Acid Base Properties of Salt Solutions

Acid Base Properties of Salt Solutions Salts are solids (at SATP) composed of cations and anions arranged in a crystalline lattice. When they dissolve in water, they dissociate into individual hydrated ions that may or may not affect the p of the solution. Sodium chloride, NaCl (s), like all sodium salts, is a strong electrolyte. When dissolved in water it dissociates quantitatively into aqueous sodium and chloride ions. 8.3 INVESTIGATION 8.3.1 The p of Salt Solutions (p. 627) Do soluble salts affect the p of aqueous solutions? NaCl (s) Na (aq) Cl (aq) When the p of a sodium chloride solution is tested with a p meter or acid base indicator, it shows a p of 7 at SATP: neutral. Sodium chloride does not produce hydronium or hydroxide ions in solution, so it is classed as a neutral electrolyte. Sodium carbonate, Na 2 CO 3(s), is another highly soluble ionic compound. It dissociates quantitatively in water to produce aqueous sodium and carbonate ions. Na 2 CO 3(s) 2Na (aq) CO 3 2 (aq) When tested with a p meter or acid base indicator, the sodium carbonate solution proves to have a p greater than 7: basic. owever, sodium carbonate cannot contribute hydronium or hydroxide ions to the solution directly. So, why is a sodium carbonate solution basic while a sodium chloride solution is neutral? The reason must lie in the action of aqueous carbonate ions, not sodium ions, because both salts release sodium ions in solution. The basic character of carbonate solutions can be explained by the Brønsted Lowry theory. As in the case of aqueous ammonia, a Brønsted Lowry base acts as a proton acceptor, and may remove a proton from water to form hydroxide ions in solution. The following equation describes how the carbonate ion acts as a Brønsted Lowry base. CO 2 3(aq) 2 O (l) O (aq) CO 3(aq) base acid Empirical studies have shown that the carbonate ion is a relatively weak base with a base ionization constant, K b, of 2.1 10 4. Ammonium chloride, N 4 Cl (s), is a soluble salt that forms an acidic solution when dissolved in water. The dissociation of the ions in aqueous solution is described by the equation N 4 Cl (s) N 4 (aq) Cl (aq) The p of an NaCl (aq) solution is neutral, so p is unaffected by the presence of Cl (aq) ions. Is it possible that the acidity of ammonium chloride is caused by the N 4 (aq) ion? We can hypothesize that a Brønsted-Lowry proton-transfer reaction occurs between ammonium ions and water molecules: N 4 (aq) 2 O (l) e 3 O (aq) N 3(aq) Laboratory tests show that the ammonium ion is indeed a weak acid with an acid ionization constant, K a, of 5.8 10 10. In general, since salts contain two different ions, the p of an aqueous salt solution may be affected by the anion, the cation, or both. NEL Acid Base Equilibrium 581

hydrolysis a reaction of an ion with water to produce an acidic or basic solution (hydronium or hydroxide ions) LEARNING TIP Do not confuse the definition of hydrolysis as it applies to acid base chemistry with the meaning of the term presented in Chapter 1. In acid base chemistry it means a reaction of an ion with water to produce an acidic or basic solution. In Chapter 1 you used it to mean the breaking of a covalent bond using the elements of water. The reaction of an ion with water to produce an acidic or basic solution is called hydrolysis (from the Greek hydro, meaning water, and lysis, meaning splitting ). In the case of sodium chloride, neither sodium nor chloride ions hydrolyze, so neither ion affects the acid base properties of an aqueous solution; the solution remains neutral. Conversely, when ammonium carbonate dissolves, carbonate ions hydrolyze to produce hydroxide ions (which may produce a basic solution) and ammonium ions hydrolyze to produce hydronium ions (which may produce an acidic solution). You must always consider both ions when assessing the effect of a salt on the p of an aqueous solution. So, can we accurately predict if a salt will produce an acidic or a basic solution? In the following analysis, we will develop models that will help you make such predictions. Salts That Form Neutral Solutions In general, salts that consist of the cations of strong bases (like Na (aq) and K (aq)) and the anions of strong acids (like Cl (aq) and NO 3 (aq) ) have no effect on the p of an aqueous solution. Examples include NaCl (aq),kcl (aq),nai (aq), and NaNO 3(aq) (Table 1). Table 1 Composition of Neutral Salts Salt Cation from strong base Anion from strong acid NaCl NaO Cl KCl KO Cl NaI NaO I NaNO 3 NaO NO 3 Table 3 Metal cation * K a for Some Metal Ions at SATP K a Zr 4 (aq) 2.1 Sn 2 (aq) Fe 3 (aq) Cr 3 (aq) Al 3 (aq) Be 2 (aq) Fe 2 (aq) Pb 2 (aq) Cu 2 (aq) 2.0 10 2 1.5 10 3 1.0 10 4 9.8 10 6 3.2 10 7 1.8 10 7 1.6 10 8 1.0 10 8 * The aqueous metal ion is a hydrated complex ion (e.g., Cu( 2 O) 2 4(aq) ). Aqueous ions of transition metals are usually written in a simplified form, without showing the number of water molecules present in the actual hydrated complex ion, as shown in Table 3. Salts That Form Acidic Solutions Cations such as the ammonium ion, N 4 (aq), and hydrazinium ion, N 2 5 (aq),act as Brønsted Lowry acids. They hydrolyze to form hydronium ions in solution. Consequently, solutions of N 4 Cl (aq) and N 2 5 Cl (aq) are acidic. In general, cations that are conjugate acids of weak molecular bases act as weak acids that have a tendency to lower the p of a solution (Table 2). Table 2 Composition and ydrolysis of Acidic Salts Salt Cation (acid) of weak base ydrolysis reaction N 4 Cl N 4 (aq) N 3 N 4 (aq) 2 O (l) e 3 O (aq) N 3(aq) N 2 5 Br N 2 5 (aq) N 2 4 N 2 5 (aq) 2 O (l) e 3 O (aq) N 2 4(aq) Certain metal cations also hydrolyze and act as acids. An aluminum nitrate solution, Al(NO 3 ) 3(aq), is acidic. owever, other solutions, such as sodium nitrate, NaNO 3(aq), and calcium nitrate, Ca(NO 3 ) 2(aq), are neutral. Why are some metal salts acidic and some neutral? We could propose a hypothesis that the aluminum ion is responsible for the acidity of the solution. Further tests with aluminum solutions indeed establish that Al 3 behaves as an acid in water (increases [ (aq)]). The research indicates that Group 1 and 2 metal ions (except for Be 2 ) do not produce acidic solutions, but that highly charged small ions do form acidic solutions. (See Table 3 and Appendix C9.) Ions such as Al 3 (aq),fe 3 (aq), and Sn 2 (aq) have large (positive) charge densities a large amount of charge in a small volume. These cations produce hydronium ions indirectly by a slightly different reaction than the ammonium ion example above. When a highly charged metal ion such as Al 3 dissolves in water, it becomes hydrated with six water molecules (waters of hydration) according to the following equation: Al 3 (aq) 6 2 O (l) e Al( 2 O) 6 3 (aq) 582 Chapter 8 NEL

Section 8.3 This means that six water molecules bond to the ion with relatively weak electrostatic forces of attraction. The high-charge density of the Al(aq) 3 ion increases the polarity of the O bonds in the 2 O molecules. That is, the 2 O molecules in Al( 2 O) 3 6(aq) are more likely to transfer a proton to the solvent ( 2 O (l) ) than are 2 O (l) molecules in pure water. Although we might imagine that several (or all) of the water molecules in Al( 2 O) 3 6(aq) could transfer a proton, experiments show that only one 2 O will. Thus, Al( 2 O) 3 6(aq) behaves as a weak monoprotic acid in aqueous solution, according to the following equation: (aq) Al( 2 O) 3 6(aq) 2 O (l) 3 O (aq) Al( 2 O) 5 (O) 2 (aq) acid base conjugate conjugate base acid Al 3+ Al 3+ + + Al( 2 O) 3+ 6 (aq) Al( 2 O) 5 (O) 2+ (aq) + + (aq) Notice that this weak acid and its acid ionization constant appear in Appendix C9. No cation with low charge density acts as an acid in this way. This includes all of the singly charged ions of Group 1 metals, Li,Na,K,Rb, and Cs. Since they cannot hydrolyze, they do not affect the p of aqueous solutions. With the notable exception of Be 2, none of the Group 2 cations acts as an acid either. Salts That Form Basic Solutions A sodium chloride solution, NaCl (aq), is neutral, but a sodium acetate solution, NaC 2 3 O 2(aq), is basic. Why? Both solutions contain aqueous sodium ions that do not hydrolyze in solution, so the change in p must be due to the effect of the acetate ion, C 2 3 O 2 (aq). It is reasonable to suspect hydrolysis to be the cause. In this case, hydrolysis caused by the acetate ion produces hydroxide ions and increases the p of the solution. The following two equations show the dissociation of sodium acetate followed by the hydrolysis of the resulting acetate ion: NaC 2 3 O 2(s) Na (aq) C 2 3 O 2(aq) C 2 3 O 2(aq) 2 O (l) e C 2 3 O 2(aq) O (aq) (dissociation) (hydrolysis) NEL Acid Base Equilibrium 583

Earlier, we explained that Group 1 anions like Cl (aq) cannot hydrolyze. So, why is it that the acetate ion is able to hydrolyze, but the chloride ion cannot? Earlier in this chapter, you learned that the stronger the acid, the weaker its conjugate base. Extremely strong acids such as Cl (aq) and NO 3(aq) possess extremely weak conjugate bases, Cl (aq) and NO 3 (aq). These bases attract protons ( ions) so poorly that they are unable to hydrolyze to any significant degree. If they did attract protons strongly, Cl and NO 3 would not be strong acids and would not ionize so readily in aqueous solution. In general, the conjugate bases of strong acids (e.g., Cl (aq),br (aq),i (aq),no 3(aq)) do not affect the p of an aqueous solution. Conversely, the acetate ion, C 2 3 O 2 (aq), is the conjugate base of acetic acid, C 2 3 O 2(aq). Acetic acid is a weak acid because the acetate ion portion of the molecule possesses a relatively high affinity for the ionizable proton ( ion). This relatively strong attraction for protons makes the acetate ion able to hydrolyze to form O (aq) ions in aqueous solution. A solution of sodium acetate is basic because the sodium ions cannot act as acids, but the acetate ions do act as weak bases according to the hydrolysis equation above. Remember that, in general, the conjugate base of a weak acid is also weak. The K a of acetic acid is 1.8 10 5 (weak), and the K b of its conjugate base, the acetate ion, is 5.6 10 10 (weak). owever, when the acetate ion is introduced as the anion of a salt containing a nonhydrolyzing cation like Na (aq) or K (aq), the acetate ion forms the major acid base equilibrium in solution. (The only other reaction is the autoionization of water.) This results in a net production of O (aq) and a basic solution. When determining whether an anion will affect the p of a solution, it helps to remember that the anion of a strong acid is too weak a base to affect the p of an aqueous solution (e.g., Cl (aq) and NO 3 (aq) ), and the anion of a weak acid, although it is also weak, is a strong enough base (when acting alone) to affect the p of an aqueous solution. It will tend to increase the p of the solution and make it more basic. In general, a salt made of a nonhydrolyzing cation (like Na (aq) or K (aq)), and an anion that is the conjugate base of a weak acid, will form a basic aqueous solution (p 7). Salts That Act as Acids and Bases Some salts contain the cation of a weak base and the anion of a weak acid both ions can hydrolyze. To obtain a precise estimate of the p of such a solution, we would have to solve two hydrolysis equilibria simultaneously. This is a difficult mathematical problem. owever, we can predict whether the solution will be acidic, basic, or (approximately) neutral without performing any calculations. Consider the ions produced when ammonium cyanide, N 4 CN (s), dissolves in water. N 4 CN (s) N 4 (aq) CN (aq) (dissociation) The cation, N 4 (aq), is the conjugate acid of the weak molecular base, N 3(aq), so it will hydrolyze to produce hydronium ions in solution. Its tendency is to lower the p. The anion, CN (aq), is the conjugate base of the weak acid, CN (aq), so will hydrolyze to produce hydroxide ions in solution. Its tendency is to increase the p. What is the net effect? That depends on the relative strengths of the two ions, as judged by the K a value for the acid and the K b value for the base. If these values are equal, then the salt will have no net effect on the p of the solution. If the K a of the acid is larger than the K b of the base, then the solution will be more acidic. Conversely, a higher K b would result in a basic solution. In the case of ammonium cyanide, 584 Chapter 8 NEL

Section 8.3 N 4(aq) 2 O (l) N 3(aq) 3 O (aq) CN (aq) 2 O (l) CN (aq) O (aq) K a 5.8 10 10 mol/l K b 1.6 10 5 mol/l Since the K b of the cyanide ion (1.6 10 5 mol/l) is much larger than the K a of the ammonium ion (5.8 10 10 mol/l), an aqueous solution of ammonium cyanide will be basic. SUMMARY Predicting the Acid Base Behaviour of a Salt The following steps will help you predict whether the ions of a salt have an effect on the p of an aqueous solution. Step 1 Step 2 Step 3 Determine if the cation is the conjugate acid of a weak base or a cation with high charge density. If so, it will make the solution more acidic. If not, it will not affect the p of the solution. Determine if the anion is the conjugate base of a weak acid. If so, it will make the solution more basic. If not, it will not affect the p of the solution. If the salt has a cation and an anion that can both hydrolyze, compare the K a and K b values of the cation and anion. If K a K b, then the solution will become more acidic. If K a K b, the solution will become more basic. If K a K b, the solution will be neutral. Predicting the Acidic or Basic Nature of Solutions SAMPLE problem (a) Predict whether a 0.10 mol/l solution of NaNO 2(aq) will be acidic, basic, or neutral. (b) Calculate the p of a 0.10 mol/l solution of NaNO 2(aq). (a) First, write the dissociation equation for NaNO 2(s). NaNO 2(s) Na (aq) NO 2 (aq) Next, examine the cation. Na (aq) is a Group 1 metal cation so will have no effect on p. Now, examine the anion, NO 2 (aq). It is the conjugate base of the weak acid, NO 2(aq), so will act as a base and increase the p. A 0.1 mol/l solution of NaNO 2(aq) will be basic. (b) Since Na (aq) cannot hydrolyze, the p of a 0.10 mol/l NaNO 2(aq) solution is determined solely by the reaction of NO 2 (aq) with water. Write the chemical equation for the hydrolysis reaction between the base, NO 2 (aq), and water, then write the corresponding K b expression. NO 2 (aq) 2 O (l) e O (aq) NO 2(aq) [O (aq) ][NO 2(aq) ] [NO2 (aq) ] K b NEL Acid Base Equilibrium 585

Calculate the value of K b for NO 2 (aq) from the value of K a for NO 2(aq) (which you can find in Appendix C9). K a K b K w K b K b w K K a 14 1. 0 10 7. 2 10 4 1.4 10 11 Now, construct an ICE table to show the changes in concentration that occur as the reaction reaches equilibrium. Note that, in the reaction, one mole of NO 2(aq) and one mole of O (aq) are produced for every mole of NO 2 (aq) that reacts. Let x represent the changes in concentration that occur as the reaction establishes equilibrium. Table 4 ICE Table for the ydrolysis of NO 2 (aq) [O (aq) ] [NO2(aq) ] [ NO2 ( aq) ] K b [O (aq)] [NO 2(aq) ] x [NO 2 (aq) ] 0.10 x 0.10 (Since the value of K b is so small, we will make the assumption that (0.10 x) 0.10. This simplification is warranted by the hundred rule. Check it yourself.) 2 x 1.4 10 11 0.10 x 1.4 10 12 x 1.2 10 6 As usual, we now justify our simplifying assumption, using the 5% rule. 1.2 1 100% 1.2 10 3 % 0.10 0 6 Since 1.2 10 3 % 5%, our assumption is justified. Therefore, x 1.2 10 6 [O (aq)] 1.2 10 6 mol/l Now, we can calculate po and p. po log [O (aq)] log(1.2 10 6 mol/l) po 5.92 p po 14.00 14.00 po 14.00 5.92 p 8.08 The p of a 0.10 mol/l NaNO 2(aq) solution is 8.08. NO 2 (aq) 2 O (l) e O (aq) NO 2(aq) Initial concentration (mol/l) 0.10 0 0 Changes in concentration (mol/l) x x x Equilibrium concentration (mol/l) 0.10 x x x 586 Chapter 8 NEL

Section 8.3 Example (a) Predict whether a 0.20 mol/l solution of ammonium chloride, N 4 Cl (aq), will be acidic, basic, or neutral. (b) Calculate the p of a 0.20 mol/l solution of N 4 Cl (aq). Solution (a) First, write the dissociation equation for N 4 Cl (aq). N 4 Cl (aq) N 4 (aq) Cl (aq) Since Cl (aq) is the conjugate base of a strong acid, it will not affect the p of the solution. N 4(aq) is the conjugate acid of the weak base, N 3(aq), so it will hydrolyze according to the following equation: N 4(aq) 2 O (l) e 3 O (aq) N 3(aq) A solution of N 4 Cl (aq) will be acidic. (b) N 4 (aq) 2 O (l) e 3 O (aq) N 3(aq) [ 3 O [ ( aq) ] N N4 [ ( aq K a 5.8 10 10 (from Appendix C9) 3(aq) ] ) ] K a Table 5 ICE Table for the ydrolysis of N 4(aq) N 4(aq) 2 O (l) e 3 O (aq) N 3(aq) Initial concentration (mol/l) 0.20 0 0 Changes in concentration (mol/l) x x x Equilibrium concentration (mol/l) 0.20 x x x [ 3 O (aq)][n 3(aq) ] [N 4 (aq) ] K a x 2 5.8 10 10 0.20 x Predicting whether 0.20 x 0.20... [A ] initial 0.20 K 5.8 10 10 a 3.4 10 8 Since 3.4 10 8 100, we assume that 0.20 x 0.20. 2 x 5.8 10 10 0.20 x 2 2.9 10 9 x 5.4 10 5 Since x [ 3 O (aq)], [ 3 O (aq)] 5.4 10 5 p log [ 3 O (aq)] log(5.4 10 5 ) p 4. The p of a 0.20 mol/l N 4 Cl (aq) solution is 4. NEL Acid Base Equilibrium 587

Practice Understanding Concepts Answers 3. 4.88 4. 4.92 1. Predict whether the following solutions are acidic, basic, or neutral. Provide explanations to support your predictions. (a) ammonium phosphate, (N 4 ) 3 PO 4(aq) (fertilizer) (b) ammonium sulfate, (N 4 ) 2 SO 4(aq) (fertilizer) (c) magnesium oxide, MgO (aq) (milk of magnesia) 2. Predict whether a solution of sodium sulfite, Na 2 SO 3(aq) (photographic developer) will be acidic, basic, or neutral. 3. Calculate the p of a 0.30 mol/l ammonium nitrate (fertilizer) solution. 4. Calculate the p of 0.25 mol/l N 4 Br (aq). 5. Predict whether a solution of N 4 C 2 3 O 2(aq) is acidic, basic, or neutral. Explain. Making Connections 6. What kind of fertilizers would be appropriate for acid-loving plants like evergreens? ydrolysis of Amphoteric Ions Remember that all polyatomic ions whose chemical formulas begin with hydrogen, (e.g., CO 3 (aq) and SO 4 (aq) ) are amphoteric. As mentioned in Section 8.1, the term amphiprotic may also be used to describe such entities because they can either donate or accept a hydrogen ion (proton). NaCO 3(s) dissolves in water, forming a conducting solution containing sodium and hydrogen carbonate ions. NaCO 3(s) Na (aq) CO 3 (aq) Because it is amphoteric, the CO 3 (aq) ion may hydrolyze as an acid or a base, according to the following equilibrium equations: CO 3(aq) 2 O (l) e 3 O + (aq) CO 2 3(aq) CO 3(aq) 2 O (l) e O (aq) 2 CO 3(aq) (acid hydrolysis) (base hydrolysis) Testing a sodium hydrogen carbonate solution with litmus paper reveals that it is basic. Therefore, we assume that the base hydrolysis equilibrium dominates in solution. Similarly, the acidic character of a sodium hydrogen sulfate, NaSO 4(s) solution is explained by the dissociation and subsequent acid hydrolysis of the hydrogen sulfate ion. NaSO 4(s) Na (aq) SO 4(aq) (dissociation) SO 4(aq) 2 O (l) e 3 O + (aq) SO 2 4(aq) (acid hydrolysis ) In both cases, the hydrolyzing ion may act as an acid or a base, but one equilibrium predominates and gives rise to the overall acidic or basic character of the aqueous solution. ow do we predict which one will win out? The following discussion describes the theory. Picture a laboratory setting in which a student tests the p of a sodium dihydrogen borate, Na 2 BO 3(aq), solution. The solution has a p greater than 7. What reaction occurs to make the solution basic? 588 Chapter 8 NEL

Section 8.3 The following equations describe the dissociation of Na 2 BO 3(s) in water, and the hydrolysis of the 2 BO 3 (aq) ions in solution. Na 2 BO 3(s) Na (aq) 2 BO 3 (aq) (dissociation) 2 BO 3(aq) 2 O (l) e 3 O (aq) BO 2 3(aq) 2 BO 3(aq) 2 O (l) e O (aq) 3 BO 3(aq) (acid hydrolysis) (base hydrolysis) Since the solution is basic when tested with an acid base indicator, 2 BO 3 (aq) must hydrolyze primarily as a base. Is there a way to predict the acid base character of a solution without testing it directly in the laboratory? Remember that when a salt containing both a hydrolyzing anion and a hydrolyzing cation, such as N 4 CN (aq), is dissolved in water, we predict the acid base characteristics of the solution by comparing K a and K b values. The following sample problem will provide you with a similar model for predicting hydrolysis in solutions of amphoteric ions. Predicting the Acidity of Amphoteric Ions SAMPLE problem Predict whether an aqueous solution of baking soda, NaCO 3(s), is acidic, basic, or neutral. First, write the dissociation equation. NaCO 3(s) Na (aq) CO 3 (aq) (dissociation) Next, find the acid ion dissociation constant (from a reference table), as if the hydrogen carbonate ion were to act as an acid: CO 3 (aq) 2 O (l) e CO 3 2 (aq) 3 O (aq) K a 4.7 10 11 (hydrolysis as an acid) Next, find the base ion dissociation constant, as if the hydrogen carbonate ion were to act as a base (use either a table of K b values, or calculate from K a ): CO 3 (aq) 2 O (l) e 2 CO 3(aq) O (aq) K b 2.7 10 8 (hydrolysis as a base) According to the relative values of K a and K b, the baking soda solution should be basic because the K b is larger than the K a. The hydrogen carbonate ion is a stronger base than it is an acid. Example Predict whether a solution of NaBO 3(aq) is acidic, basic, or neutral. Solution Na 2 BO 3(s) Na (aq) 2 BO 3 (aq) (dissociation) 2 BO 3 (aq) 2 O (l) e BO 3 2 (aq) 3 O (aq) K a 5.8 10 10 (hydrolysis as an acid) 2 BO 3 (aq) 2 O (l) e 3 BO 3(aq) O (aq) K b 1.7 10 5 (hydrolysis as a base) Since K b K a, a solution of Na 2 BO 3(aq) is basic. Practice Understanding Concepts 7. Make a list of all amphoteric ions from your acid base table, Appendix C9. 8. Predict whether the following solutions are acidic, basic, or neutral. (a) NaSO 4(aq) (b) Na 2 PO 4(aq) NEL Acid Base Equilibrium 589

ydrolysis of Metal and Nonmetal Oxides When calcium oxide, CaO (s), is dissolved in water it produces a basic solution. owever, an aqueous carbon dioxide solution, CO 2(aq), is acidic. Can we explain this evidence using the hydrolysis theory? First, we must realize that calcium oxide and carbon dioxide both have low solubility in water. Therefore, we show the pure substance reacting with water rather than dissolving in water. CaO (s) 2 O (l) Ca 2 (aq) 2O (aq) As you know, most metal oxides have low solubility in water, but the accepted theory is that the solid state oxide ions are converted quantitatively into aqueous hydroxide ions by the reaction with water to form a basic solution, according to the following equation. O 2 (s) 2 O (l) 2O (aq) The O 2 (s) ions do not exist in aqueous solution; they are quantitatively converted to O (aq) ions from the solid state (in CaO (s) ). owever, the equation may be used to explain the basic nature of the solutions of metal oxides. Now, let us try to explain the acidic character of carbon dioxide in water. A possible explanation is the two-step process presented below. CO 2(g) 2 O (l) e 2 CO 3(aq) 2 CO 3(aq) 2 O (l) e 3 O (aq) CO 3(aq) CO 2(g) 2 2 O (l) e 3 O (aq) CO 3 (aq) (net equation) Chemists have done numerous tests on metal oxides and nonmetal oxides to determine the acidic and basic character of the solutions formed. Their evidence led to the following generalizations. Metal oxides react with water to produce basic solutions. Nonmetal oxides react with water to produce acidic solutions. SAMPLE problem Predicting the Acidity of Solutions of Oxides Predict, using the hydrolysis concept, whether solutions of the following oxides will be acidic, neutral, or basic. Write an appropriate hydrolysis equation in each case. (a) magnesium oxide, MgO (s) (b) sulfur dioxide, SO 2(g) (a) Magnesium oxide, MgO (s), is a metal oxide, so will react with water to form a basic solution, according to the following hydrolysis equation: MgO (s) 2 O (l) e Mg 2 (aq) 2O (aq) The hydroxide ions are produced according to the following equation: O 2 (s) 2 O (l) e 2O (aq) (b) Sulfur dioxide, SO 2(g), is a nonmetal oxide, so will react with water to form an acidic solution, according to the following hydrolysis equation: SO 2(g) 2 2 O (l) e 3 O (aq) SO 3(aq) 590 Chapter 8 NEL

Section 8.3 Example Use hydrolysis concepts to predict whether an aqueous solution of copper(ii) oxide, CuO (s), is acidic, basic, or neutral. Solution Copper (II) oxide, CuO (s), is a metal oxide, so will react with water to form a basic solution, according to the following equation: CuO (s) 2 O (l) e Cu 2 (aq) 2O (aq) O 2 (s) 2 O (l) e 2O (aq) The acidic properties of nonmetal oxides are responsible for many natural processes such as the the weathering of minerals, the absorption of nutrients by the roots of plants, and the chemistry of tooth decay. For example, limestone, CaCO 3(s), dissolves in water that is made acidic by the dissolution of CO 2(g) from the atmosphere, according to the following two-step process: CO 2(g) 2 O (l) e + (aq) CO 3(aq) (aq) CaCO 3(s) e Ca 2+ (aq) CO 3(aq) (acidification of rain or ground water) (dissolution of CaCO 3(s) in acidic solution) Limestone caves and the stalagmites and stalactites they contain are formed by the action of acidic ground water on limestone deposits in Earth s crust. A cave is formed when the above reactions proceed to the right and calcium carbonate in underground limestone deposits is dissolved. Stalactites are formed in the cave when the aqueous solution on the ceiling evaporates. As it evaporates, carbon dioxide escapes. This shifts the equilibrium to the left, causing solid calcium carbonate to precipitate. This precipitation causes stalactites to form on the ceilings and stalagmites to form where drops of the solution drip onto the cave floor. Stalactite and stalagmite formation is very slow; the growth rate is, on average, about one millimetre per century. Tooth decay is also caused by the dissolution of minerals in acidic solutions. Tooth enamel is composed of the mineral hydroxyapatite, Ca 5 (PO 4 ) 3 O (K sp 6.8 10 37 ). Acids (in saliva, from fruits and fruit juices, or formed when sugars are metabolized by bacteria in the mouth) react with hydroxyapatite, leading to erosion of tooth enamel and eventually tooth decay (cavities). Fluoride salts (as a source of fluoride ions) are often added to toothpaste and to drinking water in treatment plants in some communities to help prevent tooth decay. The fluoride ions react with the Ca 5 (PO 4 ) 3 O in tooth enamel to form more decay-resistant fluorapatite Ca 5 (PO 4 ) 3 F (K sp 1.0 10 60 ). DID YOU KNOW? Growing Up or Growing Down? When viewing photos of stalactites and stalagmites in a cave, you may not be able to tell which way is up. Look at the tips of the formations. Stalactites almost always have pointed tips, whereas stalagmites are usually rounded or flat. Is the photo right-side up or upside down? Practice Understanding Concepts 9. For each of the following solutions of compounds, write an ionization or dissociation equation where appropriate, and then write a net equation showing reactions with water to produce either hydronium or hydroxide ions (consistent with the evidence). (a) Na 2 O (s) in solution turns red litmus blue. (b) SO 3(g) in solution turns blue litmus red. NEL Acid Base Equilibrium 591

Making Connections 10. Limestone caves are very popular tourist attractions. Some of the most popular caves attract up to 500 000 people per year. Many speleologists (cave experts) believe that cave formations such as stalactites and stalagmites are in danger of dissolving away on account of the large numbers of visitors. (a) Explain how the presence of large numbers of humans may affect the chemical stability of stalagmites and stalactites in caves. (b) Suggest possible solutions that do not include barring people from the caves. 11. Many communities are against the addition of fluoride to municipal drinking water. Conduct library and Internet research to learn more about this issue. List arguments for and against this practice and write a brief position paper. GO www.science.nelson.com SUMMARY Acid Base Characteristics of Salts Table 6 Type of Salt Examples Description p cation of strong base and NaCl (aq), KNO 3(aq) does not hydrolyze neutral anion of strong acid NaI (aq) as an acid or as a base cation of strong base and NaC 2 3 O 2(aq) anion hydrolyzes as a base; basic anion of weak acid KF (aq) cation does not hydrolyze cation is conjugate acid N 4 NO 3(aq) cation hydrolyzes as an acid; acidic of a weak base; anion does not hydrolyze anion of a strong acid N 4 Cl (aq) cation is conjugate acid N 4 C 2 3 O 2(aq) cation hydrolyzes as an acid; acidic if K a K b of a weak base; anion is N 4 F (aq) anion hydrolyzes as a base basic if K a K b conjugate base of a weak neutral if K a K b acid cation is highly AlCl 3(aq) hydrated cation hydrolyzes acidic charged metal ion; FeI 3(aq) as an acid; anion of a strong acid anion does not hydrolyze metal oxides CuO (s) solid state oxide ion reacts with basic water to form O (aq) nonmetal oxides CO 2(g) compound reacts with water acidic to form 3 O (aq) Lewis acid an electron-pair acceptor Lewis base an electron-pair donor The Lewis Model of Acids and Bases The Arrhenius and Brønsted Lowry models successfully explain much of the behaviour of acids and bases. Nevertheless, both of these models contain limitations. Remember that the Arrhenius model could not adequately explain the basic properties of an aqueous ammonia solution. In the early 1920s, G. N. Lewis expanded the Brønsted Lowry model to encompass a number of substances that would not normally be classified as Brønsted Lowry acids or bases. According to the Lewis model, a Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor. In order to act as a Lewis base, a substance must possess a non-bonded pair of electrons in one of its orbitals. Conversely, in order to act as a Lewis acid, a substance must possess an empty valence orbital that may accept (share) a pair of non-bonding valence electrons from a Lewis base. The following structural formula equation illustrates the reaction between a Lewis acid, (aq), and a Lewis base, 2 O (l), to form 3 O (aq). 592 Chapter 8 NEL

Section 8.3 + O O + + ion (Lewis acid) water (Lewis base) hydronium ion In the above reaction, the (aq) ion (proton) acts as the Lewis acid (electron pair acceptor) and the water molecule acts as a Lewis base (electron-pair donor). In this case, the Lewis base is also a Brønsted Lowry base, and the Lewis acid is also a Brønsted Lowry acid. owever, in the following example, this is not the case. F B F F boron trifluoride (Lewis acid) N ammonia (Lewis base) F F F B N boron trifluoride ammonia complex Note that the Brønsted Lowry model also accounts for ammonia as a base, but does not characterize boron trifluoride as an acid. The Lewis acid base theory explains the reaction between BF 3 (boron trifluoride) and N 3 (ammonia). Boron trifluoride is a trigonal planar molecule with sp 2 hybrid orbitals. This arrangement leaves an empty 2p orbital on the boron atom that is able to accommodate the pair of nonbonding electrons in the sp 3 hybrid orbital of N 3.A covalent bond forms between the boron and the nitrogen, forming the compound BF 3 N 3. The Lewis acid base theory can also explain why small, highly positive ions such as Al 3 form complex ions in water: Al 3 (aq) + 6 2 O (l) e Al( 2 O) 6 3 (aq) This is a Lewis acid base reaction. The water molecules each possess nonbonding pairs of electrons and so act as Lewis bases, and the Al(aq) 3 ion possesses empty 3s,3p,and 3d orbitals that may accommodate electron pairs. The electron configuration of the Al 3 ion can be represented as Al 3 : [Ne] 3s 0 3p 0. The complex Al( 2 O) 3 6(aq) is formed when an Al 3 ion, acting as a Lewis acid, bonds with six water molecules, each acting as a Lewis base. The Lewis model is a more general model of acids and bases that not only encompasses the Brønsted Lowry model, but extends it. Brønsted Lowry acids and bases are acids and bases in the Lewis model, but the reverse is not always true. Example Identify the Lewis acid and the Lewis base in the following reaction. SO 3(aq) 2 O (l) e 2 SO 4(aq) O O S O sulfur trioxide (Lewis acid) O O O S O O water sulfuric acid (Lewis base) NEL Acid Base Equilibrium 593

Solution SO 3(aq) is the Lewis acid and the water is the Lewis base. Practice Understanding Concepts 12. Identify the Lewis acid and the Lewis base in each of the following reactions. (a) (aq) O (aq) 2 O (l) (b) (aq) N 3(aq) e N 4 (aq) Section 8.3 Questions Understanding Concepts 1. Predict whether the following solutions are acidic, basic, or neutral. Provide explanations to support your predictions. (a) table salt, NaCl (aq) (saline or brine) solution (b) aluminum chloride, AlCl 3(aq) (antiperspirant) (c) Na 2 CO 3(aq) (washing soda) 2. Predict whether a solution of ammonium carbonate (a component of baking powder) will be acidic, basic, or neutral. 3. What is the strongest possible acid in an aqueous solution, and what is the strongest possible base in an aqueous solution? 4. Will an aqueous solution of BeCl 2(aq) turn litmus red or blue? Explain. 5. Predict whether the following solutions are acidic, basic, or neutral. (a) a carbonated beverage containing CO 2(aq) (pop) (b) strontium oxide, SrO (s) Applying Inquiry Skills 6. Analyze the Evidence (Table 7) to determine the kind of solutions (acidic, basic, or neutral) formed when Period 3 oxides are placed in water. Experimental Design Oxides of elements in Period 3 are tested in water, using litmus paper. To all the oxides, a strong acid (hydrochloric Table 7 acid) and a strong base (sodium hydroxide) are added to determine if a neutralization reaction occurs. Evidence Extension Litmus and Neutralization Tests on Oxides Oxide Litmus test Cl (aq) test NaO (aq) test Na 2 O (s) red to blue neutralizes no reaction MgO (s) red to blue neutralizes no reaction As 2 O 3(s) (insoluble) neutralizes neutralizes SiO 2(s) (insoluble) no reaction neutralizes P 2 O 3(s) blue to red no reaction neutralizes SO 3(g) blue to red no reaction neutralizes Cl 2 O (g) blue to red no reaction neutralizes 7. Nitrogen oxides such as nitrogen dioxide, NO 2(g), are produced in automobile engines and released into the atmosphere via exhaust fumes. In the atmosphere, nitrogen dioxide will dissolve in droplets of rain. Use hydrolysis concepts to predict what will happen to the p of rain when NO 2(g) dissolves. 594 Chapter 8 NEL