2524 kJmol1 is released when the gaseous ions come together to form the ionic solid. Therefore, Elattice = 2524 kJmol1, the amount of energy to get the ions apart. (b) From the energy diagram, the energy required to change gaseous atoms to gaseous ions, Eion form = 1490 kJmol1. (c) The energy of formation of ionic crystals from gaseous atoms, Extal form, is the difference between the two energies in parts (a) and (b): Extal form = 1034 kJmol1 as shown on the diagram. (d) This table shows the comparison between these energies for MgBr2 and those for NaCl from Figure 2.14.

Ionic compound

Elattice kJmol1

Eion form kJmol1

Extal form kJmol1

MgCl2 2524 1490 1034

NaCl 787 145 642

Note that the large difference between the lattice energies is largely a result of the much higher energy required to remove two electrons from magnesium atoms to form the doubly charged Mg2+(g) cations compared to removal of only one electron to form the Na+(g) cation. This large amount of energy is about 60% of the energy released when the coulombic attraction of the doubly-charged cations and singly-charged anions brings them together to form the ionic solid.



Problem 2.40. (a) We can use the information that the energy required to remove electrons from gaseous silver atoms to form gaseous silver cations is 731 kJmol1 and that 296 kJmol1 of energy is released when gaseous iodine atoms gain electrons to form gaseous iodide anions, to construct an energy diagram to find the energy change for the net reaction: Ag(g) + I(g) Ag+(g) + I(g). The loss of an electron by each Ag(s) atom and gain of an electron by each I(g) atom is represented on the diagram by the slanting arrow showing that the electron lost by a silver atom is gained by an iodine atom, so there is no net loss or gain of electrons in the overall reaction The diagram shows that the net reaction energy is 435 kJmol1.

(b) Use the result from part (a) and the lattice energy for AgI(s) crystals, Elattice = 887 kJmol1, to draw an energy diagram analogous to Figure 2.14 and use it to find Extal form for the formation of ionic crystals of AgI(s) from the gaseous atoms.

Problem 2.41. In the energy diagrams we have seen in the textbook or constructed for the formation of solid ionic compounds, we have seen this relationship among the energies represented on the diagrams:



Extal form = Eion form + ( Elattice) = Eion form Elattice For KBr(s) crystals, the problem statement says that Elattice = 689 kJmol1. The formation of separate gaseous atoms of potassium, K(g), and bromine, Br(g), from the ionic crystal requires 594 kJmol1, so Extal form (energy change for the reverse of process, forming the ionic crystal from the gaseous atoms) = 594 kJmol1. Substituting these values in the preceding equation, gives: 594 kJmol1 = Eion form (689 kJmol1) Eion form = 95 kJmol1 An energy diagram incorporating these data is shown here (not quite to scale, so the numeric values can be included with the arrows representing them):

Problem 2.42. As in the solution for Problem 2.41, to find the lattice energy, Elattice, for magnesium fluoride, MgF2(s), we use the energy relationship:

Extal form = Eion form + ( Elattice) = Eion form Elattice Substituting the values given in the problem statement, Extal form = 1424 kJmol1 and Eion form = 1533 kJmol1, we get:

1424 kJmol1 = (1533 kJmol1) Elattice Elattice = 2957 kJmol1 An energy diagram incorporating these data is shown here:



Problem 2.43. (a) When ammonium acetate, NH4C2H3O2 [= (NH4+)(C2H3O2)], is dissolved in water, the mixture becomes quite cold. This observation means that the dissolution process is taking thermal energy from its surroundings, the molecules in the solution, the container, and your hand, if you are holding the container. A reaction that requires an input of thermal energy, Ereaction > 0, is endothermic. (b) The energy change for the process of dissolving ionic solutes in water can be broken into two parts. Lattice energy is required to break the coulombic electrical attractions between cations and anions in the lattice while hydration energy is released as water molecules surround these ions and are attracted to them by coulombic attractions. The difference between these two energies determines whether thermal energy will be absorbed (endothermic) or released (exothermic) by the dissolving process. For ammonium acetate, the dissolving process absorbs energy, so breaking the lattice attractions must take more energy than is gained back by hydration of the ions. An energy diagram that represents this case is:

[Note that water is shown separately with the solid and gaseous ions as a reminder that the system as a whole contains the water into which the solid dissolves and which hydrates the ions. Sometimes, as in Figures 2.15 and 2.16 in the text and the solution to Problem 2.44 below, the water is omitted for simplicity, but this is probably not a good idea.] (c) The ions present in a solution of ammonium acetate are the same as the ions in the solid crystal, ammonium cation and acetate anion, which we write as NH4

+(aq) and C2H3O2(aq) to



signify that the ions are hydrated, that is, surrounded by polar water molecules that are attracted by the ionic charges. (d) The molecular level interactions of hydrated ions are represented in Figure 2.9 by showing the negative oxygen end of water molecules oriented toward the positive cations and the positive hydrogen end of water molecules oriented toward the negative cations. We expect similar orientations and interactions for the NH4

+ and C2H3O2 ions, but there is an added factor

for these ions, because they can hydrogen bond with the water molecules and form even more directed interactions. Some of these H bonds are illustrated here:

N HH

H

H

O H

H

O H

H

O H

H

OH

H

OH

H

CH3C O

O

OHH

OH

H

OH

H

Problem 2.44. (a) Use the data in Table 2.3 for LiCl, Elattice = 861 kJmol1 and Ehydration = 898 kJmol1, to sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to find that Edissolve = 37 kJmol1. Energy is released, so the dissolution reaction is exothermic.

(b) Use the data in Table 2.3 for KBr, Elattice = 689 kJmol1 and Ehydration = 670 kJmol1, to sketch an energy diagram for the dissolution process (broken into two steps). Use the diagram to find that Edissolve = 19 kJmol1. Energy is required, so the dissolution reaction is endothermic.



Problem 2.45. We are asked to try to explain why lithium sulfate, Li2SO4, is quite soluble in water (261 gL

1) while calcium sulfate, CaSO4, is essentially insoluble (4.9 mgL

1). To see what factors might be responsible for such a difference, lets compare the data for a pair calcium ionic compounds in Table 2.3:

Compound Elattice kJmol1

Ehydration kJmol1

CaCl2(s) 2260 2337

CaCO3(s) 2804 2817

In CaCl2(s) and CaCO3(s), we have the cation with a 2+ charge and anions with a 1 charge and a 2 charge, respectively. CaCl2(s) is quite soluble in water, as you found in Investigate This 2.22, and CaCO3(s), marble or chalk, is quite insoluble. As a first approximation, these compounds are rather like Li2SO4(s) and CaSO4(s), where we have the anion with a 2 charge and cations with a 1+ charge and 2+ charge, respectively. The ion-ion interactions in the solids and ion-polar solvent interactions in aqueous solution should be roughly the same for an ionic compound with 1+ cations and a 2 anion as for a compound with a 2+ cation and 1 anions. If this supposition is correct, we see from the comparison in this table that the hydration energy should substantially outweigh the lattice energy, for a 2:1 ionic compound like Li2SO4(s) compared to a 2:2 ionic compound like CaSO4(s). As a first suggestion about the factor responsible for the high solubility of Li2SO4(s) compared to CaSO4(s), we would probably say that the hydration energy (due to solvation of the ions) favors dissolution of Li2SO4(s), Edissolve = 67 kJmol1, more than CaSO4(s), Edissolve = 13 kJmol1. This is a large factor in this case, but we have to be careful making to much of this argument, since we know that there are soluble ionic compounds whose dissolution is endothermic, so energetics cannot be the whole picture.



Problem 2.46. We asked to describe or represent in three different ways what happens when sodium sulfate, Na2SO4(s), dissolves in water. First we describe the dissolution in words. Water molecules are attracted to the Na+ and SO4

2 ions in the solid crystal lattice, especially those at the edges and corners of the crystals. The negative oxygen ends of the water molecules are attracted to the Na+ and positive hydrogen ends in water are attracted to SO4

2-. These ion-dipole attractions compete with the ion-ion attractions, lattice energy, that holds Na+ and SO4

2- in the crystal. In the case of a soluble ionic compound like Na2SO4(s), the ion-dipole attractions are finally successful in the tug of war and the ions are broken away from the crystal. Once the ions are broken away from the crystal, more water molecules surround each ion creating a hydration layer. These hydration layers create a shield, making it difficult for hydrated ions that are oppositely charged to get too close to each other. In our example, hydrated Na+ and hydrated SO4

2- do not interact much with one another (until their concentrations become so high that they are forced close to one another because there are so many of them). An ionic equation that succinctly represents this process is usually written as:

Na2SO4(s) 2Na+(aq) + SO42-(aq) Although water is obviously involved in the process, it is usually left out of such equations, because its stoichiometry is not easily represented. If we just add H2O(l) to the reactant side of the equation, it might be interpreted as one molecule of water for each formula unit of Na2SO4 dissolved, and that would be misleading. We just have to remember that water is an active reactant in the dissolution process, not an inert by-stander. A molecular level representation of the dissolution process is shown in this very rough sketch showing a two-dimensional crystal interacting with solvent (water):

+2 2

222

2

+

+

+

+ + +

+++

++

2

+

S

SS S

SS S

S

S

SS

S

S

S S

S S

S S

SS

Na+ cation

SO42 anion

solvent, water molecules

For simplicity, the many water molecules that are not interacting with the ions are omitted and the polar ends of the water molecules are not represented. See Figure 2.9(b) for a representation of the orientation of the water molecules with respect to the cations and anions.



Problem 2.47. (a) The mixing process described in this problem is represented in this table, modeled after Table 2.4 in the text:

Before Mixing After Mixing

Na3PO4 solution

CaBr2 solution

Na3PO4 and CaBr2

Positive ion(s) Na+ (aq) Ca2+(aq) Na+(aq)+ Ca2+(aq)

Negative ion(s) PO4-(aq) Br(aq) PO4

- (aq)+ Br (aq)

Conductivity? yes yes yes

Precipitate? yes

(b) After mixing, two new combinations of cations and anions are possible: NaBr and Ca3(PO4)2. Our solubility rules say that ionic compounds of alkali metal cations and halide anions are soluble, so sodium bromide, NaBr, is likely to be a soluble compound. Calcium phosphate, Ca3(PO4)2, must be the precipitate and it fits our solubility rule that says that ionic compounds of multiply-charged cations and anions are likely to be insoluble. (c) This diagram is a simple molecular level representation of this mixing and reaction (precipitate formation) with only enough of each ion shown to represent the stoichiometry of the reaction:

Note that each of the individual solutions that are mixed contains ions and there are also ions in the solution remaining after the solid has precipitated. The presence of the ions in all three solutions explains why they are all conduct an electric current, as shown in the table above. (d) The complete ionic equation that represents the precipitation reaction involves all four ions: 6Na+(aq) + 2PO4

(aq) + 3Ca2+(aq) + 6 Br(aq) 6Na+(aq) + 6Br--(aq) + Ca3(PO4)2(s) (e) The net ionic equation that represents the precipitation reaction involves only the two ions that react to form the precipitate (the spectator ions are omitted):

3Ca2+(aq) + 2PO4(aq) Ca3(PO4)2(s)

Problem 2.48. (a) When aqueous solutions of potassium chloride, KCl, and sodium bromide, NaBr, are mixed, the solution contains all four ions, Na+(aq), K+(aq), Cl(aq), and Br(aq), and no precipitate is formed. We can conclude that both NaCl(s) and KBr(s) are water soluble, since the solids could



have formed from this mixture, but did not. We also note that all of these ionic compounds are combinations of an alkali metal cation and a halide anion, which our solubility rules predict will be soluble. (b) This molecular level representation, similar to Figure 2.17, illustrates the result when the two solutions are mixed.

(c) The complete ionic equation representing what happens when the two solutions are mixed involves all four ions:

K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) The ions are present in two separate solutions on the left-hand (reactant) side of this equation and together in a single solution on the right-hand (product) side. The separation on the left is not represented in this standard ionic equation, but we can amend it slightly to suggest the separation by bracketing the separate solution components on the left:

[K+(aq) + Cl-(aq)] + [Na+(aq) + Br-(aq)] K+(aq) + Cl-(aq) + Na+(aq) + Br-(aq) Since there is no net reaction (no apparent reaction of any kind), there is no net ionic equation for this mixing. In a sense, all the ions are spectator ions.

Problem 2.49. The solubility rules indicate that multiple charged cations and anions tend to form insoluble ionic compounds. Barium sulfate, BaSO4, falls into this category. The mixture that doctors use to x-ray the gastrointestinal (GI) tract is a suspension of insoluble solid barium sulfate in water. The solubility of the solid ionic compound is so low that the suspension contains only a tiny amount of Ba2+ cation. Its concentration is below the toxic level of Ba2+, so the patient is not harmed (except by having to swallow a substantial amount of chalky suspension).



Problem 2.50. We are asked to suggest a sequence of selective precipitation reactions to separate Ag+, Ba2+, and Fe3+ from solution in which all three cations are present. The information we have to work with is in this table:

Cation

Test Solution Ag+(aq) Ba2+(aq) Fe3+(aq)

NaCl ppt no ppt no ppt

NaOH ppt no ppt ppt

Na2SO4 no ppt ppt no ppt

The objective of our sequence of additions of the test solutions (which we might also call the separation reagents) is to precipitate the cations one at a time without forming any mixtures of precipitates that contain more than one cation. For example, if we were to add the NaOH test solution to the original mixture of cations, both the Ag+(aq) and Fe3+(aq) cations would react to form precipitates, so the solid product would be a mixture of solids and we would not have separated the cations from one another. However, if the Ag+(aq) cation had already been removed from the solution, addition of the NaOH test solution would result in a precipitate containing the Fe3+(aq) cations and we would have the Fe3+(aq) cations as a separate precipitate. Our strategy is to look for a test solution that will precipitate only one of the cations, use it to precipitate that cation and then use a second test solution that will precipitate one or the other of the remaining cations. There are two ways to start our sequence. (1) We can add the NaCl test solution to precipitate the Ag+(aq) cation as AgCl(s), leaving behind a solution containing the Ba2+(aq) and Fe3+(aq) cations [as well as the Na+(aq) cations added with the test solution]. Or, (2) we can begin by adding the Na2SO4 test solution to precipitate the Ba

2+(aq) cations as BaSO4(s), leaving behind a solution containing the Ag+(aq) and Fe3+(aq) cations [as well as the Na+(aq) cations added with the test solution]. To pick the next step in sequence (1), consider this table that tells us how our test solutions react with the remaining cations:

Cation

Test Solution Ba2+(aq) Fe3+(aq)

NaOH no ppt ppt

Na2SO4 ppt no ppt

We have two choices, both of which will separate these two cations from one another. In the pathway that we will label (1a), we add NaOH test solution to the solution we get after removing the AgCl(s) in order to precipitate the Fe3+(aq) as Fe(OH)3(s). The solution now contains the Ba2+(aq) [and the Na+(aq) cations added with the test solutions]. The cations are now separated as two precipitates, AgCl(s) and Fe(OH)3(s), with the remaining cation, Ba

2+(aq), in solution. If we need to have the Ba2+ separated from all the Na+(aq) also in the solution, we can add the Na2SO4 test solution to precipitate it as BaSO4(s). The net ionic reactions for the three steps of pathway/sequence (1a) [and the alternative (1b)] are:



Pathway (1a) Pathway (1b)

Ag+(aq) + Cl(aq) AgCl(s) Ag+(aq) + Cl(aq) AgCl(s) Fe3+(aq) + 3OH(aq) Fe(OH)3(s) Ba2+(aq) + SO42(aq) BaSO4(s) Ba2+(aq) + SO4

2(aq) BaSO4(s) Fe3+(aq) + 3OH(aq) Fe(OH)3(s) The net ionic reactions for the three steps of pathway (2) are given here. You can reason through the logic that leads to this sequence and the reason why there is no alternative possibility for this sequence. Pathway (2)

Ba2+(aq) + SO42(aq) BaSO4(s)

Ag+(aq) + Cl(aq) AgCl(s) Fe3+(aq) + 3OH(aq) Fe(OH)3(s)

Problem 2.51. (a) When an aluminum nitrate, Al(NO3)3, solution is mixed with a sodium oxalate, Na2C2O4, solution the solution formed contains aluminum, Al3+(aq), and sodium, Na+(aq), cations, and nitrate, NO3

(aq), and oxalate, C2O42(aq), anions. A precipitate forms from this mixed solution.

The possibilities for this solid ionic compound are Al2(C2O4)3(s) and NaNO3(s). Our solubility rules indicate that ionic compounds with an alkali metal cation and nitrate anion are likely to be soluble, so the precipitate is not NaNO3(s). The precipitate, Al2(C2O4)3(s), is an ionic compound formed from a multiply-charged cation and a multiply-charged anion, which our rules suggest is likely to be insoluble. (b) The net ionic equation for the reaction that forms Al2(C2O4)3(s) has the aqueous ions with appropriate stoichiometric coefficients as reactants and the solid as product:

2Al3+(aq) + 3C2O42-(aq) Al2(C2O4)3(s)

Problem 2.52. When a solution of lithium nitrate, LiNO3, is mixed with a solution of sodium phosphate, Na3PO4, the solution contains lithium, Li

+(aq), and sodium, Na+(aq), cations and nitrate, NO3

(aq), and phosphate, PO43(aq), anions. A white precipitate is observed to form from this

mixture. The possibilities for this solid ionic compound are Li3PO4(s) and NaNO3(s). Our solubility rules indicate that ionic compounds with an alkali metal cation and nitrate anion are likely to be soluble, so the precipitate is not NaNO3(s). The precipitate must be Li3PO4(s). Although ionic compounds with alkali metal cations are generally expected to be soluble, note that, in Worked Example 2.34, lithium is discussed as an exception to the general rule. (b) The net ionic equation for the precipitation reaction is: 3Li+(aq) + PO43(aq) Li3PO4(s)

Problem 2.53. [Note that there seems to be no symbol for the equilibrium double arrow that is common to both PC and Macintosh computer platforms, so the symbol is used in these notes to make them cross-platform compatible.] When the reactants and products are separated by a forward arrow over a backward arrow, in a reaction equation, it means that the reaction can or is going in both directions. That is, the



reactant species are combining to yield products and the product species are combining to give reactants and both processes are occurring simultaneously.

Problem 2.54. (a) When a solution of cadmium chloride, CdCl2, is mixed with a solution of ammonium sulfide, (NH4)2S, the solution contains cadmium, Cd

2+(aq), and ammonium, NH4+(aq), cations

and chloride, Cl(aq), and sulfide, S2(aq), anions. A yellow-orange precipitate is observed to form from this mixture. The possibilities for this solid ionic compound are CdS(s) and NH4Cl(s). Our solubility rules indicate that ionic compounds with a singly-charged cation and halide anion are likely to be soluble, so the precipitate is not NH4Cl(s). The precipitate must be CdS(s). (Cadmium sulfide -- cadmium yellow -- is used as a pigment in paints.) (b) The net ionic equation for the precipitation reaction is: Cd2+(aq) + S2-(aq) CdS(s)

Problem 2.55. 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 write complete and net ionic equations for those reactions that produce a precipitate.

(a) barium chloride(aq) + sodium sulfate(aq) The mixed solution contains barium, Ba2+(aq), and sodium, Na+(aq), cations and chloride, Cl

(aq), and sulfate, SO42(aq), anions. Our solubility rules indicate that an ionic compound with a

multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that BaSO4(s) is insoluble and write this complete and net ionic reaction equation:

Ba+2(aq) + 2Cl-(aq) + 2Na+(aq) + SO42-(aq) BaSO4(s) + 2Na+(aq) + 2Cl-(aq)

Ba+2(aq) + SO42-(aq) BaSO4(s)

(b) silver nitrate(aq) + magnesium chloride(aq) The mixed solution contains silver, Ag+(aq), and magnesium, Mg2+(aq), cations and nitrate, NO3

(aq), and chloride, Cl(aq), anions. Our solubility rules indicate that ionic compounds with a halide or nitrate anion are likely to be soluble, so the initial instinct is to predict that no precipitate will form. However, when we recall that the silver ion is a notable exception to the rules, we predict that silver chloride, AgCl(s), is insoluble and write this complete and net ionic reaction equation:

2Ag+(aq) + 2NO3(aq) + Mg2+(aq) + 2Cl(aq) 2AgCl(s) + Mg2+(aq) + 2NO3(aq)

2Ag+(aq) + 2Cl(aq) 2AgCl(s)

(c) strontium nitrate(aq) + potassium nitrate(aq) The mixed solution contains strontium, Sr2+(aq), and potassium, K+(aq), cations and nitrate, NO3

(aq), anions. No new products are possible, because there is only one anion that is common to both soluble starting compounds. Thus we can write:

Sr2+(aq) + 3NO3-(aq) + K+(aq) + NO3

-(aq) NO APPARENT REACTION

(d) ammonium phosphate(aq) + calcium bromide(aq) The mixed solution contains barium, NH4

+(aq), and calcium, Ca2+(aq), cations and phosphate, PO4

3(aq), and bromide, Br(aq), anions. Our solubility rules indicate that an ionic compound



with a multiply-charged cation and anion is likely to be insoluble. Thus, we can predict that Ca3(PO4)2(s) is insoluble and write this complete and net ionic reaction equation:

6NH4+(aq) + 2PO4

3-(aq) + 3Ca2+(aq) + 6Br-(aq) Ca3(PO4)2(s) + 6NH4+(aq) + 6Br-(aq) 3Ca2+(aq) + 2PO4

3-(aq) Ca3(PO4)2(s)

Problem 2.56. The objective of this problem is to use your knowledge of the solubility rules to propose a way to prepare (synthesize) several solid ionic compounds by precipitation reactions from mixtures of soluble ionic compounds.

(a) The net ionic reaction for preparing BaSO4(s) is: Ba2+(aq) + SO4

2(aq) BaSO4(s) Mixing aqueous solutions of the soluble ionic compounds Ba(NO3)2 (nitrates are soluble) and Na2SO4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction.

(b) The net ionic reaction for preparing AgCl(s) is: Ag+(aq) + Cl(aq) AgCl(s) Mixing aqueous solutions of the soluble ionic compounds AgNO3 (nitrates are soluble) and KCl (alkali metal halide salts are soluble) will provide the reactant ions necessary for this reaction.

(c) The net ionic reaction for preparing Ca3(PO4)2(s) is: 3Ca2+(aq) + 2PO4

3- (aq) Ca3(PO4)2 (s) Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are soluble) and K3PO4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction.

(d) The net ionic reaction for preparing CaC2O4(s) is: Ca2+ (aq) + C2O4

2-(aq) CaC2O4 (s) Mixing aqueous solutions of the soluble ionic compounds CaCl2 (halides, except silver, are soluble) and K2C2O4 (alkali metal salts are soluble) will provide the reactant ions necessary for this reaction.

Problem 2.57. Since the concentration of a solution (assuming it is well mixed after preparation) is uniform throughout the entire volume, spilling some of it will not change the concentration of the remaining solution. Concentration is an intensive variable (like density or temperature) that does not depend upon the amount of solution you have.

Problem 2.58. (a) If it is properly labeled, the solution in a bottle labeled, "0.5 M CaCl2" contains 0.5 moles of CaCl2(aq) per liter of solution. There are 0.5 moles of Ca

2+(aq) per liter of solution and 1 mole of Cl(aq) per liter of solution. (b) If a 0.5-L bottle is about half full, then it contains approximately 0.25 L of 0.5 M CaCl2 solution. Use the definition of molarity, moles of solute per liter of solution, to determine the number of moles of the solute in about one-quarter liter of solution:



mol CaCl2 in 0.25 L = (0.25 L)0.5 mol

1 L

= 0.13 mol CaCl2

(c) Use the molar formula mass of CaCl2, 111 gmol1, to convert the number of moles of CaCl2 from part (b) to mass:

0.13 mol CaCl2 = (0.13 mol CaCl2)111 g

1 mol CaCl2

= 14 g CaCl2

Problem 2.59. (a) Counting the number of atoms of each element in the molecular structure for vitamin C in Problem 2.15 gives the molecular formula C6H8O6. (b) Find the molar mass of vitamin C by summing the masses of each element in a mole of the compound:

(6 mol C)12.01 g1 mol C

= 72.06 g

(8 mol H)1.008 g1 mol H

= 8.06 g

(6 mol O)16.00 g1 mol O

= 96.00 g

molar mass vitamin C = 176.12 g = 176 g (accurate enough for the rest of the data)

(c) To find the number of moles of vitamin C in a tablet that contains 500-mg of the vitamin, we use the molar mass to convert this mass to moles. The actual mass of vitamin in a vitamin tablet is only accurate to a few percent, so we will assume that the mass of vitamin C in the tablet is

about 0.50 g [= (500 mg) 1 g1000 mg ], with an implied accuracy of 1 part in 50 or about 2%. 0.50 g vit C = (0.50 g vit C)

1 mol vit C176 g

= 0.0028 mol vit C = 2.8 103 mol vit C

(d) To find the number of molecules of vitamin C in a tablet that contains 500-mg of the vitamin, we use Avogadros number to convert number of moles from part (c) to number of molecules:

2.8 103 mol vit C = (2.8 103 mol vit C)6.02 1023 molec

1 mol

= 1.7 1021 molec vit C

Problem 2.60. (a) To convert moles of aspartame to mass, we need the molar mass of aspartame, C14H18N2O5:

(14 mol C)12.01 g1 mol C

= 168.14 g

(18 mol H)1.008 g1 mol H

= 18.14 g



(2 mol N)14.01 g1 mol N

= 28.02 g

(5 mol O)16.00 g1 mol O

= 80.00 g

molar mass aspartame = 294.30 g = 294 g (accurate enough for the rest of the data)

The mass of aspartame in 2.5 mol of aspartame is:

2.5 mol = (2.5 mol)294 g

1 mol aspartame

= 735 g = 7.4 102 g = 0.74 kg

(b) To convert moles of aspirin to mass, we need the molar mass of aspirin, C9H8O4:

(9 mol C)12.01 g1 mol C

= 108.09 g

(8 mol H)1.008 g1 mol H

= 8.06 g

(4 mol O)16.00 g1 mol O

= 64.00 g

molar mass aspirin = 180.15 g = 180 g (accurate enough for the rest of the data)

The mass of aspirin in 0.040 mol of aspirin is:

0.040 mol = (0.040 mol)180 g

1 mol aspirin

= 7.2 g

(c) To convert number of molecules of cholesterol, C27H46O, to mass, we first need to convert number of molecules to number of moles (using Avogadros number) and then number of moles to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in 2.5 1023 molecules of cholesterol is:

2.5 1023 molec = (2.5 1023 molec)1 mol

6.02 1023 molec

= 0.42 mol

The molar mass of cholesterol is:

(27 mol C)12.01 g1 mol C

= 324.27 g

(46 mol H)1.008 g1 mol H

= 46.37 g

(1 mol O)16.00 g1 mol O

= 16.00 g

molar mass cholesterol = 386.64 g = 387 g (accurate enough for the rest of the data)

The mass of cholesterol in 0.42 mol cholesterol is:

0.42 mol = (0.42 mol)387 g

1 mol cholesterol

= 163 g = 1.6 10

2 g = 0.16 kg



(d) To convert number of molecules of caffeine, C8H10N4O2, to mass, we first need to convert number of molecules to number of moles (using Avogadros number) and then number of moles to mass [using the molar mass, as in parts (a) and (b) of the problem]. The number of moles in 1.2 1022 molecules of caffeine is:

1.2 1022 molec = (1.2 1022 molec)1 mol

6.02 1023 molec

= 0.020 mol

The molar mass of caffeine is:

(8 mol C)12.01 g1 mol C

= 96.08 g

(10 mol H)1.008 g1 mol H

= 10.08 g

(4 mol N)14.01 g1 mol N

= 56.04 g

(2 mol O)16.00 g1 mol O

= 32.00 g

molar mass caffeine = 194.20 g = 194 g (accurate enough for the rest of the data)

The mass of caffeine in 0.020 mol of caffeine is:

0.020 mol = (0.020 mol)194 g

1 mol caffeine

= 3.9 g

Problem 2.61. To find the number of atoms of carbon in 5 mg of niacin, we need a strategy to get from the mass of a substance to the number of atoms of an element in that mass of the substance. This problem involves understanding and applying the concept of the mole. One way to plan your work is to reason backward from the desired answer, using the information given in the problem, the mass of niacin, 5 mg, and its molecular structure:

NC

CC

C

C

H

H C

HH

N

O

H

H

The molecular formula for niacin, from its structure, is C6H6N2O, so each molecule of niacin contains six atoms of carbon. (Note that you might write the formula a different way, ON2C6H6 for example, and that is OK. The format we used, C6H6N2O, is the conventional way chemists write formulas for carbon-containing compounds: carbon first, hydrogen second, and then all other elements in alphabetical order. Any format that shows the correct number of each atom in the molecule is fine for stoichiometric problems like the one here.) If we know the number of molecules of niacin in 5 mg of niacin, there are six times as many atoms of carbon in the sample. If we know the number of moles of niacin in the sample, we use Avogadros number to get the number of molecules in the sample. Finally, use the mass (in grams) of the sample of



niacin and its molar mass (from the formula) to get the number of moles of niacin. Going forward through the solution, our strategy is summarized in this sequence:

mass moles molecules atoms

To find the number of moles of niacin in a 0.005 g [= (5 mg) 1 g1000 mg ] sample, we need the molar mass of niacin:

(6 mol C)12.01 g1 mol C

= 72.06 g

(6 mol H)1.008 g1 mol H

= 6.05 g

(2 mol N)14.01 g1 mol N

= 28.02 g

(1 mol O)16.00 g1 mol O

= 16.00 g

molar mass niacin = 122.13 g = 122 g (accurate enough for the rest of the data)

The number of moles of niacin is:

0.005 g niacin = (0.005 g niacin)1 mol niacin

122 g

= 4 105 mol niacin

The number of molecules of niacin is:

4 105 mol niacin = (4 105 mol niacin)6.02 1023 molec

1 mol

= 2.4 1019 molec niacin

The number on atoms of carbon in the sample is:

2.4 1019 molec niacin = (2.4 1019 molec niacin)6 atoms carbon1 molec niacin

= 1.4 1020 atoms carbon

Problem 2.62. In order to determine the molar concentration, molL1, of DNA in the given bacterium, we need to know the number of moles of DNA and the volume of the bacterium in which it is found. We are told that there is one molecule of DNA in the bacterium, so we can use Avogadros number to find the number of moles of DNA:

1 molec DNA = (1 molec DNA)1 mol

6.02 1023 molec

= 1.7 10

24 mol

We are told that the bacterium is spherical and has a diameter, d, of 1 106 m. The formula for

the volume of a sphere is 4 3 r3 = 16 d3 (where radius, r, = d/2). We want the volume in liters and one way to get it is to recall that 1 L = 1 dm3. Therefore, if we express the diameter of the bacterium in decimeters, 1 dm = 101 m, the volume we get will be in dm3 (= liters).



(Another approach is to calculate the volume in cubic centimeters, which are equal to milliliters, and thence to liters.) The required diameter of the bacterium is:

d = 1 106 m = (1 106 m)1 dm

101 m

= 1 10

5 dm

The volume, V, of the bacterium is:

V = 16 d3 = 16 (1 105 dm)3 = 5 1016 dm3 = 5 1016 L The DNA concentration is:

[DNA] = mol DNA

V =

1.7 1024 mol5 1016 L

= 3 109 M

Problem 2.63. To find the number of sodium ions in 50 mL of blood serum, we need a strategy to get from the volume of a solution to the number of ions of an element in that volume of the solution. One way to plan your work is to reason backward from the desired answer, using the information given in the problem, the volume of solution (serum), 50 mL, and its molarity, 0.14 M in NaCl. If we know the number of moles of sodium ion in the sample, we use Avogadros number to get the number of sodium ions in the sample. We know that every mole of NaCl dissolved in the serum produces a mole of sodium ions, Na+(aq). Finally, use the molarity of the serum sample and its volume (in liters) to get the number of moles of NaCl dissolved. Going forward through the solution, our strategy is summarized in this sequence:

volume solution moles NaCl moles ions number ions

The volume of the serum sample is:

volume = 50 mL = (50 mL)1 L

1000 mL

= 0.050 L

The number of moles of NaCl in the serum sample is:

0.050 L = ( 0.050 L)0.14 mol

1 L

= 7.0 10

3 mol

The number of moles of sodium ions in the serum sample is:

7.0 103 mol NaCl = (7.0 103 mol NaCl)1 mol Na+

1 mol NaCl

= 7.0 103 mol Na+

The number of molecules of sodium ions in the serum sample is:

7.0 103 mol Na+ = (7.0 103 mol Na+)6.02 1023 molec

1 mol

= 4.2 1021 Na+

Problem 2.64. For each solution in this problem we are asked to calculate the mass of solute present in a known volume of solution of a known molar concentration. The route map shown in Worked Example 2.50 (read from bottom to top) shows the sequence of steps required to get these masses: volume solution moles solute mass solute



These steps can be combined:

mass solute, g = (volume solution, L)mol solute

1 L solution

g solute1 mol solute

In each case, we will need the molar mass of the solute and will calculate this first.

(a) The molar mass of K2Cr2O7 is:

(2 mol K)39.10 g1 mol K

= 78.20 g

(2 mol Cr)52.00 g

1 mol Cr

= 104.00 g

(7 mol O)16.00 g1 mol O

= 112.00 g

molar mass K2Cr2O7 = 294.20 g = 294 g (accurate enough for the rest of the data)

The mass of K2Cr2O7 in 350 mL of 0.105 M K2Cr2O7 is:

mass K2Cr2O7 = (0.350 L)0.105 mol K2Cr2O 7

1 L solution

294 g1 mol K2Cr2O 7

= 10.8 g

(b) The molar mass of FeCl36H2O is:

(1 mol Fe)55.85 g

1 mol Fe

= 55.85 g

(3 mol Cl)35.45 g

1 mol Cl

= 106.35 g

(12 mol H)1.008 g1 mol H

= 12.10 g

(6 mol O)16.00 g1 mol O

= 112.00 g

molar mass FeCl36H2O = 286.30 g = 286 g (accurate enough for the rest of the data)

The mass of FeCl36H2O in 50 mL of 1.0 M FeCl36H2O is:

mass FeCl36H2O = (0.050 L)1.0 mol FeCl3 6H2O

1 L solution

286 g1 mol FeCl3 6H2O

= 14.3 g

(c) The molar mass of KCl is:

(1 mol K)39.10 g1 mol K

= 39.10 g

(1 mol Cl)35.45 g

1 mol Cl

= 35.45 g

molar mass KCl = 74.55 g = 74.6 g (accurate enough for the rest of the data)

The mass of KCl in 0.3 L of 1.70 M KCl is:

mass KCl = (0.3 L)1.70 mol KCl1 L solution

74.6 g1 mol KCl

= 38.0 g



Or we get about 4 102 g, if the volume is really only known to about 1 part in 3 ( 33%).

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:

molarity solute, molL1 = (mass solute, g)1 mol solute

g solute

1volume solution, L

In each case, we will need the molar mass of the solute and will calculate this first.

(a) The molar mass of NaCl is:

(1 mol Na)22.99 g

1 mol Na

= 22.99 g

(1 mol Cl)35.45 g

1 mol Cl

= 35.45 g

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

10.120 L

= 0.64 M

(b) The molar mass of NH4Cl is:

(1 mol N)14.01 g1 mol N

= 14.01 g

(4 mol H)1.008 g1 mol H

= 4.03 g

(1 mol Cl)35.45 g

1 mol Cl

= 35.45 g

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:

molarity NH4Cl = (1.3 g NH4Cl)1 mol NH 4Cl

53.5 g

10.25 L

= 0.097 M

Or we get about 0.10 M, if the mass is only known to a precision of about 1 part in 13.

(c) The molar mass of AgNO3 is:

(1 mol Ag)107.9 g

1 mol Ag

= 107.9 g

(1 mol N)14.01 g1 mol N

= 14.01 g



(3 mol O)16.00 g1 mol O

= 48.00 g

molar mass AgNO3 = 169.9 g = 170 g (accurate enough for the rest of the data)

The molarity of AgNO3 in 1.3 L of solution containing 1.85 g AgNO3 is:

molarity AgNO3 = (1.85 g AgNO3)1 mol AgNO3

170 g

11.3 L

= 0.0084 M

Or we get about 0.008 M, if the volume is only known to a precision of about 1 part in 13.

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:

molarity glucose, molL1 = (mass glucose, g)1 mol glucose

g glucose

1volume solution, L

We will need the molar mass of glucose, C6H12O6, and will calculate this first.

(6 mol C)12.01 g1 mol C

= 72.06 g

(12 mol H)1.008 g1 mol H

= 12.10 g

(6 mol O)16.00 g1 mol O

= 96.00 g

molar mass glucose = 180.16 g

molarity glucose = (5.405 g glucose)1 mol glucose

180.16 g

11.000 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 mmolmL1 equals the concentration in molL1. 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:

0.950 mmol glucose = (0.950 mmol glucose)1.000 mL

0.03000 mmol

= 31.67 mL



You may have chosen a different route to calculate this volume, and, as we said at the beginning, thats 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:

(1 mol Cu)63.55 g

1 mol Cu

= 63.55 g

(1 mol S)32.07 g1 mol S

= 32.07 g

(4 mol O)16.00 g1 mol O

= 64.00 g

molar mass CuSO4 = 159.62 g The molar mass of CuSO45H2O adds the mass of the five moles of water, 90.08 g, to the mass of the CuSO4 for a total of 249.70 gmol

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:

159.60 g CuSO45H2O = (159.60 g CuSO45H2O)1 mol

249.70 g CuSO4 5H2O

= 0.639 mol He prepared a solution of too low a concentration: 0.639 molL1 = 0.639 M. The female student used a solid of the correct composition, CuSO4, and prepared a solution of the correct composition: 1.000 molL1 = 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 gmol1) to get the mass of NaOH that must be dissolved. The calculation may be done in a single step by combining the two conversions:

0.250 L soln = (0.250 L solution)0.10 mol NaOH

1 L solution

40.0 g1 mol NaOH

= 1.00 g NaOH

You may not have known the standard sizes of volumetric flasks and have chosen a different volume of solution to prepare. Thats 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:

250. mL solution = (250. mL solution)0.90 g NaCl

100. mL solution

= 2.25 g NaCl

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 gmol1), and then divide by the volume to get molarity (molL1). 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). Lets use the former:

0.90 g NaCl = (0.90 g NaCl)1 mol

58.4 g NaCl

10.100 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 gmol1), and then divide by the volume to get molarity (molL1):

25 g urea = (25 g urea)1 mol

60 g urea

12.5 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 gmol1) and of KH2PO4 (136 gmol

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+

= 7.7 104 mol K+ = 7.7 104 mol KH2PO4

7.7 104 mol KH2PO4 = (7.7 104 mol KH2PO4)136 g KH2PO 4

1 mol

= 0.10 g

KH2PO4

7.7 104 mol KH2PO4 = (7.7 104 mol KH2PO4)1

0.240 L

= 3.2 10

3 M KH2PO4

In the second approach we have:

0.030 g K+ = (0.030 g K+)136 (g KH 2PO4 ) (mol KH2PO4 )

1

39.1 (g K+) (mol KH2PO 4 )1

= 0.10 g KH2PO4

0.030 g K+ = (0.030 g K+)1 mol

39.1 g K+

10.240 L

= 3.2 10

3 M K+

= 3.2 103 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, thats 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 4C2H 3O2

= 4 mol C

2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)7 mol H

1 mol NH 4C2H 3O2

= 14 mol H

2 mol NH4C2H3O2 = (2 mol NH4C2H3O2)2 mol O

1 mol NH 4C2H 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 4C2H 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, thats 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 dont 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 PO4

3(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

1 mol Ca2+

1 mol CaCl2

= 0.100 M Ca2+(aq)

0.100 M Ca2+(aq) = 0.100 mol Ca2+

1 L solution

(0.125 L solution) = 1.25 102 mol Ca2+(aq)

The same chain of reasoning gives the concentration and number of moles of phosphate anion:

0.100 M Na3PO4 = 0.100 mol Na2PO4

1 L solution

1 mol PO43

1 mol Na2 PO4

= 0.100 M PO43(aq)

0.100 M PO43(aq) =

0.100 mol PO43

1 L solution

(0.125 L solution) = 1.25 102 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

(= 32 ) mol of Ca2+(aq) is required to react completely with 1 mol of PO4

3(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 PO4

3(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:

1.25 102 mol Ca2+ = (1.25 102 mol Ca2+)1 mol Ca3(PO 4 )2

3 mol Ca2+

= 4.17 103 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:

4.17 103 mol Ca3(PO4)2 = (4.17 103 mol Ca3(PO4)2)310.2 g Ca 3(PO4 )21 mol Ca3 (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 SO3

2(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

Fe3+)0.200 mmol Fe3+

1.000 mL Fe3+

1 mmol SO 32

2 mmol Fe3+

1.000 mL SO32

0.100 mmol SO 32

= 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:

0.050 L Na2SO4 = (0.050 L Na2SO4)0.45 mol Na2SO 4

1 L

2 mol Na+

1 mol Na2SO4

= 0.045 mol Na+

0.050 L Na2SO4 = (0.050 L Na2SO4)0.45 mol Na2SO 4

1 L

1 mol SO42

1 mol Na2SO4

= 0.023 mol SO42

0.050 L BaCl2 = (0.050 L BaCl2)0.36 mol BaCl2

1 L

1 mol Ba2+

1 mol BaCl2

= 0.018 mol Ba2+

Chapter 2 Aqueous Soluti

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