656 CHAPTER 16 Acids and Bases Remember that the hy d ron iu m io n can be expressed as either H+ or H 30 + These are two e qu iv alent ways to represent the autoionization of water. Think About It Bec ause the conjugates of weak ac id s a nd ba ses have ionization constants, salts containing these ions have an effect on the pH of a solution. In Section 16.10 we will use the ionization constants of conjugate acids and conjugate ba ses to calculate pH for solutions containing dissolved salts. ' -' ' ' -~-~ - • • As for any chemical equations, we ca n add these two equilibria and cancel identical terms: _ CH 3 COO II (aq) :;:::, ====:' H+ (a q) + _CH 3 COO (aq) DA "'1(-;;'a q;;) ) + H) O(l ) , ' S:: H3 COOII(aq) + OH - Caq) The sum is the autoionization of wate r. In fact, this is the case for any weak acid and its conjugate ba se: 1M :;:::, ====:' H+ + Pi" + K + H 2 0 , ' 1M + OH . . , . . . . . . . . . . . . . . . . . . . . . . . . . . . . . H 2 0 :;::, =::z:' H + + OH - ar for any weak base and its conjugate acid . .B + H 2 0 :;::, ====:' .wr + 0 H- + ~ + H 2 0. ' J1 + H3 0 + . . . . . . . . 2H 2 0 :;::. ====:' H30 + + OH- Recall that when we add two equilibria, the equilibrium constant for the net reaction is the product of the equilibrium constants for the individual equations [ ~~ Section 15 .3]. Thus, for any conjugate acid-base pair, Equation 16.7 Equation 16.7 gives th e quantitative basis for the reciprocal relationship between the strength of an acid and that of its conjugate base Co r between the strength of a base and that of it s conjugate acid). Because Kw is a constant, Kb mu st decrease if Ka increase s, and vice versa. Sample Problem 16.16 shows how to determine ionization constants for conjugates. Sample Problem 16.16 Determine (a) Kb of the acetate ion (CH 3 COO- ), (b) Ka of the methylammo nium ion (CH 3 NH ~ ) , (c) Kb of the fluoride ion (F -), and (d) Ka of t he anunonium ion (NHt) . Strategy Each species listed is eith er a conjugate base or a conjugate acid. Determine the identity of the acid corresponding to eac h conjugate base and the identity of the ba se corresponding to each con jug ate ac id; then, consult Tables 16.6 and 16.7 for their ioni za tion constants. Use the tabulated ionization constants and Equation 16.7 to calculate eac h indicated K value. Setup (a) A Kb value is requested, indicating that the acetate ion is a conjugate ba se. To identify the corresponding Br Sil nsted acid, add a proton to th e formula to get CH 3 COOH (acetic acid). The Ka of acetic acid (from Table 16.6) is 1.8 X 10 -'. (b) A Ka value is requested, indicating that the me thy lamm onium ion is a conjugate acid. Determine the identity of the corresponding BrSiln sted base by removing a proton from the formula to get CH 3 NH 2 (methylamine). The Kb of methylamine (from Table 16.7) is 4.4 X 10- 4 (c) F- is the conjugate base of HF; Ka = 7 .1 X 10- 4 (d) NHt is the conjugate acid ofN H 3 ; Kb = 1. 8 X 10- 5 Solving Equation 16.7 separately for Ka and Kb gives, re sp ec ti vely, and Solution (a) Conjugate base CH 3 COO - : Kb = 1.0 X 10- 14 = 5.6 X 10 - 10 1.8 X 10- 5 (b) Conjugate acid CH 3 NH i : Ka = 1.0 X 10 - 14 = 2.3 X 10 - 11 4.4 X 10 - 4 (c) Conjugate base F- : Kb = 1.0 X 10 - 14 = 1.4 X 10- 11 7 .1 X 10 - 4 (d) Conjugate acid NHt : Ka = 1. 0 X 1O - 1~ = 5.6 X 10- 10 1 .8 X 10 - ) SECTION 16.8 Diprotic and Polyprotic Acids 657 , I Practice Problem A Determine (a) Kb of the benzoate ion (C 6 H s COO- ), (b) Kb of the ascorbate ion I (HC 6 H 6 0 6 ), and (c) Ka of the ethyl ammonium ion (C 2 HsNH ~ ). Practice Problem B Determine (a) Kb of the weak base B whose conjugate acid HB + has Ka . = 8.9 X 10- 4 and (b) Ka of the weak acid HA whose conjugate base has Kb = 2.1 X 10- 8 I ~i ____________________________________________________________________ ~EI Checkpoint 16.7 Conjugate Acid-Base Pairs 16.7.1 Calculate the Kb of the cyanide ion 16.7.2 Which of the anions listed is the (CW). strongest base? (See Table 16.6.) a) 4.9 X 10- 10 a) Ascorbate ion (HC6H60 (j ) b) 2.0 X lO- s b) Benzoate ion (C 6 H s COO- ) c) 4.9 X 10- 24 c) Nitrite ion (N0 2 ) d) 1.0 X 10- 7 d) Phenolate ion (C 6 H s O- ) e) 2.2 X lO- s e) Formate ion (HCOO- ) Diprotic and Polyprotic Acids Diprotic and polyprotic acids undergo successive ionizations, losing one proton at a time [ ~~ Sec - . ,. . . . . . . . . . , . . . . . . . . . tion 4.3], and each ionization has a Ka associated with it Ionization constants for a diprotic acid are designated Ka and Ka . We write a separate equilibrium expression for each ionization, and we J 2 may need two or more equilibrium expressions to calculate the concentrations of species in solu- tion at equilibrium. For carbonic acid (H 2 C0 3 ), for example, we write HC0 3 (aq) +.==' H+(aq) + CO~ - (aq) [H + ][HC0 3 ] K = a J [H 2 C0 3 ] K = a, [H + ][CO ~ -] [HC0 3 ] Note that the conjugate base in the first ionization is the acid in the second ionization, Table 16,8 shows the ionization constants of several diprotic acids and one polyprotic acid, For a given acid, the first ionization constant is much larger than the second ionization constant, and so on, This trend makes sense because it is easier to remove a proton from a neutral species than from one that is negatively charged, and it is easier to remove a proton from a species with a single negative charge than from one with a double negative charge, Sample Problem 16.17 shows how to calculate equilibrium concentrations of all species in solution for an aqueous solution of a diprotic acid. Oxalic acid (H 2 C 2 0 4 ) is a poisonous substance used mainly as a bleaching agent. Calculate the concentrations of all species present at equilibrium in a 0.10 M solution at 25°C. Strategy Follow the same procedure for each ionization as for the determination of equilibrium concentrations for a monoprotic acid. The conjugate base resulting from the first ionization is the acid for the second ionization, and its starting concentration is the equilibrium concentration from the first ionization. Setup The ionizations of oxalic acid and the corre s ponding ionization constants are H 2 C 2 0 4 (aq) +.==' H+(aq) + HC 2 0 4 (aq) HC 2 0 4 (aq)' ' H+(aq) + C 2 0~-(aq) K aJ = 6.5 X 10- 2 Ka = 6,1 X 10- 5 2 Construct an equilibrium table for each ionization, using x as the unknown in the first ionization and y as the unknown in the second ionization. (Continued) , A triprotic acid h as Ka K a , and Ka . I' 1 3 658 CHAPTER 16 Acids and Bases Name of Acid Formula Structure Ka 1 Ka 2 Ka 3 Sulfuric acid Oxalic acid Sulfurous acid Ascorbic acid (vitamin C) Carbonic acid Hydrosulfuric acid* Phosphoric acid • H 2 SO 4 0 II H-O - S-O-H II 0 o 0 II II H-O-C-C-O -H o II H-O-S-O-H H-O O-H "c C/ H" / \ iC""-O'/C=O CHOH I CH?OH - o II H-O-C-O-"H H-S-H o II H-O-P-O-H I o I H Very large 6.5 X 10- 2 1.3 X 10- 2 8.0 X 10- 5 4.2 X 10 - 7 9.5 X 10 - 8 7.5 X 10- 3 1.3 x 10 - 2 6.1 X 10- 5 6.3 X 10- 8 1.6 X 10- 12 4.8 X 10- 11 1 X 10- 19 6.2 X 10- 8 4.2 X 10- 13 *Th e second ionization constant of H 2 5 is very low and difficult to measure. The va lue in this table is an estimate. H 2 C 2 0iaq) :;:.=::!:' H+(aq) + HC 2 0 4 (aq) Initial concentration ( M): 0.10 o o Change in concentration (M) : -x +x +x Equilibrium concentration ( M): 0.10 - x x x The equilibrium concentration of the hydrogen oxalate ion (HC 2 0 4 ) after the first ionization becomes the starting concentration for the second ionization. Additionally, the equilibrium concentration of H+ is the starting concentration for the second ionization. Initial concentration (M): x x o Change in concentration (M): - y +y +y Equilibrium concentration ( M): x - y x+y y Solution K = [H +][HC 2 0 4" l a, [H 2 C 2 0 4 l . 2 6.5 X 10 - 2 = x 0.10 - x Applying the approximation and neglecting x in the denominator of the expression gives 2 6.5 X 10 - 2 = Ox 10 X 2 = 6.5 X 10 - 3 x = 8.1 X 10 - 2 M SECTION 16.8 Diprotic and Po l yprotic Acids 659 Testing the appr ox imation, 8.1 X 10- 2 M X 100% = 81% O.lOM Clearly the approximation is not valid, so we must solve the following quadratic equation: X Z + 6.5 X lO- z x - 6.5 X 10- 3 = 0 The result is x = 0.054 M. Thus, after the first io ni zation, the concentrations of species in solution are [H+l = 0.054 M [HC 2 0 4 l = 0.054M [H Z C 2 0 4 l = (0.10 - 0.054) M = 0.046 M Rewriting the equilibrium table for the second ionization, using the calculated value of x, gives the following: HC Z 0 4 (aq) +. =::!:' H+ (a q) + C20 ~ - (aq) Initial concentration (M): Change in concentration (M) : Equilibrium concentration (M): 0.054 0.054 -y +y 0.054 - y 0.054 + Y K = [H+ ][C 20~ - l a, [HC 2 0 4 l 6 .1 X 10 - 5 = _( 0_.0_S4 _+-= y) ,(y_) 0.054 - y o +y y Assuming that y is very small and applying the approximations 0.054 + Y = 0.054 and 0.054 - y = 0.054 gives ( 0.O S4 )(y) - = =-=, '-'- = = 6. 1 X 10 - ) M 0.054 y We must test the approximation as follows to see if it is valid: 6.1 X 1O- 5 M X 100 '7l = 011 '7l 0.054 M O. 0 This time, because the ionization constant is much smaller, the approximation is valid. At equilibrium, the co ncentrations of all species are [H 2 C 2 0 4 l = 0.046 M [HC 2 0 4 l = (0.054 - 6.1 X 10 - 5 ) M = 0.054 M [H+l = (0.054 + 6 .1 X 10- 5 ) M = 0.054 M [ Cz O ~- l = 6. 1 X 10- 5 M Practice Problem A Calculate the concentrations of H Z C 2 0 4 , HC 2 0 4 , C20 ~- , and H+ ions in a 0.20 M oxalic acid solution at 2S°C. Practice Problem B Calculate the co ncentrations of H 2 S0 4 , HS0 4 , SO ~- , and H+ ions in a 0.14 M sulfuric acid solution at 25 ° C. Checkpoint 16.8 Diprotic and Polyprotic Acids 16 .8.1 Calculate the equilibrium concentration of CO ~ - in a 0.050 M solution of carbonic acid at 25 ° C. a) 4.2 X 10- 7 M b) 4.8 X 10- 11 M c) 1.5 X 10- 4 M d) O.OSOM e) 0.049 M 16 . 8.2 What is the pH of a 0.40 M solution of phosphoric acid at 25 °C? a) 5.48 b) 1.26 c) 3.98 d) 12.74 e) 0.80 Think About It Note that the second ionization did not contribute significantly to the H+ concentration. Therefore, we could determine the pH of this solution by considering only the first ionization. This is true in general for polyprotic acids where K a , is at least 1000 X K a , . [It is necessary to consider the second ionization to determine the concentration of oxalate ion (Cz O ~-). l 660 CHAPTER 16 Ac ids and Bases The polarity of the H - X bo nd actually decreases from H - F to H -I , large ly be c ause F is the most electronegative element. This would suggest tha t HF would be the str o nge st of the hydrohalic acid s. Ba s ed on th e data in Table 16 . 9, however, bond e ntha l py is t he more important factor in determining the stre n gths of these acids. Figure 16.2 Lewis structures of some common oxoacids. Molecular Structure and Acid Strength The strength of an acid is mea sured by its tendency to ionize: HX • H+ + X- Two factors influence the extent to which the acid undergoes ionization. One is the strength of the H- X bond. The stronger the bond, the more difficult it is for the HX molecule to break up and hence the weaker the acid. The other factor is the polarity of the H - X bond. The difference in the electronegativities between H and X results in a polar bond like 0+ 0- H-X If the bond is highly polarized (i.e., if there is a large accumulation of positive and negative charges on the H and X atoms, respectively), HX will tend to break up into H+ and X- ions. A high degree of polarity, therefore, gives rise to a stronger acid. In this section, we consider the roles of bond stren gt h and bond polarity in determining the strength of an acid. Hydrohalic Acids The halogens form a series of binary acids called the hydrohalic acids (HF, HCI, HBr, and HI). Table 16.9 shows that of this series only HF is a weak acid (Ka = 7.1 X 10- 4 ) . The data in the table indicate that the predominant factor in determining the strength of the hydrohalic acids is . bond strength. HF has the largest bond enthalpy, making its bond the most difficult to break. In this series of binary acids, acid strength increases as bond strength decreases. The strength of the acids increases as follows: HF < < HCI < HBr < HI Oxoacids An oxoacid, as we learned in Chapter 2, contains hydrogen, oxygen, and a central, nonmetal atom [ ~~ Section 2.7] . As the Lewis structures in Figure 16.2 show, oxoacids contain one or more 0- H bonds. If the central atom is an electronegative element, or is in a high oxidation state, it will attract electron s, causing the O- H bond to be more polar. This makes it easier for the hydrogen to be lost as H+, making the acid stronge r. Acid Strengtlis Bond Bond Enthalpy (kJ/mol) Acid Strength H-F 562.8 Weak H-CI 431.9 Strong H-Br 366.1 Strong H-I 298.3 Strong •• • H -O-N =O: oo • • • '0' oo II oo H-O-N -O : •• • • • • '0 ' oo II oo H-O- C -O-H • • • • Carbonic ac id N itrous ac id N itri c acid • • • • '0' '0 ' • • '0' oo II oo H-O -P -O-H oo II oo H -O -S-O - H oo II oo . . I .• oo II oo H- O -P-O - H '0' • • .0 . • . I . • I • • H H Ph osphor ou s acid Ph osphoric acid Sulfuric ac id SECTION 16.9 Molecular Structure and Acid Strength 661 •• • • H-O-CI: •• •• •• H-O-C1-0: •• •• •• •• •• Hypochlorous acid (+ 1) Chlorous acid (+3) •• :0: •• :0: • . I H-O-CI-O: . . I H-O-Cl-O: . . I :0: •• •• •• • • Chloric acid (+5) Perchloric acid (+7) To compare their strengths, it is convenient to divide the oxoacids into two groups: 1. Oxoacids having different central atoms that are from the same group of the periodic table and that have the same oxidation number. Two examples are • • • • '0' • • '0' • • •• I •• • • I • • H-O Cl 0: H-O Br 0: • • • •• •• • • • • • • Within this group, acid strength increases with increasing electronegativity of the central atom. Cl and Br have the same oxidation number in these acids, + S. However, because Cl is more electronegative than Br, it attracts the electron pair it shares with oxygen (in the CI-O-H group) to a greater extent than Br does (in the corresponding Br-O-H group). Consequently, the O-H bond is more polar in chloric acid than in bromic acid and ionizes more readily. The relative acid strengths are HCI0 3 > HBr0 3 2. Oxoacids having the same central atom but different numbers of oxygen atoms. Within this • group, acid strength increases with increasing oxidation number of the central atom. Con- sider the oxoacids of chlorine shown in Figure 16.3. In this series the ability of chlorine to draw electrons away from the OH group (thus making the O-H bond more polar) increases with the number of electronegative 0 atoms attached to Cl. Thus, HCI0 4 is the strongest acid because it has the largest number of oxygen atoms attached to Cl. The acid strength decreases as follows: HCIQ4 > HCI0 3 > HCI0 2 > HCIO Sample Problem 16.18 compares acid strengths based on molecular structure. , Predict the relative strengths of the oxoacids in each of the following groups: (a) HClO, HErO, and HIO; (b) HNO} and RN0 2 . Strategy In each group, compare the electronegativities or oxidation numbers of the central atoms to determine which O-H bonds are the most polar. The more polar the O-H bond, the more readily it is broken and the stronger the acid. , Setup (a) In a group with different central atoms, we must compare electronegativitie s, The electronegativities of the central atoms in this group decrease as follows: Cl > Br > L (b) These two acids have the same central atom but differ in the number of attached oxygen atoms. In a group such as this, the greater the number of attached oxygen atoms, the higher the oxidation number of the central atom and the stronger the acid, Solution (a) Acid strength decreases as follows: HCIO > HErO> RIO, . . . . . . . . . · . . . . . . . . . . . , . . . . . (b) RN0 3 is a stronger acid than HN0 2 . Practice Problem Indicate which is the stronger acid: (a) HBr03 or HBr04; (b) H2 Se04 or H 2 S0 4 , • .' , , , , • , Figure 16.3 Lewis structures of the oxoacids of chlorine. The oxidation number of the Cl atom is shown in parentheses. Note that although hypochlorous acid is written as HClO , the H atom is bonded to the ° atom. As the number of attached oxygen atoms increases. the oxidat i on number of the central atom also increases [ ~~ Section 4.4]. • Another way to compare the strengths of these two is to remember that HN0 3 is one of the seven strong acids, HN0 2 is not. Think About It Four of the strong acids are oxoacids: RN0 3 , HCI0 4 , HCI0 3 , and H 2 S0 4 , 662 CHAPTER 16 Acids and Bases You learned in Chapter 4 [ ~~ Section 4. 1] that carboxylic acid formulas are often written with the ionizable H atom last, in order to keep the functional group together. You should recognize the formulas for organic acid s written either way. For example , acetic acid may be written as HC 2 H 3 0 2 or as CH 3 COOH. Recall that a salt is an ionic compound formed by the reaction between an acid and a base [ ~~ Section 4.3] . Salts are strong electrolytes that dissociate completely into ions. Carboxylic Acids So far our discussion has focused on inorganic acids. A particularly important group of organic acids is the carboxylic acids, whose Lewis structures can be represented by • • "0' II R-C-O-H where ' ids ' part ' Of the ' acid ' iiioiecuie ' aiid ' the ' shaded portioii represents ' fue ' carijoxyi" group : -COOH. The conjugate base of a carboxylic acid, called a carboxylate anion; RCOO - , can be repre- sented by more than one resonance structure: • • - U '0' • • - • R C 0: • , R C 0: • In terms of molecular orbital theory [ ~~ S ecti on 9.6], we attribute the stability of the anion to its ability to spread out or delocalize the electron density over several atoms. The greater the extent of electron delocalization, the more stable the anion and the greater the tendency for the acid to undergo ionization that is, the stronger the acid. The strength of carboxylic acids depends on the nature of the R group. Consider, for exam- ple, acetic acid and chloroacetic acid: • • • • H '0' CI "0' I II I II H C C 0 H H-C C 0 H I I H H Acetic acid (Ka = l.8 x 10 - 5 ) Chloroacetic acid (Ka = l.4 x 10- 3 ) The presence of the electronegative CI atom in chloroacetic acid shifts the electron density toward the R group, thereby making the 0- H bond more polar. Consequently, there is a greater tendency for chloroacetic acid to ionize: • • • • :CI: U :CI: U I II I II H- C C O-H , H C C 0: + H+ • I I H H Chloroacetic acid is the stronger of the two acids. • Acid-Base Properties of Salt Solutions In Section 16.7, we saw that the conjugate base of a weak acid acts as a weak Br\?lnsted base in . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . water. Consider a solution of the salt sodium fluoride (NaF). Because NaF is a strong electrolyte, it dissociates completely in water to give a solution of sodium cations (Na +) and fluoride anions (F- ). The fluoride ion, which is the conjugate base of hydrofluoric acid, reacts with water to pro- duce hydrofluoric acid and hydroxide ion: F- (aq) + H 2 0(l) :;:. :=:z; ' HF(aq) + OH - (aq) This is a specific example of salt hydrolysis, in whi ch ions produced by the dissociation of a salt react with water to produce either hydroxide ions or hydronium ions thus impacting pH . Using our knowledge of how ions from a dissolved salt interact with water, we can determine (based on the identity of the dissolved salt) whether a solution will be neutral, basic, or acidic. Note in the preceding example that sodium ions (Na +) do not hydrolyze and thus have no impact on the pH of the solution. Basic Salt Solutions Sodium fluoride is a salt that dissolves to give a basic solution. In general, an anion that is the con - jugate base of a weak acid reacts with water to produce hydroxide ion. Other examples include the acetate ion (CH 3 COO - ), the nitrite ion (N0 2 ), the sulfite ion (SO~ - ), and the hydrogen carbonate SECTION 16.10 Acid-Base Properties of Salt Solutions 663 . . . . . . . . ion (HCO)). Each of these anions undergoes hydrolysis to produce the corresponding weak acid and hydroxide ion: A - (aq) + H 2 0(l) :;::. =::=:" HA(aq) + OH - (aq) We can therefore make the qualitative prediction that a solution of a salt in which the anion is the conjugate base of a weak acid will be basic. We calculate the pH of a basic salt solution the same way we calculate the pH of any weak base solution, using the Kb value for the anion. The necessary Kb value is calculated u si ng the tabulated Ka value of the corresponding weak acid (see Table 16.6). Sample Problem 16.19 shows how to calculate the pH of a basic salt solution. Calculate the pH of a 0.10 M solution of sodium fluoride (NaF) at 25°C. Strategy A solution of NaF contains Na + ions and F- ions. The F- ion is the conjugate base of the weak acid, HF. Use the Ka value for HF (7.1 X 10- 4 , from Table 16.6) and Equation 16.7 to determine Kb for F- : 1.0 X 10- 14 = 1.4 X 10- 11 7.1 X 10- 4 , Then, solve this pH problem like any equilibrium problem, using an equilibrium table. Setup It's always a good idea to write the equation corresponding to the reaction that takes plac e along with the equilibrium expression: F-(aq) + H zO( I) ::;:. ==" HF(aq) + OW(aq ) Construct an equilibrium table, and determine, in terms of the unkn own x, the equilibrium concentrations of the species in the equilibrium expression: F- (aq) + HzO(l) ::;:. ==" HF(aq) + OH - (aq) Initial concentration (M): 0.10 o o Change in concentration (M): -x +x +x Equilibrium concentration (M): 0.10 - x x x Solution Substituting the equilibrium concentrations into the eq uilibrium expression and using the shortcut to solve for x, we get z 1.4 X 10- 11 = X = 0.10 - x 0.10 x = ~( 1.4 X 10- 11 )(0.10) = 1.2 X 10- 6 M According to our equilibrium table, x = [OH- j. In this case, the autoionization of water makes a significant contribution to the hydroxide ion concentration so the total concentration will be the s um of 1.2 X 10- 6 M (from the ionization of F-) and 1.0 X 10- 7 M (from the autoionization of water). Therefore, we calculate the pOH first as pOH = -log (1.2 X 10- 6 + 1.0 X 10- 7 ) = 5.95 and then the pH , pH = 14.00 - pOH = 14.00 - 5.95 = 8.05 The pH of a 0.10 M solution of NaF at 25°C is 8.05. Practice Problem A Determine the pH of a 0.15 M so lution of sod ium acetate (CH 3 COONa) at 25°C. Practice Problem B Determine the concentration of a solution of sodium fluoride (NaF) that has pH 8.51 at 25 °C. HC0 3" has an ionizable proton and can also act as a Br0nsted acid. H owever, its tendency to ac ce pt a proton is stronger than its tendency to donate a proton: HC0 3 + H 2 0 :;:. =z· H 2 C0 3 + OW Kb = 10- 8 Remember that for any conjugate acid -ba se pair ( Equation 16.7): Think About It It' s easy to mix up pH and pOH in this type of problem. Always make a qualitative prediction regarding the pH of a salt solution first, and then check to make sure that your calculated pH agrees with your prediction. In this case, we would predict a basic pH becau se the anion in the salt (F - ) is the conjugate base of a weak acid (HF). The calculated pH, 8.05, is indeed basic. 664 CHAPTER 16 Acids and Bases Think About It In this case, we would predict an acidic pH because the cation in the salt (NHt) is the conjugate acid of a weak base (NH 3 ). The calculated pH is acidic. • Acidic Salt Solutions When the cation of a salt is the conjugate acid of a weak base, a solution of the salt will be acidic. For example , when ammonium chloride dissolves in water, it dissociates to give a solution of ammonium ion s and chloride ions: The ammonium ion is the conjugate acid of the weak base ammonia (NH3)' It acts as a weak Br0n- sted acid, reacting with water to produce hydronium ion: We would therefore predict that a solution containing the ammonium ion is acidic. To calculate the pH, we must deterllline the Ka for NHt using the tabulated Kb value for NH 3 and Equation 16.7. Because Cl - is the weak conjugate base ofthe strong acid HCI, Cl - does not hydrolyze and there- fore has no impact on the pH of the solution . Sample Problem 16.20 shows how to calculate the pH of an acidic salt solution. Sample Problem 16.20 ·. Calculate the pH of a 0.10 M so lution of ammonium chloride (NH 4 Cl) at 25°e. Strategy A solution ofNH 4 Cl contains NHt cations and Cl- anions. The NH t ion is the conjugate acid of the weak base NH 3 . Use the Kb value for NH3 (1.8 X lO - s from Table 16.7) and Equation 16.7 to determine Ka for NH t . 1.0 X 10- 14 = 5.6 X 10- 10 1.8 X lO - s Setup Again, we write the balanced chemical equation and the equilibrium expression: [NH 3 ] [H3 0 +] K = ~ == ,: =. a [NH t ] Next, construct a table to determine the equilibrium concentrations of the species in the equilibrium • expre ss lOn: NH t(aq) + Hz O(l) ::;:, =~. NH3(aq) + H3 0+(aq) Initial concentration (M) : 0.10 o o Change in concentration (M): +x • -rx -x x Equilibrium concentration (M): 0.10 - x x ,_._ Solution Substituting the equilibrium concentrations into the equilibrium expression and using the shortcut to solve for x, we get 2 5.6 X 10- 10 = X = 0.10 -x 0.10 x = ~(5 .6 X 10 10)( 0.10) = 7.5 X 10- 6 M According to the equilibrium table, x = [H 30 +]. The pH can be calculated as follows: pH = - log (7 .5 X 10- 6 ) = 5.12 The pH of a 0.10 M solution of ammonium chloride (at 25°C) is 5.12. . Practice Problem A Determine the pH of a 0.25 M solution of pyridinium nitrate (C s H 6 NN0 3 ) at 2Ye. [Pyridinium nitrate dissociates in water to give pyridinium ions (CsH6N+), the conjugate acid of pyridinium (see Table 16.7), and nitrate ions (N0 3 ).] Practice Problem B Determine the concentration of a solution of ammonium chloride (NH 4 Cl) that has pH 5.37 at 25° e. The metal ion in a dissolved salt can also react with water to produce an acidic solution. The extent of hydrolysis is greatest for the small and highly charged metal y ations such as AI3+, Cr 3 +, Fe 3+ , Bi 3 +, and Be2+. For example, when aluminum chloride dissolves in water, each AI 3+ ion becomes associated with six water molecules (Figure 16.4). SECTION 16.10 Acid-Base Properties of Salt Solutions 665 + Al(OH)(H20)~ + + Consider one of the bonds that forms between the metal ion and an oxygen atom from one of the six water molecules in Al(H20)~+: • H AI ' I cf l ~ H The positively charged A1 3 + ion draws electron density toward itself, increasing the polarity of the 0- H bonds. Consequently, the H atoms have a greater tendency to ionize than those in water molecules not associated with the Al3+ ion. The resulting ionization process can be written as or as The equilibrium constant for the metal cation hydrolysis is given by [Al(OH)(H 2 0) ~ + ][H + ] -5 Ka = = 1.3 X 10 [Al(H20)~ + ] Al(OH)(H20)~+ can undergo further ionization: and so on. It is generally sufficient, however, to take into account only the first stage of hydrolysis when determining the pH of a solution that contains metal ions. Neutral Salt Solutions The extent of hydrolysis is greatest for the smallest and mo st highly charged metal ions because a compact, highly charged ion is more effective in polarizing the O-H bond and facilitating ioniza- tion. This is why relatively large ions of low charge, including the metal cations of Groups lA and . . . . ·· 2 :.f ·· ··························· 2A (the cations of the strong bases), do not undergo significant hydrolysis (Be is an exception). Thus, most metal cations of Groups lA and 2A do not impact the pH of a solution. Similarly, anions that are conjugate bases of strong acids do not hydrolyze to any significant degree. Consequently, a salt composed of the cation of a strong base and the anion of a strong acid, such as NaCl, produces a neutral solution. To summarize, the pH of a salt solution can be predicted qualitatively by identifying the ions in solution and determining which of them, if any, undergoes significant hydrolysis. Examples A cation that will make a solution acidic is • The conjugate acid of a weak base • A small, highly charged metal ion (other than from Group lA or 2A) An anion that will make a solution basic is • The conjugate base of a weak acid CN - , N0 2 , CH 3 COO- A cation that will not affect the pH of a solution is • A Group lA or heavy Group 2A cation (except Be 2 +) An anion that will not affect the pH of a solution is • The conjugate base of a strong acid Cl - , N0 3 , CI0 4 • Figure 16.4 The six H 2 0 molecules surround the AIH ion in an octahedral arrangement. The attraction of the small AIH ion for the lone pairs on the oxygen atoms is so great that the 0- H bonds in an H 2 0 molecule attached to the metal cation are weakened, allowing the loss of a proton (H +) to an incoming H 2 0 molecule. This hydrolysis of the metal cation makes the so lution acidic. The metal cations of the strong bases are those ofthe alkali metals: (Li +, Na +, K+, Rb +, and (s+) and those of the heavy alkaline earth metals (51"'+ and Ba 2+) . [...]... [OH- ] in the solution in part (b) d) Calculate the pOH of the solution in part (b) e) Sample Problem 16.3] [~ Sample Problem 16.6] What concentration of HCI would have the same pH as the solution in part (b)? [ ~~ SampleProblem 16.9] f) Detelmine Kb for the ascorbate ion (C6H60~-) [ ~~ Sample Problem 16.16] g) Calculate the concentrations of all species in the solution in part (b) [ ~~ Sample Problem... 3COOH)2 • • 2CH 3COOH (b) Calculate the equilibrium constant Kp for the reaction in part (a) 16.142 Calculate the concentrations of all the species in a 0.100 M Na2C03 solution 16.143 Henry's law constant for CO 2 at 38°C is 2.28 X 10- 3 mol/L atm Calculate the pH of a solution of CO 2 at 38°C in equilibrium with the gas at a partial pressure of 3.20 atm 16.144 Hydrocyanic acid (HCN) is a weak acid and... of nonmetals generally are acidic • Metal hydroxides may be basic or amphoteric Section 16.12 Section 16.6 • Lewis theory provides more general definitions of acids and bases • A weak base ionizes only partially The base ionization constant, K", is the equilibrium constant that indicates to what extent a weak base ionizes • A Lewis acid accepts a pair of electrons; a Lewis base donates a pair of electrons... acid H 2S04 is complete.) • The strong bases are the Group IA and the heaviest Group 2A hydroxides: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OHh, Sr(OHh, and Ba(OH)z Section 16.5 • • • A weak acid ionizes only partially The acid ionization constant, Ka , is the equilibrium constant that indicates to what extent a weak acid ionizes We solve for the pH of a solution of weak acid using the concentration of the . A particularly important group of organic acids is the carboxylic acids, whose Lewis structures can be represented by • • "0' II R-C-O-H where ' ids ' part ' Of. Problem 16.12] c) Calculate [OH- ] in the solution in part (b). [ ~~ Sample Problem 16.3] d) Calculate the pOH of the solution in part (b). [ ~ Sample Problem 16.6] e) What concentration. Sample Problem 16.6] e) What concentration of HCI would have the same pH as the solution in part (b)? [ ~~ SampleProblem 16.9] f) Detelmine Kb for the ascorbate ion (C6H60~-). [ ~~