Aqueous Solutions and Solubility Ch 4

Chapter 4 Aqueous Solutions and Solubility Problem 2.94. The two movies in the Web Companion, Chapter 2, Section 2.12, page 3, are an attempt to show in animation what the figure at the bottom of that page and Figure 2.26 in the textbook try to illustrate. What the figures and movies try to show is how hydronium cations and hydroxide anions “migrate” in aqueous solution by transfer of a hydrogen ion, H+ (a proton), from one molecule or ion to another ion or molecule. Four frames from the hydronium ion movie are shown here: In the first frame (on the left), one of the hydrogens from the hydronium at the far left of the frame is hydrogen bonded to the adjacent water molecule and by a small movement of this hydrogen atomic core, the hydronium ion has transferred a proton to form a new hydronium ion (center of second frame). This new hydronium ion goes on in frames three and four to transfer a proton (not the one it got in the previous transfer) to a third water molecule. Although the water molecules and hydronium are moving about, they don’t move far during these transfers, but a hydronium ion has moved almost all the way across the frame and, in the movie, is shown finally bonded to the oxygen atom on the far right of the frame. A similar sequence of proton transfers is shown in the hydroxide ion movie where transfers of a proton from water to hydroxide result in the movement of a hydroxide ion. [Theoretical calculations and experimental reaction rates are consistent with this mechanism for the migration of hydronium and hydroxide ions through aqueous solutions. It is only a matter of time until very fast spectroscopic methods will be able to observe events like these “directly” to see if they actually occur as we imagine.] Problem 2.95. (a) The balanced chemical reaction equation for the reaction of phosphorus pentoxide, P2O5(s), with water to form a solution of phosphoric acid, (HO)3PO(aq) (or H3PO4(aq)) is: P2O5(s) + 3H2O(l) 2H3PO4(aq) (b) To find the molarity of the phosphoric acid solution formed when 1.42 g of phosphorus pentoxide is mixed with 250. mL of water, we need to convert moles of phosphorus pentoxide (molar mass = 142 g) to moles of phosphoric acid, using the reaction equation in part (a), and then divide by the volume of solution, in liters, to get the molarity. We assume that the volume of liquid does not change significantly when this small amount of phosphorus pentoxide is added and reacted. 1.42 g P2O5 = (1.42 g P2O5) = 8.0 1 mol P2 O5 142 g P2 O5 10-2 M H3PO4(aq) 2 mol H3 PO 4 (aq) 1 mol P2 O5 1 0.250 L Problem 2.96. The name is given for each of these ionic compounds with oxyanions. The arsenate oxyanion, AsO43–, is named by analogy with its phosphorus analog, the phosphate oxyanion, PO43–. (a) Ca(HSO4)2 calcium hydrogen sulfate (b) Na2CO3 sodium carbonate (c) Al2(HPO4)3 aluminum hydrogen phosphate (d) Ce2SO4 cesium sulfate (e) KHCO3 potassium hydrogen carbonate (f) Na3AsO4 sodium arsenate Problem 2.97. The Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent bonds as lines) for the nitrate, ethanoate (acetate), and hydrogen sulfate oxyanions are: O O O H O C C N H O O H O S O H O nitrate ethanoate hydrogen sulfate The hydrogen sulfate oxyanion is shown with six bonding electron pairs around the central sulfur atomic core, as in Table 2.7. The marginal note on page 126 of the textbook indicates that, in Chapters 5 and 6, we will find that third (and higher) period elements often accommodate more than four bonding electron pairs and that the structure with six bonding electron pairs and several double bonds seems to fit the properties of the oxyanion better than a structure with four bonding pairs (an octet). Problem 2.98. Each Brønsted-Lowry acid and base is identified with an A or B, respectively, below the species in these reactions. When necessary, the reaction in the problem is written out in is complete ionic form to make identification of the acid and base possible. (a) H2S(g) + H2O(l) A B HS–(aq) + H3O+(aq) B A NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) + – Na (aq) + OH (aq) + H3O+(aq) + Cl–(aq) Na+(aq) + Cl–(aq) + H2O(l) + H2O(l) B A B A The Na+(aq) and Cl–(aq) ions are spectator ions in this reaction, which, in aqueous solution, is just a reaction between hydronium cation and hydroxide anion to produce two molecules of water. Water can act as either a Brønsted-Lowry acid or base, depending upon whether it is a proton donor or proton acceptor in the reaction. (b) NH3(g) + HCl(g) NH4+Cl–(s) B A A B In this case, the molecule, HCl(g) is the acid (proton donor). This is unlike part (b), where HCl(aq) does not exist as a molecule in aqueous solution, but as the hydronium cation and chloride anion. (c) Problem 2.99. Each Brønsted-Lowry acid and base is identified with an A or B, respectively, below the species in these reactions. When necessary, the reaction in the problem is written out in is complete ionic form to make identification of the acid and base possible. (a) NO2-(aq) + H3O+(aq) B A HNO2(aq) + H2O(l) B A (b) 2H3O+(aq) + 2ClO4–(aq) + Mg2+(OH–)2(s) Mg2+(aq) + 2ClO4–(aq) + 4H2O(l) 2H3O+(aq) + 2ClO4–(aq) + Mg2+(OH–)2(s) Mg2+(aq) + 2ClO4–(aq) + 2H2O(l) + 2H2O(l) A A B B In this reaction, the hydronium ion from a perchloric acid, HOClO3(aq), solution reacts with the hydroxide ion in solid magnesium hydroxide, Mg2+(OH–)2(s). The magnesium hydroxide formula is written out this way to make explicit the presence of the ions in the solid. Like the perchlorate anion, the magnesium cation is essentially a spectator ion in this reaction. See the solution for Problem 2.98(b) for a discussion of the role of water molecules. HNO3(aq) + Al3+(OH–)3(s) Al3+(aq) + 3NO–4(aq) + 3H2O(l) + 3H2O(l) 3H3O+(aq) + 3NO3–(aq) + Al3+(OH–)3(s) A A B B This reaction is modeled after the one in part (b), which was written out in full as a model. (c) (d) HCN(aq) + NaOH(aq) Assuming that hydrocyanic acid, HCN(aq), like hydrochloric acid, transfers a proton to a water molecule to form H3O+(aq) and CN–(aq), we can write the reaction equation as: H3O+(aq) + CN–(aq) + Na+(aq) + OH–(aq) Na+(aq) + CN–(aq) + H2O(l) + H2O(l) A A B B [At this point in this chapter, weak acids have not been introduced, so we are not forced to consider whether the reaction could also be written with the aquated hydrogen cyanide molecule, HCN(aq), as the acid: Na+(aq) + CN–(aq) + H2O(l) HCN(aq) + Na+(aq) + OH–(aq) A A B B This, of course, changes the interpretation of the reaction somewhat and makes it clearer, as will be discussed in Chapter 6, that there is a competition between OH–(aq) a CN–(aq) for protons in this solution.] Problem 2.100. The Lewis structures (showing all nonbonding electron pairs as a pair of dots and all covalent bonds as lines) for HOCO2–, (HO)2PO2–, and HOPO32– (modeled on the structures for the corresponding oxyacids in Table 2.7) are: O O C H O H hydrogen carbonate O P 2 O O O H O dihydrogen phosphate O P O H O monohydrogen phosphate Problem 2.101. (a) The Brønsted-Lowry acids and bases in two possible acid-base reactions of ammonia and water are identified by placing an A below each acid and a B below each base. H2O(l) + NH3(g) H3O+(aq) + NH2–(aq) B A B A OH–(aq) + NH4+(aq) H2O(l) + NH3(g) A B B A The ammonia and water molecules switch roles in these two possible reactions. (b) We predict that the second equation represents the correct acid-base reaction. Because oxygen is more electronegative than nitrogen, the O–H bonding electrons in water are held more closely by the oxygen than the N–H bonding electrons in ammonia. Thus, in comparison with the H+ in the N–H bond, the H+ in the O–H bond can interact more effectively with non-bonding electron pairs on other molecules. The result is that H+ from a water molecule is transferred to a Brønsted-Lowry base more readily than the H+ from an ammonia molecule. If the second reaction is the correct one, then solutions of NH3 in water will have extra OH–(aq) and will be basic. The fact that the pH > 7 in aqueous ammonia solutions indicates that there is hydroxide present in concentrations greater than found in neutral water (pH = 7) and confirms our prediction. Problem 2.102. To explain the electrical conductivity observed in a solution of methylamine, CH3NH2(g), dissolved in water, we need to consider how ions are formed in the interaction of the methylamine and water molecules. By analogy with the reaction of ammonia with water, we can write: CH3NH3+(aq) + OH–(aq) CH3NH2(aq) + H2O(l) Since the electrical conduction by the solution is weak, we conclude that only a few ions are produced, that is, the reaction does not proceed far toward products. We indicate this with an equilibrium arrow in the balanced equation to show that the system comes to equilibrium before all the methylamine has reacted. Problem 2.103. To predict/decide whether an aqueous solution of ethylene glycol, HOCH2CH2OH will be basic, acidic, or neutral and whether it will conduct an electric current, we need to consider how the molecules will interact with water molecules. Other compounds we know about with –OH groups bonded to carbon are alcohols, like methanol (CH3OH) and ethanol (CH3CH2OH), and compounds with multiple –OH groups, like glucose (se Figure 2.6 in the textbook). These compounds are miscible with water (the alcohols) or very soluble (glucose), so it is not surprising that ethylene glycol, a small molecule with two –OH groups, should be miscible with water. The reason all these compounds are so soluble is that they can form several hydrogen bonds with water and the non-polar part of the molecule does not disrupt the water structure enough to be a major factor in their solubility. For all these molecules, including ethylene glycol, the interactions with water do not involve any proton transfers that would form ionic products, so the solutions do not conduct an electric current and are not acidic or basic. Problem 2.104. Our task in this problem is to use Le Chatelier’s principle to predict and clearly explain the outcome of changes in the reaction conditions for this equilibrium reaction system: CH3C(O)OH + HOCH2CH3 CH3C(O)OCH2CH3 + H2O acetic acid ethanol ethyl acetate (a) Starting with 0.1 mole of acetic acid and 0.1 mole of ethanol, would more, less, or the same amount of ethyl acetate be formed, if water is added to the reaction mixture? Addition of water is a disturbance to the equilibrium system, which will respond by reacting to use up some of the added water. The reaction will proceed from the right to the left, using up both water and ethyl acetate. The amount of ethyl acetate will be less when the extra water is added. (b) Would a mixture of 0.2 mole of acetic acid and 0.1 mole of ethanol form more, less, or the same amount of ethyl acetate as a mixture of 0.1 mole of acetic acid and 0.1 mole of ethanol? Addition of acetic acid (0.2 mol instead of 0.1 mol) is a disturbance to the equilibrium system, which will respond by reacting to use up some of the added acetic acid. The reaction will proceed from the left to the right, using up both acetic acid and ethanol and forming more ethyl acetate. The amount of ethyl acetate will be more when the extra water is added. Problem 2.105. Our task in this problem is to use Le Chatelier’s principle to predict and/or clearly explain the outcome of the changes in the reaction conditions for two equilibrium reaction systems: (a) To explain why the solubility of carbon dioxide is greater in an aqueous sodium hydroxide solution, Na+(aq) + OH–(aq), than in water itself, we need to examine the reactions of dissolved carbon dioxide and see how increasing the hydroxide anion concentration would affect them. The dissolution equilibrium reactions are: CO2(g) + H2O(l) (HO)2CO(aq) H3O+(aq) + HOCO2–(aq) (HO)2CO(aq) + H2O(l) Both of the products of the second reaction, H3O+(aq) and HOCO2–(aq), can react with OH–(aq) to “use them up”: H3O+(aq) + OH–(aq) 2H2O(l) HOCO2–(aq) + OH–(aq) H2O(l) + CO32–(aq) These two equations are also reversible and could be shown with equilibrium arrows, but the directional arrow is used to show how the H3O+(aq) and HOCO2–(aq) are used up by added OH– (aq) anion. Removing the two products of the dissolution reactions is a disturbance to the system, which reacts by forming more of these products (which are, in turn, used up until all the hydroxide anion has reacted). The direction of the response to the disturbance is shown by this series of reactions with directional arrows: CO2(g) + H2O(l) (HO)2CO(aq) (HO)2CO(aq) + H2O(l) H3O+(aq) + HOCO2–(aq) H3O+(aq) + OH–(aq) 2H2O(l) HOCO2–(aq) + OH–(aq) H2O(l) + CO32–(aq) The overall reaction, in the presence hydroxide anion is the sum of these four reactions: CO2(g) + 2OH–(aq) H2O(l) + CO32–(aq) (b) To predict what will happen to the concentration of calcium cation, [Ca2+(aq)], if sodium sulfate (solid), Na2SO4(s), is added to a saturated aqueous solution of sparingly soluble calcium sulfate, we need know what equilibrium reaction might be disturbed and what happens to the added sodium sulfate. Sodium sulfate is a soluble ionic compound, so it will dissolve in the solution to give Na+(aq) and SO42–(aq) ions. The equilibrium reaction in a saturated solution of calcium sulfate is the equilibrium dissolution of the solid ionic compound: CaSO4(s) Ca2+(aq) + SO42–(aq) Sodium sulfate is a soluble ionic compound, so it will dissolve in the solution to give Na+(aq) and SO42–(aq) ions. The added SO42–(aq) anions are a disturbance to the equilibrium system and the system will respond by trying to use up some of the ions by reacting to form more CaSO4(s). This requires using up some of the Ca2+(aq) cation, so we predict that [Ca2+(aq)] will decrease when Na2SO4(s) is added to a saturated aqueous solution of CaSO4(s). Problem 2.106. We are to represent each of these statements as a complete balanced chemical equation: (a) Carbonic acid is formed when carbon dioxide reacts with water. CO2(g) + H2O(l) (HO)2CO(aq) (b) Calcium carbonate (limestone) reacts with carbonic acid to form an aqueous solution of calcium hydrogen carbonate. CaCO3(s) + (HO) 2CO(aq) Ca2+(aq) + 2HOCO2–(aq) (c) Calcium hydrogen carbonate reacts with calcium hydroxide to form calcium carbonate precipitate. [The implication of this statement is that one or both of the reactants is in aqueous solution. Calcium hydroxide is a sparingly soluble ionic compound, so we might write the reaction of solid calcium hydroxide with a solution of calcium hydrogen carbonate. We’ll do this as well as write the reaction for both ionic compounds in solution.] Ca2+(aq) + 2HOCO2–(aq) + Ca(OH)2(s) 2CaCO3(s) + 2H2O(l) Ca2+(aq) + 2HOCO2–(aq) + Ca2+(aq) + 2OH–(aq) 2CaCO3(s) + 2H2O(l) (d) Calcium hydrogen carbonate reacts with sodium hydroxide to form calcium carbonate precipitate and the water-soluble salt sodium carbonate. [Assume both ionic compounds are in aqueous solution.] Ca2+(aq) + 2HOCO2–(aq) + 2Na+(aq) + 2OH–(aq) 2CaCO3(s) + 2Na+(aq) + CO32–(aq) + 2H2O(l) (e) The mixing of aqueous solutions of sodium hydrogen carbonate and sodium hydroxide is exothermic. A reaction has occurred but there is no precipitate. Na+(aq) + HOCO2–(aq) + Na+(aq) + OH–(aq) 2Na+(aq) + CO32–(aq) + H2O(l) Since the reaction is exothermic, you may show “energy” as a product, if you wish. In this case, we are given this information, because there is no visible evidence (precipitate, gas formation, color change, etc.) that a reaction has occurred. Problem 2.107. (a) As a droplet of solution containing calcium cations and hydrogen carbonate anions evaporates from the tip of a stalactite, the concentration of these solute ions increases. The volume of solution is less, so the number of ions per unit volume is greater, that is, a higher concentration. (b) The ions present in the droplet of solution in part (a) were a result of reaction (2.40): Ca2+(aq) + 2HOCO22–(aq) CaCO3(s) + H2O(l) + CO2(g) The reaction is reversible and is shown here as an equilibrium. The increased concentrations of the products of reaction (2.40), which were noted in part (a), are a disturbance to the system. Le Chatelier’s Principle states that the system will respond by minimizing the disturbance, that is, by reacting in a way that decreases the concentrations. The reaction that does this is the reverse of reaction (2.40), which will proceed to precipitate solid calcium carbonate and release carbon dioxide gas into the surrounding air in the cave. (c) The result of the reaction in part (b) is to precipitate some solid calcium carbonate, which causes the stalactite to grow downward (very slowly). If the drop of liquid should drop off the tip of the stalactite, it will end up directly below and the same evaporation and precipitation process on the floor of the cave will build stalagmites upward toward the stalactites. If they meet, they form a column, which is another common structure in limestone caves. Problem 2.108. (a) We are asked what gas is evolved when calcium carbonate, a solid, is placed in an aqueous solution of HCl(g). We know that HCl(g) dissolves and reacts completely with water to produce hydronium cations and chloride anions (hydrochloric acid solution): HCl(g) + H2O(l) H3O+(aq) + Cl–(aq) Calcium carbonate is a sparingly soluble solid whose dissolution in water is represented by this equilibrium reaction: CaCO3(s) (i) Ca2+(aq) + CO32–(aq) The carbonate ion product of this dissolution is a Brønsted-Lowry acid that can react with the hydronium cation in the solution [from the dissolved HCl(g)] to form hydrogen carbonate anion and this anion can react further with another hydronium ion to form carbonic acid: CO32–(aq) + 2H3O+(aq) (ii) (HO)2CO(aq) + 2H2O(l) We know that carbonic acid is formed in water when carbon dioxide gas dissolves and we can write this equilibrium reaction in reverse to show how carbon dioxide gas is released by the decomposition of carbonic acid: (iii) (HO)2CO(aq) CO2(g) + H2O(l) This series of reactions shows that carbon dioxide is likely to be the gas evolved when calcium carbonate is placed in hydrochloric acid solution. (b) The net ionic equation for the reaction that produces the gas when solid calcium carbonate is placed in hydrochloric acid solution is the sum of reactions (i), (ii), and (iii) in part (a): CaCO3(s) + 2H3O+(aq) Ca2+(aq) + CO2(g) + 3H2O(l) (c) Only a tiny amount of calcium carbonate (marble, chalk) dissolves in water, which is represented by equation (i) in part (a). The addition of hydronium ion [from the dissolved HCl(g)] is a disturbance to this solubility equilibrium because the hydronium ion reacts with one of the products, carbonate anion and reduces its concentration. In response, LeChatelier’s principle, more of the solid calcium carbonate dissolves to try to increase the carbonate anion concentration. If there is a good deal of hydronium present, it continues to react (and the calcium carbonate continues to dissolve) forming more and more carbonic acid, because the presence of a lot of hydronium ion displaces equilibrium equation (ii) toward products. The increase in concentration of carbonic acid is a disturbance to equilibrium reaction (iii), which responds, LeChatelier’s principle, by trying to use it up to form products, one of which, carbon dioxide gas, we observe as it leaves the reaction solution. Problem 2.109. The molecular level representation of sucrose in the Web Companion Chapter 2, Section 2.2, page 3 shows that sugar molecules have several regions of positive and negative charge, which are due to the presence of a number of –OH groups that are shown in the molecular structure on this page. Thus, in its solid crystals, the sugar molecules are oriented in specific directions, as shown in the movie, that maximize their positive-negative attractions and hydrogen bonding between molecules. When you put pressure on the crystal (by trying to crush it, for example), these many oriented interactions strongly resist being reoriented and the crystal breaks into smaller pieces as some of the interactions are finally broken. The molecular level representation of grease on page 5, shows that the molecules are long chains of hydrocarbons that have essentially no polarity. Attractions of these molecules for one another are through nondirectional dispersion (induced dipole) interactions and their packing in the solid (or semisolid) form is rather random, as shown in the movie. Thus, when pressure is applied, the hydrocarbon molecules can slide over each other because their attractions are non-directional and any orientation of the molecules with respect to one another is as good as any other. Problem 2.110. (a) The dissolution of calcium sulfate, CaSO4(s), in water can be represented as: Ca2+(aq) + SO42–(aq) CaSO4(s) About 2 g of calcium sulfate dissolve in a liter of water, so we can calculate the number of moles that dissolve in a liter, using the molar mass of CaSO4(s) (= 136 g·mol–1), and from the stoichiometry of the dissolution reaction find the molarities of the ions in the saturated solution: 1 mol CaSO 4 2 (g CaSO4)·L = [2 (g CaSO4)·L ] 136 g CaSO4 –1 –1 2+ 1 mol Ca 1 mol CaSO4 = [Ca2+(aq)] = 0.015 mol·L–1 The stoichiometry of the dissolution shows that [Ca2+(aq)] = [SO42–(aq)] = 0.015 mol·L–1. Thus, a saturated solution of calcium sulfate is 0.015 M in both ions. From the table in Consider This 2.54, we find that seawater contains 0.010 M Ca2+(aq) and 0.028 M SO42–(aq). Seawater contain less calcium ion than in a saturated solution of calcium sulfate, but almost twice as much sulfate ion as in the saturated calcium sulfate solution. It’s hard to tell from these data whether seawater is saturated with calcium sulfate, because, at this point in the textbook, our analysis of solubility has not provided a way to handle a situation like this. [The remainder of the problem is designed to help get a qualitative feeling for what is going on and Chapter 9 will provide a quantitative way to analyze solubilities in a solution like seawater.] (b) Solubility equilibria, like the one represented in the equation in part (a), should respond (Le Chatelier’s principle) to the disturbance caused by adding more of the cation or anion (by adding a soluble ionic compound containing the cation or anion, for example). Since the ions are on the product side of the solubility equilibrium, the system will react by trying to remove some of the added ion by formation of more solid. Removing some of the added ion also, of course, removes some of the other ion, the counterion. In the case of the calcium sulfate in this problem, addition of SO42–(aq) (by adding some sodium sulfate, Na2SO4, for example) to a saturated solution of calcium sulfate will result in formation of a little more solid calcium sulfate and a decreased concentration of Ca2+(aq). Looked at from another point of view, the amount of calcium sulfate that will dissolve in a solution that already contains SO42–(aq) will be lower, the solid will be less soluble, than in pure water. A saturated solution under these conditions will contain a lower concentration of Ca2+(aq) than a saturated solution in pure water. The ionic compound will be less soluble than it would be in pure water because Le Chatelier’s principle favors the precipitation (or decreased solubility). (c) Seawater contains more sulfate ion than is present in the saturated solution of calcium sulfate we calculated in part (a). Since seawater contains more of the anion than could have been present from CaSO4(s) alone, we can imagine that the solution is formed when CaSO4(s) dissolves in a solution that already contains some of the anion. From part (b), we find that the solubility of the CaSO4(s) will be lower in such a solution. This is the direction of the effect we observe; there is less calcium cation in seawater than would be present in a saturated solution in pure water. Although we still cannot be certain, this analysis makes it more likely that we can conclude that seawater is saturated with CaSO4(s), in the presence of extra sulfate anion from some other source. [In Chapter 9, we will find that the product of the concentrations of the cation and anion characterizes the solubility equilibrium for a saturated solution of a sparingly soluble salt like CaSO4. For the saturated solution in water, we have [Ca2+(aq)][SO42–(aq)] = (0.015 M)(0.015 M) = 2.2 10–4 M2 For seawater, we have [Ca2+(aq)][SO42–(aq)] = (0.010 M)(0.028 M) = 2.8 10–4 M2 These two products are almost the same and reinforce the conclusion that seawater is saturated with CaSO4, in the presence of extra sulfate anion.] (d) Since seawater already contains a substantial concentration of calcium anion, the solubility of calcium carbonate will be lower than it would be in pure water. The solubility of calcium carbonate is very low in pure water, so lowering it even further helps explain why seashells do not redissolve (at least not rapidly) in the sea. Problem 2.111. We are asked to find the molarity of nitrate, NO3–(aq), anion in a sample of salt water with a density of 1.02 g·mL–1 that contains 17.8 ppm (parts per million, by mass) of NO3–(aq). If we can find the mass of NO3–(aq) in a known volume of the solution, we can use the molar mass of NO3–(aq) (= 62.01 g) to get the number of moles and then the molarity. The concentration, 17.8 ppm (by mass) means that exactly one million grams of solution contain 17.8 g of NO3–(aq). We can use this information to convert the density (mass per milliliter) to grams and then to moles and molarity: 1.02 g·mL–1 sol’n = (1.02 g·mL–1 sol’n) = 2.93 10–4 mol·L–1 17.8 g NO–3 1 10 6 g sol'n – 1 mol NO3 – 62.01 g NO3 1000 mL 1L Problem 2.112. (a) We are asked to find the molarity of sodium chloride, NaCl, in a solution of salt water (representing seawater) that has a density of 1.025 g·mL–1 and contains 3.50% by weight (mass) of NaCl. If we can find the mass of NaCl in a known volume of the solution, we can use the molar mass of NaCl (= 58.44 g) to get the number of moles and then the molarity. The concentration, 3.50% by mass means that exactly one hundred grams of solution contain 3.50 g of NaCl. We can use this information to convert the density (mass per milliliter) to grams and then to moles and molarity: 1.025 g·mL–1 sol’n = (1.025 g·mL–1 sol’n) 3.50 g NaCl 1 10 2 g sol'n 1 mol NaCl 58.44 g NaCl 1000 mL 1L = 0.614 mol·L–1 (b) In Consider This 2.54(b) we found that the sum of the concentrations of positive charges in seawater is 0.606 M (and for negative charges 0.603 M). The result in part (a) is consistent (within about 1.3%) with the answer to Consider This 2.54(b), because the solution in part (a) has a 0.614 M concentration of positive charge from the Na+ cation (and the same concentration of negative charge from the Cl– anion). Problem 2.113. We are given a good deal of information about a solution prepared by a student. The solution contains 5.15 g of a compound dissolved in 10.0 g of water, has a density of 1.34 g·mL–1, and is 2.7 M. We are asked to determine which of these ionic compounds is the solute: (NH4)2SO4, KI, CsCl, or Na2S2O3. If we can find out how many moles of solute are dissolved, we can determine its molar mass and compare the result with the molar masses of the possible solutes. The known molarity of the solution, 2.7 M, is equal to the number of moles dissolved divided by the volume (liters) of the solution. To get the volume of solution, we divide the total mass of the solution 15.15 g (= 5.15 g + 10.0 g) by its density: 15.15 g sol’n = (15.15 g sol’n) 1 mL sol'n 1.34 g sol'n 1L 1000 mL Therefore, using the molarity of the solution, we have: 2.7 mol solute 0.0113 L sol’n = (0.0113 L sol’n) 1 L sol'n = 0.0113 L sol’n = 0.0305 mol solute It is not really legitimate to carry the third significant figure in this result, since the molarity has an uncertainly of about 3% (1 part in 27), but we’ll carry it along and check whether it makes a difference later. Now we know that 5.15 g of the solute is 0.0305 mol of the solute, so we can find its molar mass: 5.15 g solute = 169 g·mol–1 molar mass = 0.0305 mol solute The molar masses of the four possible compounds are: (NH4)2SO4 , 132 g·mol–1; KI, 166 g·mol– ; CsCl, 168 g·mol–1; and Na2S2O3, 158 g·mol–1. The molar mass we calculated has about a 3% uncertainty from the uncertainty in the second significant figure of molarity, so the range of possibilities is about 164 to 174 g·mol–1. The possibilities, therefore, are KI or CsCl for the compound the student dissolved. The calculation fits CsCl best, but the uncertainty makes this identification unsure. We would need a better value for the molarity to be sure. 1 Problem 2.114. We are asked to find the approximate percentage of the water molecules that is protonated in aqueous solutions with different pHs. We can use Figure 2.24, or the relationship, [H3O+(aq)] = 10–pH, to find the molarity of hydronium ion (protonated water molecules) in each solution and compare this molarity to the molarity of the water, 55.5 M, to get the fraction/percentage of the water molecules that are protonated. (a) At pH 7, we have [H3O+(aq)] = 10–7 M, so the percent of molecules protonated is: % H2O protonated = 10 –7 + mol [H3 O (aq)] 100% = 2 55.5 mol H2 O(l ) 10–7 % (b) At pH 6, we have [H3O+(aq)] = 10–6 M, so the percent of molecules protonated is: % H2O protonated = 10 –6 + mol [H3 O (aq)] 100% = 2 55.5 mol H2 O(l ) 10–6 % (a) At pH 4, we have [H3O+(aq)] = 10–4 M, so the percent of molecules protonated is: % H2O protonated = 10 –4 + mol [H 3O (aq )] 100% = 2 55.5 mol H2 O(l ) 10–4 % Problem 2.115. (a) Figure 2.24 shows that [OH–(aq)] = 10–11 M and [H3O+(aq)] = 10–3 M in aqueous solution at pH 3. At this pH, the numeric value of the mathematical product [H3O+(aq)]·[OH–(aq)] = (10– 3 M)(10–11 M) = 10–14 M2. (b) For any pH you choose, the mathematical product [H3O+(aq)]·[OH–(aq)] = 10–14 M2. This observation certainly suggests that there must be some underlying molecular phenomenon that is responsible for the constancy of this product of concentrations. In pure water, the only source of hydronium and hydroxide ions is the reaction of water with itself: H2O(l) + H2O(l) H3O+(aq) + OH–(aq) Recall that in pure water, [H3O+(aq)] = [OH–(aq)] = 10–7 M and, therefore, as in any aqueous solution, [H3O+(aq)]·[OH–(aq)] = 10–14 M2. If the concentration of hydroxide anion is increased (by adding a tiny bit of sodium hydroxide, for example) in this system, the equilibrium is disturbed and the system will respond (Le Chatelier’s principle) by trying to use up some of the added hydroxide. This will, of necessity require reaction of some of the hydronium cation to form some water, that is, the equilibrium reaction, as written, will go toward reactants. The result will be a solution with [H3O+(aq)] < 10–7 M and [OH–(aq)] > 10–7 M. Apparently, given our observation about the product of concentrations, [H3O+(aq)]·[OH–(aq)], the quantitative decrease in [H3O+(aq)] and increase in [OH–(aq)] exactly compensate one another. Problem 2.116. (a) If the sulfur dioxide, SO2(g), molecule has a permanent dipole moment, the S–O bond dipoles must not cancel out. Sulfur dioxide must have a bent structure, similar to water. See Chapter 1, Section 1.6, for a discussion of bond dipoles and molecular dipole moments. (b) If a solution of SO2 in water conducts an electric current, the dissolved gas must have formed ions in the solution. From Section 2.13, we know that the oxides of nonmetals often dissolve and react with water to give a solution of an oxyacid. For sulfur dioxide, the reaction is: SO2(g) + H2O(l) (HO)2SO(aq) The product, sulfurous acid, can transfer one or both protons to water to form ions: (HO)2SO(aq) + H2O(l) H3O+(aq) + HOSO2–(aq) H3O+(aq) + SO32–(aq) HOSO2–(aq) + H2O(l) Although sulfurous acid transfers only about 10% of its protons to water, enough ions will be present in the solution to conduct electricity. Problem 2.117. (a) The name “bicarbonate” differentiates the HCO3– ion from the carbonate ion, CO32–. It could be that the “bi” refers to the fact that it takes two HCO3– ions to be “equivalent” to one CO32– ion when forming ionic compounds and in their reactions. For example, when calcium carbonate is reacted with an acid, carbon dioxide gas is formed and one mole of gas is formed for every two moles of hydronium ion that react: CaCO3(s) + 2H3O+(aq) Ca2+(aq) + CO2(g) + 2H2O(l) However, when calcium bicarbonate is reacted with an acid two moles of carbon dioxide gas are formed for every two moles of hydronium ion that react: Ca(HOCO2)2(s) + 2H3O+(aq) Ca2+(aq) + 2CO2(g) + 4H2O(l) Ratios like these, amount of gas formed for a given amount of acid used, for example, were the only kind of data scientists had when molecular formulas and atomic masses were not known. Thus, the names for the constituents of compounds had names reflecting such observations and results. (b) Sodium ions are monopositive, Na+, and phosphate anions have a 3– charge, PO43–, so the ionic compound they form must be Na3PO4, trisodium phosphate = TSP. Problem 2.118. (a) The equivalence point in a reaction occurs when the moles of one reactant are exactly stoichiometrically equivalent to the moles of the other. In this problem, we are reacting an acid, (HO)2SO2(aq), with a base dissolved in water, KOH(aq). The reaction that occurs is between hydronium cations from the acid and hydroxide anions from the base, but we can also write the reaction in terms of the two compounds: (HO)2SO2(aq) + 2KOH(aq) 2H2O(l) + 2K+(aq) + SO42–(aq) We want to know the volume of 0.075 M sulfuric acid solution, (HO)2SO2(aq), required to reach the equivalence point of the reaction with 1.00 g of KOH(s) dissolved in 75 mL of water. Use the molar mass of KOH (= 56.1 g) to find the number of moles of KOH we have. Then determine the number of moles of (HO)2SO2 required to react with this amount of KOH and, finally, convert moles of (HO)2SO2 to volume of sulfuric acid solution, using the molarity of the sulfuric acid solution. A single equation that accomplishes these three steps is: 1 mol KOH 56.1 g KOH 1.00 g KOH = (1.00 g KOH) 1 mol (HO)2 SO2 2 mol KOH 1 L sol'n 0.075 mol (HO)2 SO2 = 0.119 L = 119 mL sulfuric acid solution (b) In the second part of this problem, we are to find the volume of 0.075 M sulfuric acid solution required to reach the equivalence point of the reaction with 1.00 g of KOH(s) dissolved in 150 mL of water. The amount of KOH here is exactly the same as in part (a). Note that the amount of water the KOH is dissolved in was never considered in part (a), because this volume is irrelevant. All that matters for the calculation is the mass/moles of KOH. Therefore, the volume of 0.075 M sulfuric acid solution required here is the same as in part (a), 119 mL. Problem 2.119. (a) The reaction between drain cleaner, NaOH(s), and vinegar, a 0.9M aqueous solution of ethanoic acid, CH3C(O)OH(aq), actually occurs between hydroxide anions as they dissolve in the aqueous vinegar and the hydronium cations from the acid, but we can write a balanced chemical reaction equation for the reaction between the drain cleaner and vinegar in terms of the compounds: CH3C(O)OH(aq) + NaOH(s) H2O(l) +Na+(aq) + CH3C(O)O–(aq) (b) To find the minimum volume of vinegar required to react completely with one pound (454 g) of drain cleaner [that is, for the reactants to be equivalent by the stoichiometric equation in part (a)], use the molar mass of NaOH (= 40 g) to find the number of moles of NaOH we have. Then determine the number of moles of CH3C(O)OH(aq) required to react with this amount of NaOH and, finally, convert moles of CH3C(O)OH(aq) to volume of vinegar, ethanoic acid solution, using the molarity of the ethanoic acid solution. A single equation that accomplishes these three steps is: 454 g NaOH = (454 g NaOH) 1 mol NaOH 40 g NaOH 1 mol CH2 C(O)OH 1 mol NaOH 1 L sol'n 0.9 mol CH2 C(O)OH = 13 L vinegar (ethanoic acid solution) Problem 2.120. (a) We are asked to find the number of moles of hydronium ion, H3O+(aq), in an acid-rain acidified pond with a volume of 4.5 104 m3 (1 m3 = 1000 L) and [H3O+(aq)] = 5.0 10–5 M (pH = 4.30). Use the volume and molarity to calculate number of moles: 4.5 4 3 10 m = (4.5 1000 L 10 m ) 1 m3 4 3 5.0 10 –5 mol H3 O+ (aq) 1L = 2.25 103 mol H3O+(aq) 2.3 103 mol H3O+(aq) The experimental data (volume of the pond and its pH) do not justify three significant figures (better than 1% uncertainty) in the moles of hydronium, but we will carry the three in the next calculation and round off at the end. (b) You wish to explain how much lime, CaO(s), will be required to react with 90% of the hydronium ion in the pond by this reaction: 2H3O+(aq) + CaO(s) 3H2O(l) + Ca2+(aq) This equation shows us that we need one mole of lime for every two moles of hydronium ion that we want to react. We want to react 90% of the H3O+(aq): [2.25 103 mol H3O+(aq)](.90) = 2.03 103 mol H3O+(aq). Thus, the number of moles of CaO(s) needed is: 103 mol H3O+) (2.03 1 mol CaO + 2 mol H3 O = 1.0 103 mol CaO(s) [about 1000 mol CaO] One mole of CaO has a mass of 56 g, so 1.0 103 mol CaO has a mass of 56 103 g = 56 kg. A mass of one kilogram is 2.2 pounds, so we need (56 kg)(2.2 pound·kg–1) = 123 pounds of CaO(s) to react with 90% of the hydronium ion in the pond. Since the lime is purchased in 40-pound bags, we will need to purchase three bags and use all of two bags and about half of the third. (c) After the lime has been added to react with 90% of the hydronium ion, 10% of the original number of moles of H3O+(aq) will remain, that is, [2.3 103 mol H3O+(aq)](0.10) = 2.3 102 mol H3O+(aq) are still present in the pond. The molar concentration, [H3O+(aq)], is: 2.3 10 2 mol H3O+ [H3O (aq)] = = 5.1 4.5 10 7 L + 10–6 M The pH of the pond is now about 5.3 [–log(5.1 10–6 M) = 5.29], that is, about one pH unit higher (less acidic) than before the addition of the lime. Note that reduction of the [H3O+(aq)] by a factor of 10 increases the pH by one unit. You can also see this relationship in Figure 2.24. Problem 2.121. (a) When chickens pant in warm weather, the shells of their eggs are less sturdy than usual because they contain too little calcium carbonate, which is what make an eggshell hard. The precipitation of calcium carbonate is being affected because the concentration of carbonate in the chicken’s bloodstream goes down when more than the normal amount of carbon dioxide is excreted. The directional flow of the reactions (by La Chatelier’s principle) is: CaCO3(s) Ca2+(aq) + CO32–(aq) CO32–(aq) + H2O(l) HOCO2–(aq) + H2O(l) HOCO2–(aq) + OH–(aq) (HO)2CO(aq) + OH–(aq) (HO)2CO(aq) CO2(g) + H2O(l) As the chicken pants and more CO2(g) leaves, CaCO3(s) tends to dissolve, as shown, rather than precipitate to form strong eggshells. (b) To counteract the loss of CO2(g) in the chickens’ breath, the farmers give them carbonated water, seltzer water, to drink. Chickens apparently like the seltzer water better than plain water and drink even more than usual, so the method is even more effective than might have been expected. Problem 2.122. The animated movie in the Web Companion Chapter 2, Section 2.6, page 3, showing the interaction of chloride and silver ions provides one way to visualize the precipitation process that we usually write like this: Ag+(aq) + Cl–(aq) AgCl(s) The movie tries to suggest some of the complexity of the molecular/ionic interactions that result in the formation of a precipitate when appropriate cations and anions are mixed. The ions are surrounded by water molecules in their hydration layers and these interactions with the water molecules compete with the interactions of the cations with the anions. However, the cations and anions sometimes come together to form an ion pair, AgCl(aq), in which some of the waters of hydration from the individual ions are lost. When these ion pairs meet one another in a favorable orientation, larger groups of ions (aggregates) can form, Ag2Cl2(aq), with the likely loss of more waters of hydration. This process continues with the aggregates growing as more ion pairs and individual ions join, until the aggregate is no longer solvated enough to hold the mass in solution and it falls out of solution as a solid, AgxClx(s). Ion pairs and individual ions continue to join this solid and ultimately form the visible crystal that we represent as AgCl(s), with no indication of the unknown (very large) number of each ion present. All these processes are reversible, but go in the direction of precipitation until the solution reaches saturation and the concentrations of the ions are in equilibrium with the solid crystals. It is not easy to translate the molecular level animation to a symbolic representation, particularly because the number of waters of hydration on all the species is unknown. Assuming that each ion has six waters of hydration and that about half are lost by the aggregates at each step, we can write these molecular level reactions: Ag+(H2O)6 + Cl–(H2O)6 AgCl(H2O)6 + 6H2O(l) AgCl(H2O)6 + AgCl(H2O)6 Ag2Cl2(H2O)6 + 6H2O(l) Ag2Cl2(H2O)6 + AgCl(H2O)6 Ag3Cl3(H2O)6 + 6H2O(l) And so on and on the process continues until the aggregates are too large to be held in solution by their waters of hydration and, finally, we might write: AgnCln(aq) AgCl(s) + some H2O(l) molecules leaving the solid surface Remember that these symbolic representations still simplify what is going on, since many of each of these kinds of reactions are happening simultaneously in the solution. A “simple” reaction like precipitation is really not so simple.