Relative Strengths Of Acids And Bases

OpenStax-CNX module: m511188Figure 4: pHindicates that a 0.l-M solution of HCl (beaker on left) has a pH of 1.ˆ paper ionized and 0.0013M2H3 O 0.1M. A 0.1-M3232) because the weak acid CH CO H is only partially ionized.[CH CO H]. (credit: modi cation of work by Sahar Atwa)http://cnx.org/content/m51118/1.10/The acid is fullyˆsolution of CH CO H (beaker on right) is a pH of 3 ( H3 O ˆIn this solution,H3 O

OpenStax-CNX module: m511189Ionization Constants of Some Weak AcidsIonization ReactionK a at 25 CHSO H O H O SOHF H O H O FHNO H O H O NOHCNO H O H O NCOHCO H H O H O HCOCH CO H H O H O CH COHCIO H O H O CIOHBrO H O H O BrOHCN H O H O CN4 232232232322323 32 3222 31.2 10 23.5 10 44.6 10 42 10 41.8 10 41.8 10 52.9 10 82.8 10 94.9 10 10 2 3234 2 Table 1Table 1 gives the ionization constants for several weak acids; additional ionization constants can be foundin Appendix H6.At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acidof the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus,a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the sameamount of a strong base).For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to theequation:(CH ) N (aq) H O (l)(CH ) NH (aq) OH (aq) ,(26)giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium isanalogous to that described for weak acids.We can con rm by measuring the pH of an aqueous solution of a weak base of known concentration thatonly a fraction of the base reacts with water (Figure 5). The remaining weak base is present as the unreactedform. The equilibrium constant for the ionization of a weak base, b, is called the ionization constant of theweak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, atequilibrium: (CH ) NH OHKb (27)[(CH ) N]3 32 3 3K 3 33 36 "Ionization Constants of Weak Acids" http://cnx.org/content/m51225/latest/ http://cnx.org/content/m51118/1.10/

OpenStax-CNX module: m5111810Figure 5:M3pH paper indicates that a 0.1solution of NH (left) is weakly basic. The solution has a pOH of 3 ([OH ] 0.001) because the weak base NH only partially reacts with water. A 0.1-M3Msolution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modi cation of workby Sahar Atwa)The ionization constants of several weak bases are given in Table 2 and in Appendix I7.Ionization Constants of Some Weak BasesIonization ReactionK b at 25 CCH ) NH H O (CH ) NH OHCH NH H O CH NH OH(CH ) N H O(CH ) NH OHNH H O NH OHC H NH H O C N NH OH(3 232223 3363222 42 33 3253 2653 5.9 104.4 106.3 101.8 104.3 104 4 5 5 10 Table 2Example 3Determination of K a from Equilibrium ConcentrationsAcetic acid is the principal ingredient in vinegar (Figure 6); that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] 0.0787 and H O [CH CO ] 0.00118 M. Whatis the value of a for acetic acid?M3K7 "Ionization Constants of Weak Bases" http://cnx.org/content/m51226/latest/ http://cnx.org/content/m51118/1.10/32

OpenStax-CNX module: m51118Figure 6:11Vinegar is a solution of acetic acid, a weak acid. (credit: modi cation of work by HomeSpotHQ /Flickr)SolutionWe are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium,the value of the equilibrium constant is equal to the reaction quotient for the reaction:CH CO H (aq) H O (l) H O (aq) CH CO (aq)(28) [CH CO ]HO(0.00118) (0.00118)Ka 1.77 10(29)[CH CO 2332 5

OpenStax-CNX module: m5111812Check Your LearningWhat is the equilibrium constant for the ionization of the HSO ion, the weak acid used in somehousehold cleansers:HSO (aq) H O (l) H O (aq) SO(aq)(30) In oneof NaHSO4 and Na2SO4 at equilibrium, H O 0.027 ; [HSO ] 0.29 M ; mixture and SO 0.13 M.a for HSO 1.2 10 244 2 342 M34note:4 2 K4 Example 4Determination of K b from Equilibrium ConcentrationsCa eine, C8H10N4O2 is a weak base. What is the value of b for ca eine if a solution at equilibriumhas [C8H10N4O2] 0.050 , C H N O H 5.0 10 3 , and [OH ] 2.5 10 3 ?SolutionAt equilibrium, the value of the equilibrium constant is equal to the reaction quotient for thereaction:C H N O (aq) H O (l) C H N O H (aq) OH (aq)(31) OH 5.0 102.5 10CHNOH 2.5 10(32)Kb [C H N O ]0.050Check Your LearningWhat is the equilibrium constant for the ionization of the HPO ion, a weak base:HPO (aq) H O (l) H PO (aq) OH (aq)(33)In a solution containing a mixture of NaH2PO4 and Na2HPO4 at equilibrium, [OH ] 1.3 10 6 ; [H PO ] 0.042 M ; and HPO 0.341 M.b for HPO 1.6 10KM810428104288104 22810 4 MM 2 3 3 4104244 2 2242 Mnote:2K4 442 72 Example 5Determination of K a or K b from pHThe pH of a 0.0516- solution of nitrous acid, HNO2, is 2.34. What is its a?HNO (aq) H O (l) H O (aq) NO (aq)MK2232(34) SolutionWe determine an equilibrium constant starting with the initial concentrations of HNO2, H O , andNO as well as one of the nal concentrations, the concentration of hydronium ion at equilibrium.(Remember that pH is simply another way to express the concentration of hydronium ion.)We can solve this problem with the following steps in which is a change in concentration of aspecies in the reaction:32 xhttp://cnx.org/content/m51118/1.10/

OpenStax-CNX module: m5111813We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to considerits concentration): ,To get the various values in the ICE the equilibrium concentration of H O , from the pH: H O 10 0.0046 M(35)concenThe change in concentration of H O ,x[H O ], is the di erence between the equilibrium tration of H3O , which we determinedfromthepH,andtheinitialconcentration,HO. Theinitial concentration of H O is its concentration in pure water, which is so much less thani thenal concentration that we approximate it as zero ( 0). . ForThe change in ofHOeach 1 mol of H O that forms, 1 mol of NO forms. The equilibrium concentration of HNO2 isequal to its initial concentration plus the change in its concentration.Now we can ll in the ICE table with the concentrations at equilibrium, as shown here:33 2.343333323http://cnx.org/content/m51118/1.10/ 32

OpenStax-CNX module: m5111814Finally, we calculate the value of the equilibrium constant using the data in the table: [NO ]HO(0.0046) (0.0046)Ka 4.5 10(36)[HNO ](0.0470)Check Your Learning.The pH of a solution of household ammonia, a 0.950- solution of NH3, is 11.612. What is bfor NH3.b 1.8 10 523 42Mnote:K Example 6Equilibrium Concentrations in a Solution of a Weak AcidFormic acid, HCO2H, is the irritant that causes the body's reaction to ant stings (Figure 7).Figure 7:The pain of an ant's sting is caused by formic acid. (credit: John Tann)http://cnx.org/content/m51118/1.10/K

OpenStax-CNX module: m5111815What is the concentration of hydronium ion and the pH in a 0.534- solution of formic acid?HCO H (aq) H O (l) H O (aq) HCO (aq)Ka 1.8 10(37)M2232 4SolutionStep 1. The equilibrium expression is:HCO H (aq) H O (l) H O (aq) HCO (aq)(38)The concentration of water does not appear in the expression for the equilibrium constant,so we do not need to consider its change in concentration when setting up the ICE table.The table shows initial concentrations (concentrations before the acid ionizes), changes inconcentration, and equilibrium concentrations follows (the data given in the problem appearin color):Determine x and equilibrium concentrations2Step 2.232At equilibrium: [HCOHO [HCO H] Solve for x and the equilibrium concentrations.Ka 1.8 10 4 32 (39)]2(x) (x) 1.8 10 40.534 xNow solve for . Because the initial concentration of acid is reasonably large and a is very small,we assume that 0.534, whichus to simplify the denominator term as (0.534 ) 0.534. This gives:x(40)KxpermitsxKa 1.8 10 4 Solve for as follows:x2 0.534(41)xx2 0.534 1.8 10 4 9.6 10 5x p9.6 10 5 9.8 10 3To check the assumption that is small compared to 0.534, we calculate:x9.8 10 1.8 10 (1.8% of 0.534)0.5340.534(42)(43)(44)x 3 2http://cnx.org/content/m51118/1.10/(45)

OpenStax-CNX module: m5111816is less than 5% of the initial concentration; the assumption is valid.We nd the equilibrium concentration of hydronium ion in this formic acid solution from itsinitial concentration and the change in that concentration as indicated in the last line of the table: H O 0 x 0 9.8 10 M.(46)x 33 9.8 10 3 M(47)The pH of the solution can be found by taking the negative log of the H O so: log 9.8 10 2.01 3 ,(48) 3Check Your LearningOnly a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization ofacetic acid in a 0.100- solution of acetic acid, CH3CO2H?CH CO H (aq) H O (l) H O (aq) CH CO (aq)Ka 1.8 10(49)(Hint: Determine [CH CO ] at equilibrium.) Recall that the percent ionization is the fraction of] 100.acetic acid that is ionized 100, or CH[CHCOCOH initialpercent ionization 1.3%The following example shows that the concentration of products produced by the ionization of a weak basecan be determined by the same series of steps used with a weak acid.M32233232 5 3[322 ]note:Example 7Equilibrium Concentrations in a Solution of a Weak BaseFind the concentration of hydroxide ion in a 0.25- solution of trimethylamine, a weak base:(CH ) N (aq) H O (l)(CH ) NH (aq) OH (aq)Kb 6.3 10(50)M3 32Solution3 3 5This problem requires that we calculate an equilibrium concentration by determining concentrationchanges as the ionization of a base goes to equilibrium. The solution is approached in the same wayas that for the ionization of formic acid in Example 6. The reactants and products will be di erentand the numbers will be di erent, but the logic will be the same:http://cnx.org/content/m51118/1.10/

OpenStax-CNX module: m51118Step 1.Step 2.17Determine x and equilibrium concentrations. The table shows the changes and concentrations:. At equilibrium:Solve for x and the equilibrium concentrationsCH ) NH OH (x) (x) 6.3 10(51)[(CH ) N]0.25 xIf we assume that is small relative to 0.25, then we can replace (0.25 ) in the precedingequation with 0.25. Solving the simpli ed equation gives:x 4.0 10(52)This change is less than 5% of the initial concentration (0.25), so the assumption is justi ed.Recall that, for this computation, is equal to the equilibrium concentration ofin the solution (see earlier tabulation): OH 0 x x 4.0 10 M(53)Kb ( 3 3 53 3xx 3xhydroxideion 3(54) 4.0 10 3 MThen calculate pOH as follows:pOH log 4.0 10 2.40Using the relation introduced in the previous section of this chapter:pH pOH pKw 14.00permits the computation of pH:pH 14.00 pOH 14.00 2.40 11.60Step 3. A check of our arithmetic shows that b 6.3 10 5. 3Check the work KCheck Your Learning(a) Show that the calculation in Step 2 of this example gives an of 4.0 10 3 and the calculationin Step 3 shows b 6.3 10 5.(b) Find the concentration of hydroxide ion in a 0.0325- solution of ammonia, a weak basewithNHa b of 1.76 10 5. Calculate the percent ionization of ammonia, the fraction ionized 100,or [ NH ] 100x KMK 4[3]http://cnx.org/content/m51118/1.10/ (55)(56)(57)

OpenStax-CNX module: m51118187.56 10 4 , 2.33%Some weak acids and weak bases ionize to such an extent that the simplifying assumption that is smallrelative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibriumconcentrations in such cases, we will see that we cannot neglect the change in the initial concentration ofthe acid or base, and we must solve the equilibrium equations by using the quadratic equation. note:MxExample 8Equilibrium Concentrations in a Solution of a Weak AcidSodium bisulfate, NaHSO4, is used in some household cleansers because it contains the HSOion, a weak acid. What is the pH of a 0.50- solution of HSO ?HSO (aq) H O (l) H O (aq) SO (aq)Ka 1.2 10(58)4M4 2434 22 SolutionWe need to determine the equilibrium concentration of the hydronium ion that results from theionization of HSO so that we can use H O to determine the pH. As in the previous examples,we can approach the solution by the following steps:4Step 1.Step 2. 3tions:Determine x and equilibrium concentrations. This table shows the changes and concentra-. As we begin solving for , we will nd this is morecomplicated than in previous examples. As we discuss these complications we should not losetrack of the fact that it is still the purpose of this step to determine the value of .At equilibrium: SO HO(x) (x)Ka 1.2 10 (59)[HSO ]0.50 xIf we assume that is small and approximate (0.50 ) as 0.50, we nd:x 7.7 10(60)Solve for x and the concentrationsxx 2344 xx 2http://cnx.org/content/m51118/1.10/2

OpenStax-CNX module: m5111819When we check the assumption, we calculate:(61)x[HSO4 ]i7.7 10 2x 0.15 (15%)0.500.50(62)The value of is not less than 5% of 0.50, so the assumption is not valid. We need thequadratic formula to nd .The equation:xxKa 1.2 10 2 (63)(x) (x)0.50 xgives6.0 10 3 1.2 10 2 x x2 (64)x2 1.2 10 2 x 6.0 10 3 0(65)orThis equation can be solved using the quadratic formula. For an equation of the formax2 bx c 0,xis given by the equation:x b b2 4ac2a(67)In this problem, 1, 1.2 10 3, and 6.0 10 3.Solving for gives a negative root (which cannot be correct since concentration cannot benegative) and a positive root:a b(66) cxx 7.2 10 2Now determine the hydronium ion concentration and the pH: H O 0 x 0 7.2 10 23MThe pH of this solution is:pH log H O log 7.2 10http://cnx.org/content/m51118/1.10/(69)(70) 7.2 10 2 M3(68) 2 1.14(71)

OpenStax-CNX module: m5111820Check Your Learning(a) Show that the quadratic formula gives 7.2 10 2.(b) Calculate the pH in a 0.010- solution of ca eine, a weak base:C H N O (aq) H O (l) C H N O H (aq) OH (aq)Kb 2.5 10(72)(Hint: It will be necessary to convert [OH ] to H O or pOH to pH toward the end of thecalculation.)pH 11.16 xM8104228104 2 43note:2 The Relative Strengths of Strong Acids and BasesStrong acids, such as HCl, HBr, and HI, all exhibit the same strength in water. The water molecule issuch a strong base compared to the conjugate bases Cl , Br , and I that ionization of these strong acidsis essentially complete in aqueous solutions. In solvents less basic than water, we nd HCl, HBr, and HIdi er markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol(a weaker base than water), the extent of ionization increases in the order HCl HBr HI, and so HIis demonstrated to be the strongest of these acids. The inability to discern di erences in strength amongstrong acids dissolved in water is known as the leveling e ect of water.Water also exerts a leveling e ect on the strengths of strong bases. For example, the oxide ion, O2 , andthe amide ion, NH , are such strong bases that they react completely with water:O (aq) H O (l) [U 27F6] OH (aq) OH (aq)(73) 2 2 2 NH (aq) H O (l) [U 27F6] NH (aq) OH (aq)(74)Thus, O2 and NH appear to have the same base strength in water; they both give a 100% yield ofhydroxide ion.2 2 23 3 E ect of Molecular Structure on Acid-Base StrengthIn the absence of any leveling e ect, the acid strength of binary compounds of hydrogen with nonmetals (A)increases as the H-A bond strength decreases down a group in the periodic table. For group 7A, the orderof increasing acidity is HF HCl HBr HI. Likewise, for group 6A, the order of increasing acid strengthis H2O H2S H2Se H2Te.Across a row in the periodic table, the acid strength of binary hydrogen compounds increases withincreasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, theorder of increasing acidity (for removal of one proton) across the second row is CH4 NH3 H2O HF;across the third row, it is SiH4 PH3 H2S HCl (see Figure 8).http://cnx.org/content/m51118/1.10/

OpenStax-CNX module: m51118Figure 8:21As you move from left to right and down the periodic table, the acid strength increases. Asyou move from right to left and up, the base strength increases.Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric,depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group.Such compounds have the general formula OnE(OH)m, and include sulfuric acid, O2S(OH)2, sulfurous acid,OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide,Ca(OH)2, and potassium hydroxide, KOH:If the central atom, E, has alow electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to forma strong covalent bond with the oxygen atom, and bond between the element and oxygen is more readilybroken than bond between oxygen and hydrogen. Hence bond is ionic, hydroxide ions are releasedabhttp://cnx.org/content/m51118/1.10/a

OpenStax-CNX module: m5111822to the solution, and the material behaves as a base this is the case with Ca(OH)2 and KOH. Lowerelectronegativity is characteristic of the more metallic elements; hence, the

Figure 1: Some of the common strong acids and bases are listed here. The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations