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Relative Strengths of Acids and

Bases∗

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This work is produced by OpenStax-CNX and licensed under the

Creative Commons Attribution License 4.0†

Abstract

By the end of this section, you will be able to:

• Assess the relative strengths of acids and bases according to their ionization constants• Rationalize trends in acid�base strength in relation to molecular structure• Carry out equilibrium calculations for weak acid�base systems

We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reactionof an acid with water is given by the general expression:

HA (aq) + H2O (l) H3O+ (aq) + A− (aq) (1)

Water is the base that reacts with the acid HA, A− is the conjugate base of the acid HA, and the hydroniumion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of H3O

+ and A− when theacid ionizes in water; Figure 1 lists several strong acids. A weak acid gives small amounts of H3O

+ and A−.

∗Version 1.10: Apr 13, 2016 3:55 pm -0500†http://creativecommons.org/licenses/by/4.0/

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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 aqueoussolutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yieldhigher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is calledthe acid-ionization constant, Ka. For the reaction of an acid HA:

HA (aq) + H2O (l) H3O+ (aq) + A− (aq) , (2)

we write the equation for the ionization constant as:

Ka =

[H3O

+] [A−]

[HA](3)

where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is thesolvent as well, so we do not include [H2O] in the equation. The larger the Ka of an acid, the larger theconcentration of H3O

+ and A− relative to the concentration of the nonionized acid, HA. Thus a stronger acidhas a larger ionization constant than does a weaker acid. The ionization constants increase as the strengthsof the acids increase. (A table of ionization constants of weak acids appears in Appendix H1, with a partiallisting in Table 1.)

The following data on acid-ionization constants indicate the order of acid strength CH3CO2H < HNO2

< HSO4− :

CH3CO2H (aq) + H2O (l) H3O+ (aq) + CH3CO2

− (aq) Ka = 1.8 × 10−5 (4)

HNO2 (aq) + H2O (l) H3O+ (aq) + NO2

− (aq) Ka = 4.6 × 10−4 (5)

HSO4− (aq) + H2O (aq) H3O

+ (aq) + SO42− (aq) Ka = 1.2 × 10−2 (6)

1"Ionization Constants of Weak Acids" <http://cnx.org/content/m51225/latest/>



Another measure of the strength of an acid is its percent ionization. The percent ionization of a weakacid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:

% ionization =

[H3O

+]eq

[HA]0 × 100(7)

Because the ratio includes the initial concentration, the percent ionization for a solution of a given weakacid varies depending on the original concentration of the acid, and actually decreases with increasing acidconcentration.

Example 1Calculation of Percent Ionization from pHCalculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of2.09.SolutionThe percent ionization for an acid is: [

H3O+]eq

[HNO2]0× 100 (8)

The chemical equation for the dissociation of the nitrous acid is: HNO2 (aq) + H2O (l) NO2

− (aq) + H3O+ (aq) . Since 10−pH =

[H3O

+], we �nd that 10−2.09 = 8.1 × 10−3M, so that

percent ionization is:

8.1 × 10−3

0.125× 100 = 6.5% (9)

Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two signi�cant�gures.Check Your LearningCalculate the percent ionization of a 0.10-M solution of acetic acid with a pH of 2.89.

note: 1.3% ionized

We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. Thereaction of a Brønsted-Lowry base with water is given by:

B (aq) + H2O (l) HB+ (aq) + OH− (aq) (10)

Water is the acid that reacts with the base, HB+ is the conjugate acid of the base B, and the hydroxideion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH− and HB+ whenit reacts with water; Figure 1 lists several strong bases. A weak base yields a small proportion of hydroxideions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completelywhen dissolved in water.

note: View the simulation2 of strong and weak acids and bases at the molec-ular level.

2http://openstaxcollege.org/l/16AcidBase



As we did with acids, we can measure the relative strengths of bases by measuring their base-ionizationconstant (Kb) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to agreater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base hasa larger ionization constant than does a weaker base. For the reaction of a base, B:

B (aq) + H2O (l) HB+ (aq) + OH− (aq) , (11)

we write the equation for the ionization constant as:

Kb =

[HB+

] [OH−

][B]

(12)

where the concentrations are those at equilibrium. Again, we do not include [H2O] in the equation becausewater is the solvent. The chemical reactions and ionization constants of the three bases shown are:

NO2− (aq) + H2O (l) HNO2 (aq) + OH− (aq) Kb = 2.22 × 10−11 (13)

CH3CO2− (aq) + H2O (l) CH3CO2H (aq) + OH− (aq) Kb = 5.6 × 10−10 (14)

NH3 (aq) + H2O (l) NH4+ (aq) + OH− (aq) Kb = 1.8 × 10−5 (15)

A table of ionization constants of weak bases appears in Appendix I3 (with a partial list in Table 2). As withacids, percent ionization can be measured for basic solutions, but will vary depending on the base ionizationconstant and the initial concentration of the solution.

Consider the ionization reactions for a conjugate acid-base pair, HA − A−:

HA (aq) + H2O (l) H3O+ (aq) + A− (aq) Ka =

[H3O

+] [A−]

[HA](16)

A− (aq) + H2O (l) OH− (aq) + HA (aq) Kb =[HA] [OH][

A−] (17)

Adding these two chemical equations yields the equation for the autoionization for water:

)HA(aq + H2O (l) + )A− (aq) + H2O (l) H3O+ (aq) + )A− (aq) + OH− (aq) + )HA(aq (18)

2H2O (l) H3O+ (aq) + OH− (aq) (19)

As shown in the previous chapter on equilibrium, the K expression for a chemical equation derived fromadding two or more other equations is the mathematical product of the other equations' K expressions.Multiplying the mass-action expressions together and cancelling common terms, we see that:

Ka × Kb =

[H3O

+] [A−]

[HA]×

[HA][OH−

][A−] =

[H3O

+] [OH−

]= Kw (20)

For example, the acid ionization constant of acetic acid (CH3COOH) is 1.8 × 10−5, and the base ionizationconstant of its conjugate base, acetate ion

(CH3COO

−) , is 5.6 × 10−10. The product of these two constantsis indeed equal to Kw:

Ka × Kb =(1.8 × 10−5

)×(5.6 × 10−10

)= 1.0 × 10−14 = Kw (21)

3"Ionization Constants of Weak Bases" <http://cnx.org/content/m51226/latest/>



The extent to which an acid, HA, donates protons to water molecules depends on the strength of theconjugate base, A−, of the acid. If A− is a strong base, any protons that are donated to water moleculesare recaptured by A−. Thus there is relatively little A− and H3O

+ in solution, and the acid, HA, is weak.If A− is a weak base, water binds the protons more strongly, and the solution contains primarily A− andH3O

+�the acid is strong. Strong acids form very weak conjugate bases, and weak acids form strongerconjugate bases (Figure 2).

Figure 2: This diagram shows the relative strengths of conjugate acid-base pairs, as indicated by theirionization constants in aqueous solution.

Figure 3 lists a series of acids and bases in order of the decreasing strengths of the acids and thecorresponding increasing strengths of the bases. The acid and base in a given row are conjugate to eachother.



Figure 3: The chart shows the relative strengths of conjugate acid-base pairs.

The �rst six acids in Figure 3 are the most common strong acids. These acids are completely dissociatedin aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of theseacids dissolves in water, their protons are completely transferred to water, the stronger base.

Those acids that lie between the hydronium ion and water in Figure 3 form conjugate bases that cancompete with water for possession of a proton. Both hydronium ions and nonionized acid molecules arepresent in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water(those found below water in the column of acids) in Figure 3 exhibit no observable acidic behavior whendissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate basewere formed, it would react with water to re-form the acid.

The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the baserelative to that of the hydroxide ion, as shown in the last column in Figure 3. A strong base, such as one



of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid andhydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixtureof the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water inthe column of bases) show no observable basic behavior in aqueous solution.

Example 2The Product Ka×Kb = Kw

Use the Kb for the nitrite ion, NO2−, to calculate the Ka for its conjugate acid.

SolutionKb for NO2

− is given in this section as 2.17 × 10−11. The conjugate acid of NO2− is HNO2; Ka

for HNO2 can be calculated using the relationship:

Ka × Kb = 1.0 × 10−14 = Kw (22)

Solving for Ka, we get:

Ka =Kw

Kb=

1.0 × 10−14

2.17 × 10−11 = 4.6 × 10−4 (23)

This answer can be veri�ed by �nding the Ka for HNO2 in Appendix H4.Check Your LearningWe can determine the relative acid strengths of NH4

+ and HCN by comparing their ionizationconstants. The ionization constant of HCN is given in Appendix H5 as 4.9 × 10−10. The ionizationconstant of NH4

+ is not listed, but the ionization constant of its conjugate base, NH3, is listed as1.8 × 10−5. Determine the ionization constant of NH4

+, and decide which is the stronger acid,HCN or NH4

+.

note: NH4+ is the slightly stronger acid (Ka for NH4

+ = 5.6 × 10−10).

1 The Ionization of Weak Acids and Weak Bases

Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weakacid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, withthe nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ionconcentration in an aqueous solution (but not as much as the same amount of a strong acid).

Acetic acid, CH3CO2H, is a weak acid. When we add acetic acid to water, it ionizes to a small extentaccording to the equation:


− (aq) , (24)

giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. Thisequilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water moleculesand form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions toacetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of anaqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment(Figure 4). The remaining weak acid is present in the nonionized form.

For acetic acid, at equilibrium:

Ka =

[H3O

+][CH3CO2

−][CH3CO2H]

= 1.8 × 10−5 (25)

4"Ionization Constants of Weak Acids" <http://cnx.org/content/m51225/latest/>5"Ionization Constants of Weak Acids" <http://cnx.org/content/m51225/latest/>



Figure 4: pH paper indicates that a 0.l-M solution of HCl (beaker on left) has a pH of 1. The acid is fullyionized and

ˆH3O

+˜= 0.1 M. A 0.1-M solution of CH3CO2H (beaker on right) is a pH of 3 (

ˆH3O

+˜

= 0.001 M) because the weak acid CH3CO2H is only partially ionized. In this solution,ˆH3O

+˜

<[CH3CO2H]. (credit: modi�cation of work by Sahar Atwa)



Ionization Constants of Some Weak Acids

Ionization Reaction Ka at 25◦C

HSO4− + H2O H3O

+ + SO42− 1.2 × 10−2

HF + H2O H3O+ + F− 3.5 × 10−4

HNO2 + H2O H3O+ + NO2

− 4.6 × 10−4

HCNO + H2O H3O+ + NCO− 2 × 10−4

HCO2H + H2O H3O+ + HCO2

− 1.8 × 10−4

CH3CO2H + H2O H3O+ + CH3CO2

− 1.8 × 10−5

HCIO + H2O H3O+ + CIO− 2.9 × 10−8

HBrO + H2O H3O+ + BrO− 2.8 × 10−9

HCN + H2O H3O+ + CN− 4.9 × 10−10

Table 1

Table 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:

(CH3)3N (aq) + H2O (l) (CH3)3NH+ (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, Kb, is called the ionization constant of theweak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, atequilibrium:

Kb =

[(CH3)3NH

+] [OH−

][(CH3)3N]

(27)

6"Ionization Constants of Weak Acids" <http://cnx.org/content/m51225/latest/>



Figure 5: pH paper indicates that a 0.1-M solution of NH3 (left) is weakly basic. The solution has apOH of 3 ([OH−] = 0.001 M) because the weak base NH3 only partially reacts with water. A 0.1-Msolution 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 Bases

Ionization Reaction Kb at 25 ◦C

(CH3)2NH + H2O (CH3)2NH2+ + OH− 5.9 × 10−4

CH3NH2 + H2O CH3NH3+ + OH− 4.4 × 10−4

(CH3)3N + H2O (CH3)3NH+ + OH− 6.3 × 10−5

NH3 + H2O NH4+ + OH− 1.8 × 10−5

C6H5NH2 + H2O C6N5NH3+ + OH− 4.3 × 10−10

Table 2

Example 3Determination of Ka from Equilibrium ConcentrationsAcetic acid is the principal ingredient in vinegar (Figure 6); that's why it tastes sour. At equilib-rium, a solution contains [CH3CO2H] = 0.0787 M and

[H3O

+]

= [CH3CO2−] = 0.00118 M. What

is the value of Ka for acetic acid?

7"Ionization Constants of Weak Bases" <http://cnx.org/content/m51226/latest/>



Figure 6: Vinegar 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:


− (aq) (28)

Ka =

[H3O

+][CH3CO2

−][CH3CO2H]

=(0.00118) (0.00118)

0.0787= 1.77 × 10−5 (29)



Check Your LearningWhat is the equilibrium constant for the ionization of the HSO4

− ion, the weak acid used in somehousehold cleansers:

HSO4− (aq) + H2O (l) H3O

+ (aq) + SO42− (aq) (30)

In one mixture of NaHSO4 and Na2SO4 at equilibrium,[H3O

+]= 0.027 M ; [HSO4

−] = 0.29 M ;and

[SO4

2−] = 0.13 M.

note: Ka for HSO4− = 1.2 × 10−2

Example 4Determination of Kb from Equilibrium ConcentrationsCa�eine, C8H10N4O2 is a weak base. What is the value of Kb for ca�eine if a solution at equilibriumhas [C8H10N4O2] = 0.050 M,

[C8H10N4O2H

+]= 5.0 × 10−3M, and [OH−] = 2.5 × 10−3M?

SolutionAt equilibrium, the value of the equilibrium constant is equal to the reaction quotient for thereaction:

C8H10N4O2 (aq) + H2O (l) C8H10N4O2H+ (aq) + OH− (aq) (31)

Kb =

[C8H10N4O2H

+] [OH−

][C8H10N4O2]

=

(5.0 × 10−3

) (2.5 × 10−3

)0.050

= 2.5 × 10−4 (32)

Check Your LearningWhat is the equilibrium constant for the ionization of the HPO4

2− ion, a weak base:

HPO42− (aq) + H2O (l) H2PO4

− (aq) + OH− (aq) (33)

In a solution containing a mixture of NaH2PO4 and Na2HPO4 at equilibrium, [OH−] = 1.3 ×10−6M ; [H2PO4

−] = 0.042 M ; and[HPO4

2−] = 0.341 M.

note: Kb for HPO42− = 1.6 × 10−7

Example 5Determination of Ka or Kb from pHThe pH of a 0.0516-M solution of nitrous acid, HNO2, is 2.34. What is its Ka?

HNO2 (aq) + H2O (l) H3O+ (aq) + NO2

− (aq) (34)

SolutionWe determine an equilibrium constant starting with the initial concentrations of HNO2, H3O

+, andNO2

− 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 x is a change in concentration of aspecies in the reaction:



We can summarize the various concentrations and changes as shown here (the concentration of wa-ter 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 (Initial, Change, Equilibrium) table, we �rst calculate[H3O

+],

the equilibrium concentration of H3O+, from the pH:[

H3O+]

= 10−2.34 = 0.0046 M (35)

The change in concentration of H3O+,x[H3O+], is the di�erence between the equilibrium concen-

tration of H3O+, which we determined from the pH, and the initial concentration,

[H3O

+]i. The

initial concentration of H3O+ is its concentration in pure water, which is so much less than the

�nal concentration that we approximate it as zero (∼0).The change in concentration of NO2

− is equal to the change in concentration of[H3O

+]. For

each 1 mol of H3O+ that forms, 1 mol of NO2

− 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:



Finally, we calculate the value of the equilibrium constant using the data in the table:

Ka =

[H3O

+][NO2

−][HNO2]

=(0.0046) (0.0046)

(0.0470)= 4.5 × 10−4 (36)

Check Your Learning.The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb

for NH3.

note: Kb = 1.8 × 10−5

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)



What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?

HCO2H (aq) + H2O (l) H3O+ (aq) + HCO2

− (aq) Ka = 1.8 × 10−4 (37)

Solution

Step 1. Determine x and equilibrium concentrations. The equilibrium expression is:

HCO2H (aq) + H2O (l) H3O+ (aq) + HCO2

− (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):

Step 2. Solve for x and the equilibrium concentrations. At equilibrium:

Ka = 1.8 × 10−4 =

[H3O

+][HCO2

−][HCO2H]

(39)

=(x) (x)

0.534− x= 1.8 × 10−4 (40)

Now solve for x. Because the initial concentration of acid is reasonably large and Ka is very small,we assume that x � 0.534, which permits us to simplify the denominator term as (0.534 − x) =0.534. This gives:

Ka = 1.8 × 10−4 =x2+

0.534(41)

Solve for x as follows:

x2+ = 0.534 ×(1.8 × 10−4

)= 9.6 × 10−5 (42)

x =√

9.6 × 10−5 (43)

= 9.8 × 10−3 (44)

To check the assumption that x is small compared to 0.534, we calculate:

x

0.534=

9.8 × 10−3

0.534= 1.8 × 10−2 (1.8% of 0.534) (45)



x is 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 its

initial concentration and the change in that concentration as indicated in the last line of the table:[H3O

+]

=∼ 0 + x = 0 + 9.8 × 10−3 M. (46)

= 9.8 × 10−3 M (47)

The pH of the solution can be found by taking the negative log of the[H3O

+], so:

−log(9.8 × 10−3

)= 2.01 (48)

Check Your LearningOnly a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization ofacetic acid in a 0.100-M solution of acetic acid, CH3CO2H?


− (aq) Ka = 1.8 × 10−5 (49)

(Hint: Determine [CH3CO2−] at equilibrium.) Recall that the percent ionization is the fraction of

acetic acid that is ionized × 100, or[CH3CO2

−][CH3CO2H]

initial

× 100.

note: percent 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.

Example 7Equilibrium Concentrations in a Solution of a Weak BaseFind the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:

(CH3)3N (aq) + H2O (l) (CH3)3NH+ (aq) + OH− (aq) Kb = 6.3 × 10−5 (50)

SolutionThis 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:



Step 1. Determine x and equilibrium concentrations. The table shows the changes and concentrations:

Step 2. Solve for x and the equilibrium concentrations. At equilibrium:

Kb =

[(CH3)3NH

+] [OH−

][(CH3)3N]

=(x) (x)

0.25− x= 6.3 × 10−5 (51)

If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the precedingequation with 0.25. Solving the simpli�ed equation gives:

x = 4.0 × 10−3 (52)

This change is less than 5% of the initial concentration (0.25), so the assumption is justi�ed.Recall that, for this computation, x is equal to the equilibrium concentration of hydroxideion in the solution (see earlier tabulation):[

OH−]

=∼ 0 + x = x = 4.0 × 10−3 M (53)

= 4.0 × 10−3 M (54)

Then calculate pOH as follows:

pOH = −log(4.0 × 10−3

)= 2.40 (55)

Using the relation introduced in the previous section of this chapter:

pH + pOH = pKw = 14.00 (56)

permits the computation of pH:

pH = 14.00− pOH = 14.00− 2.40 = 11.60 (57)

Step 3. Check the work. A check of our arithmetic shows that Kb = 6.3 × 10−5.

Check Your Learning(a) Show that the calculation in Step 2 of this example gives an x of 4.0 × 10−3 and the calculationin Step 3 shows Kb = 6.3 × 10−5.

(b) Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak basewith a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100,

or[NH4

+][NH3]

× 100



note: 7.56 × 10−4M, 2.33%

Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x 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.

Example 8Equilibrium Concentrations in a Solution of a Weak AcidSodium bisulfate, NaHSO4, is used in some household cleansers because it contains the HSO4

−

ion, a weak acid. What is the pH of a 0.50-M solution of HSO4−?

HSO4− (aq) + H2O (l) H3O

+ (aq) + SO42− (aq) Ka = 1.2 × 10−2 (58)

SolutionWe need to determine the equilibrium concentration of the hydronium ion that results from theionization of HSO4

− so that we can use[H3O

+]to determine the pH. As in the previous examples,

we can approach the solution by the following steps:

Step 1. Determine x and equilibrium concentrations. This table shows the changes and concentra-tions:

Step 2. Solve for x and the concentrations. As we begin solving for x, 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 x.At equilibrium:

Ka = 1.2 × 10−2 =

[H3O

+] [SO4

2−][HSO4

−]=

(x) (x)0.50− x

(59)

If we assume that x is small and approximate (0.50 − x) as 0.50, we �nd:

x = 7.7 × 10−2 (60)



When we check the assumption, we calculate:

x

[HSO4−]i

(61)

x

0.50=

7.7 × 10−2

0.50= 0.15 (15%) (62)

The value of x is not less than 5% of 0.50, so the assumption is not valid. We need thequadratic formula to �nd x.The equation:

Ka = 1.2 × 10−2 =(x) (x)

0.50− x(63)

gives

6.0 × 10−3 − 1.2 × 10−2x = x2+ (64)

or

x2+ + 1.2 × 10−2x− 6.0 × 10−3 = 0 (65)

This equation can be solved using the quadratic formula. For an equation of the form

ax2+ + bx + c = 0, (66)

x is given by the equation:

x =−b ±

√b2+ − 4ac

2a(67)

In this problem, a = 1, b = 1.2 × 10−3, and c = −6.0 × 10−3.Solving for x gives a negative root (which cannot be correct since concentration cannot benegative) and a positive root:

x = 7.2 × 10−2 (68)

Now determine the hydronium ion concentration and the pH:[H3O

+]

=∼ 0 + x = 0 + 7.2 × 10−2 M (69)

= 7.2 × 10−2 M (70)

The pH of this solution is:

pH = −log[H3O

+]

= −log 7.2 × 10−2 = 1.14 (71)



Check Your Learning(a) Show that the quadratic formula gives x = 7.2 × 10−2.

(b) Calculate the pH in a 0.010-M solution of ca�eine, a weak base:

C8H10N4O2 (aq) + H2O (l) C8H10N4O2H+ (aq) + OH− (aq) Kb = 2.5 × 10−4 (72)

(Hint: It will be necessary to convert [OH−] to[H3O

+]or pOH to pH toward the end of the

calculation.)

note: pH 11.16

2 The Relative Strengths of Strong Acids and Bases

Strong 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, NH2

−, are such strong bases that they react completely with water:

O2− (aq) + H2O (l) [U+27F6]OH− (aq) + OH− (aq) (73)

NH2− (aq) + H2O (l) [U+27F6]NH3 (aq) + OH− (aq) (74)

Thus, O2− and NH2− appear to have the same base strength in water; they both give a 100% yield of

hydroxide ion.

3 E�ect of Molecular Structure on Acid-Base Strength

In 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).



Figure 8: As 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 a between the element and oxygen is more readilybroken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released



to 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 metallic elements form ionichydroxides that are by de�nition basic compounds.

If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electronsit shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond,bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releaseshydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristicof the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic−OH groups that are called oxyacids.

Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid becausethis increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond.Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurousacid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH(N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3).In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure 9).

Figure 9: As the oxidation number of the central atom E increases, the acidity also increases.

Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation num-bers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodictable) are usually amphoteric. This means that the hydroxy compounds act as acids when they react withstrong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide,which commonly exists as the hydrate Al(H2O)3(OH)3, is re�ected in its solubility in both strong acids andstrong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, Al(H2O)3(OH)3, is



converted into the soluble ion, [Al(H2O)2 (OH)4]−

, by reaction with hydroxide ion:

Al(H2O)3 (OH)3 (aq) + OH− (aq) H2O (l) + [Al(H2O)2 (OH)4]− (aq) (75)

In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ionin solution. The Al(H2O)3(OH)3 compound thus acts as an acid under these conditions. On the other hand,

when dissolved in strong acids, it is converted to the soluble ion [Al(H2O)6]3+

by reaction with hydroniumion:

3H3O+ (aq) + Al(H2O)3 (OH)3 (aq) Al(H2O)6

3+ (aq) + 3H2O (l) (76)

In this case, protons are transferred from hydronium ions in solution to Al(H2O)3(OH)3, and the compoundfunctions as a base.

4 Key Concepts and Summary

The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or baseionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugatebases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weakerbases than water. Weak acids are only partially ionized because their conjugate bases are strong enough tocompete successfully with water for possession of protons. Strong bases react with water to quantitativelyform hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binaryacids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and theyincrease down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same centralelement increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths ofoxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].

5 Key Equations

� Ka = [H3O+][A−]

[HA]

� Kb = [HB+][OH−][B]

� Ka×Kb = 1.0 × 10−14 = Kw

� Percent ionization =[H3O

+]eq

[HA]0 × 100

6 Chemistry End of Chapter Exercises

Exercise 1Explain why the neutralization reaction of a strong acid and a weak base gives a weakly acidicsolution.

Exercise 2 (Solution on p. 30.)

Explain why the neutralization reaction of a weak acid and a strong base gives a weakly basicsolution.

Exercise 3Use this list of important industrial compounds (and Figure 3) to answer the following questionsregarding: CaO, Ca(OH)2, CH3CO2H, CO2, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4,NH3, NaOH, Na2CO3.

(a) Identify the strong Brønsted-Lowry acids and strong Brønsted-Lowry bases.(b) List those compounds in (a) that can behave as Brønsted-Lowry acids with strengths lying

between those of H3O+ and H2O.



(c) List those compounds in (a) that can behave as Brønsted-Lowry bases with strengths lyingbetween those of H2O and OH−.


The odor of vinegar is due to the presence of acetic acid, CH3CO2H, a weak acid. List, in order ofdescending concentration, all of the ionic and molecular species present in a 1-M aqueous solutionof this acid.

Exercise 5Household ammonia is a solution of the weak base NH3 in water. List, in order of descendingconcentration, all of the ionic and molecular species present in a 1-M aqueous solution of this base.


Explain why the ionization constant, Ka, for H2SO4 is larger than the ionization constant forH2SO3.

Exercise 7Explain why the ionization constant, Ka, for HI is larger than the ionization constant for HF.


Gastric juice, the digestive �uid produced in the stomach, contains hydrochloric acid, HCl. Milk ofMagnesia, a suspension of solid Mg(OH)2 in an aqueous medium, is sometimes used to neutralizeexcess stomach acid. Write a complete balanced equation for the neutralization reaction, andidentify the conjugate acid-base pairs.

Exercise 9Nitric acid reacts with insoluble copper(II) oxide to form soluble copper(II) nitrate, Cu(NO3)2,a compound that has been used to prevent the growth of algae in swimming pools. Write thebalanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.


What is the ionization constant at 25 ◦C for the weak acid CH3NH3+, the conjugate acid of the

weak base CH3NH2, Kb = 4.4 × 10−4.

Exercise 11What is the ionization constant at 25 ◦C for the weak acid (CH3)2NH2

+, the conjugate acid ofthe weak base (CH3)2NH, Kb = 5.9 × 10−4?


Which base, CH3NH2 or (CH3)2NH, is the stronger base? Which conjugate acid, (CH3)2NH2+ or

(CH3)2NH, is the stronger acid?

Exercise 13Which is the stronger acid, NH4

+ or HBrO?


Which is the stronger base, (CH3)3N or H2BO3−?

Exercise 15Predict which acid in each of the following pairs is the stronger and explain your reasoning foreach.

(a) H2O or HF(b) B(OH)3 or Al(OH)3(c) HSO3

− or HSO4−

(d) NH3 or H2S(e) H2O or H2Te


Predict which compound in each of the following pairs of compounds is more acidic and explainyour reasoning for each.

(a) HSO4− or HSeO4

−



(b) NH3 or H2O(c) PH3 or HI(d) NH3 or PH3

(e) H2S or HBr

Exercise 17Rank the compounds in each of the following groups in order of increasing acidity or basicity, asindicated, and explain the order you assign.

(a) acidity: HCl, HBr, HI(b) basicity: H2O, OH

−, H−, Cl−

(c) basicity: Mg(OH)2, Si(OH)4, ClO3(OH) (Hint: Formula could also be written as HClO4).(d) acidity: HF, H2O, NH3, CH4


Rank the compounds in each of the following groups in order of increasing acidity or basicity, asindicated, and explain the order you assign.

(a) acidity: NaHSO3, NaHSeO3, NaHSO4

(b) basicity: BrO2−,ClO2

−,IO2−

(c) acidity: HOCl, HOBr, HOI(d) acidity: HOCl, HOClO, HOClO2, HOClO3

(e) basicity: NH2−, HS−, HTe−, PH2

−

(f) basicity: BrO−, BrO2−,BrO3

−,BrO4−

Exercise 19Both HF and HCN ionize in water to a limited extent. Which of the conjugate bases, F− or CN−,is the stronger base? See Table 2.


The active ingredient formed by aspirin in the body is salicylic acid, C6H4OH(CO2H). The carboxylgroup (−CO2H) acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring)also acts as an acid but a much weaker acid. List, in order of descending concentration, all of theionic and molecular species present in a 0.001-M aqueous solution of C6H4OH(CO2H).

Exercise 21What do we represent when we write:CH3CO2H (aq) + H2O (l) H3O

+ (aq) + CH3CO2− (aq)?


Explain why equilibrium calculations are not necessary to determine ionic concentrations in solu-tions of certain strong electrolytes such as NaOH and HCl. Under what conditions are equilibriumcalculations necessary as part of the determination of the concentrations of all ions of some otherstrong electrolytes in solution?

Exercise 23Are the concentrations of hydronium ion and hydroxide ion in a solution of an acid or a base inwater directly proportional or inversely proportional? Explain your answer.


What two common assumptions can simplify calculation of equilibrium concentrations in a solutionof a weak acid?

Exercise 25What two common assumptions can simplify calculation of equilibrium concentrations in a solutionof a weak base?


Which of the following will increase the percent of NH3 that is converted to the ammonium ion inwater (Hint: Use LeChâtelier's principle.)?



(a) addition of NaOH(b) addition of HCl(c) addition of NH4Cl

Exercise 27Which of the following will increase the percent of HF that is converted to the �uoride ion inwater?

(a) addition of NaOH(b) addition of HCl(c) addition of NaF


What is the e�ect on the concentrations of NO2−, HNO2, and OH− when the following are added

to a solution of KNO2 in water:(a) HCl(b) HNO2

(c) NaOH(d) NaCl(e) KNOThe equation for the equilibrium is:

NO2− (aq) + H2O (l) HNO2 (aq) + OH− (aq)

Exercise 29What is the e�ect on the concentration of hydro�uoric acid, hydronium ion, and �uoride ion whenthe following are added to separate solutions of hydro�uoric acid?

(a) HCl(b) KF(c) NaCl(d) KOH(e) HFThe equation for the equilibrium is:

HF (aq) + H2O (l) H3O+ (aq) + F− (aq)


Why is the hydronium ion concentration in a solution that is 0.10M in HCl and 0.10M in HCOOHdetermined by the concentration of HCl?

Exercise 31From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb foreach of the weak bases.

(a) CH3CO2H:[H3O

+]= 1.34 × 10−3M ;

[CH3CO2−] = 1.34 × 10−3M ;

[CH3CO2H] = 9.866 × 10−2M ;(b) ClO−: [OH−] = 4.0 × 10−4M ;[HClO] = 2.38 × 10−5M ;[ClO−] = 0.273 M ;(c) HCO2H: [HCO2H] = 0.524 M ;[

H3O+]= 9.8 × 10−3M ;

[HCO2−] = 9.8 × 10−3M ;

(d) C6H5NH3+ :[C6H5NH3

+] = 0.233 M ;[C6H5NH2] = 2.3 × 10−3M ;[

H3O+]= 2.3 × 10−3M




From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb foreach of the weak bases.

(a) NH3: [OH−] = 3.1 × 10−3M ;

[NH4+] = 3.1 × 10−3M ;

[NH3] = 0.533 M ;(b) HNO2:

[H3O

+]= 0.011 M ;

[NO2−] = 0.0438 M ;

[HNO2] = 1.07 M ;(c) (CH3)3N: [(CH3)3N] = 0.25 M ;

[(CH3)3NH+] = 4.3 × 10−3M ;

[OH−] = 4.3 × 10−3M ;(d) NH4

+ :[NH4+] = 0.100 M ;

[NH3] = 7.5 × 10−6M ;[H3O

+] = 7.5 × 10−6M

Exercise 33Determine Kb for the nitrite ion, NO2

−. In a 0.10-M solution this base is 0.0015% ionized.


Determine Ka for hydrogen sulfate ion, HSO4−. In a 0.10-M solution the acid is 29% ionized.

Exercise 35Calculate the ionization constant for each of the following acids or bases from the ionizationconstant of its conjugate base or conjugate acid:

(a) F−

(b) NH4+

(c) AsO43−

(d) (CH3)2NH2+

(e) NO2−

(f) HC2O4− (as a base)


Calculate the ionization constant for each of the following acids or bases from the ionizationconstant of its conjugate base or conjugate acid:

(a) HTe− (as a base)(b) (CH3)3NH

+

(c) HAsO43− (as a base)

(d) HO2− (as a base)

(e) C6H5NH3+

(f) HSO3− (as a base)

Exercise 37For which of the following solutions must we consider the ionization of water when calculating thepH or pOH?

(a) 3 × 10−8M HNO3

(b) 0.10 g HCl in 1.0 L of solution(c) 0.00080 g NaOH in 0.50 L of solution(d) 1 × 10−7M Ca(OH)2(e) 0.0245 M KNO3


Even though both NH3 and C6H5NH2 are weak bases, NH3 is a much stronger acid than C6H5NH2.Which of the following is correct at equilibrium for a solution that is initially 0.10 M in NH3 and0.10 M in C6H5NH2?



(a)[OH−

]= [NH4

+](b) [NH4

+] = [C6H5NH3+]

(c)[OH−

]= [C6H5NH3

+](d) [NH3] = [C6H5NH2](e) both a and b are correct

Exercise 39Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is0.25 M in HCO2H and 0.10 M in HClO.


Calculate the equilibrium concentration of the nonionized acids and all ions in a solution that is0.134 M in HNO2 and 0.120 M in HBrO.

Exercise 41Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is0.25 M in CH3NH2 and 0.10 M in C5H5N (Kb = 1.7 × 10−9).


Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is0.115 M in NH3 and 0.100 M in C6H5NH2.

Exercise 43Using the Ka value of 1.4 × 10−5, place Al(H2O)6

3+ in the correct location in Figure 3.


Calculate the concentration of all solute species in each of the following solutions of acids or bases.Assume that the ionization of water can be neglected, and show that the change in the initialconcentrations can be neglected. Ionization constants can be found in Appendix H8 and AppendixI9.

(a) 0.0092 M HClO, a weak acid(b) 0.0784 M C6H5NH2, a weak base(c) 0.0810 M HCN, a weak acid(d) 0.11 M (CH3)3N, a weak base(e) 0.120 MFe(H2O)6

2+ a weak acid, Ka = 1.6 × 10−7

Exercise 45Propionic acid, C2H5CO2H (Ka = 1.34 × 10−5), is used in the manufacture of calcium propionate,a food preservative. What is the hydronium ion concentration in a 0.698-M solution of C2H5CO2H?


White vinegar is a 5.0% by mass solution of acetic acid in water. If the density of white vinegaris 1.007 g/cm3, what is the pH?

Exercise 47The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid found in the blood after stren-uous exercise, is 1.36 × 10−4. If 20.0 g of lactic acid is used to make a solution with a volume of1.00 L, what is the concentration of hydronium ion in the solution?


Nicotine, C10H14N2, is a base that will accept two protons (K1 = 7 × 10−7, K2 = 1.4 × 10−11).What is the concentration of each species present in a 0.050-M solution of nicotine?

Exercise 49The pH of a 0.20-M solution of HF is 1.92. Determine Ka for HF from these data.


The pH of a 0.15-M solution of HSO4− is 1.43. Determine Ka for HSO4

− from these data.

8"Ionization Constants of Weak Acids" <http://cnx.org/content/m51225/latest/>9"Ionization Constants of Weak Bases" <http://cnx.org/content/m51226/latest/>



Exercise 51The pH of a 0.10-M solution of ca�eine is 11.16. Determine Kb for ca�eine from these data:C8H10N4O2 (aq) + H2O (l) C8H10N4O2H

+ (aq) + OH− (aq)Exercise 52 (Solution on p. 32.)

The pH of a solution of household ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb

for NH3 from these data.



Solutions to Exercises in this Module

Solution to Exercise (p. 23)The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reactionalso forms OH−, which causes the solution to be basic.Solution to Exercise (p. 24)[H2O] > [CH3CO2H] >

[H3O

+]≈ [CH3CO2

−] > [OH−]Solution to Exercise (p. 24)The oxidation state of the sulfur in H2SO4 is greater than the oxidation state of the sulfur in H2SO3.Solution to Exercise (p. 24)

Mg(OH)2 (s) + 2HCl (aq) [U+27F6] Mg2+ (aq) + 2Cl− (aq) + 2H2O (l)

BB BA CB CA

Solution to Exercise (p. 24)Ka = 2.3 × 10−11

Solution to Exercise (p. 24)The stronger base or stronger acid is the one with the larger Kb or Ka, respectively. In these two examples,they are (CH3)2NH and CH3NH3

+.Solution to Exercise (p. 24)triethylamine.Solution to Exercise (p. 24)(a) HSO4

−; higher electronegativity of the central ion. (b) H2O; NH3 is a base and water is neutral, ordecide on the basis of Ka values. (c) HI; PH3 is weaker than HCl; HCl is weaker than HI. Thus, PH3 isweaker than HI. (d) PH3; in binary compounds of hydrogen with nonmetals, the acidity increases for theelement lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increasesfrom top to bottom. Br is to the left and below S, so HBr is the stronger acid.Solution to Exercise (p. 25)(a) NaHSeO3 < NaHSO3 < NaHSO4; in polyoxy acids, the more electronegative central element�S, in thiscase�forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higheroxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, theacidity increases in the same manner. (b) ClO2

− < BrO2− < IO2

−; the basicity of the anions in a seriesof acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativityof the central atom increases. Cl is more electronegative than Br, and I is the least electronegative ofthe three. (c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases asthe electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the leastelectronegative of the three. (d) HOCl < HOClO < HOClO2 < HOClO3; in a series of oxyacids of thesame central element, the acidity increases as the number of oxygen atoms increases (or as the oxidationstate of the central atom increases). (e) HTe− < HS− � PH2

− < NH2−;PH2

− and NH2− are anions of

weak bases, so they act as strong bases toward H+. HTe− and HS− are anions of weak acids, so they haveless basic character. In a periodic group, the more electronegative element has the more basic anion. (f)BrO4

− < BrO3− < BrO2

− < BrO−; with a larger number of oxygen atoms (that is, as the oxidation stateof the central ion increases), the corresponding acid becomes more acidic and the anion consequently lessbasic.Solution to Exercise (p. 25)

[H2O] > [C6H4OH (CO2H)] > [H+] 0 > [C6H4OH(CO2)

− �[C6H4O(CO2H)−

]>[OH−

]Solution to Exercise (p. 25)Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weakbases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculationsare necessary when one (or more) of the ions is a weak acid or a weak base.



Solution to Exercise (p. 25)1. Assume that the change in initial concentration of the acid as the equilibrium is established can beneglected, so this concentration can be assumed constant and equal to the initial value of the total acidconcentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration ofH3O

+.Solution to Exercise (p. 25)(b) The addition of HClSolution to Exercise (p. 26)(a) Adding HCl will add H3O

+ ions, which will then react with the OH− ions, lowering their concentration.The equilibrium will shift to the right, increasing the concentration of HNO2, and decreasing the concentra-tion of NO2

− ions. (b) Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to theleft, increasing the concentration of NO2

− ions and decreasing the concentration of OH− ions. (c) AddingNaOH adds OH− ions, which shifts the equilibrium to the left, increasing the concentration of NO2

− ionsand decreasing the concentrations of HNO2. (d) Adding NaCl has no e�ect on the concentrations of theions. (e) Adding KNO2 adds NO2

− ions and shifts the equilibrium to the right, increasing the HNO2 andOH− ion concentrations.Solution to Exercise (p. 26)This is a case in which the solution contains a mixture of acids of di�erent ionization strengths. In solution,the HCO2H exists primarily as HCO2H molecules because the ionization of the weak acid is suppressed bythe strong acid. Therefore, the HCO2H contributes a negligible amount of hydronium ions to the solution.The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. Insuch a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of theweaker acid is �xed by the [H3O

+] produced by the stronger acid.Solution to Exercise (p. 26)(a) Kb = 1.8 × 10−5;(b) Ka = 4.5 × 10−4;(c) Kb = 7.4 × 10−5;(d) Ka = 5.6 × 10−10

Solution to Exercise (p. 27)Ka = 1.2 × 10−2

Solution to Exercise (p. 27)(a) Kb = 4.3 × 10−12;(b) Ka = 1.6 × 10−8;(c) Kb = 5.9 × 10−7;(d) Kb = 4.2 × 10−3;(e) Kb = 2.3 × 10−3;(f) Kb = 6.3 × 10−13

Solution to Exercise (p. 27)(a) is the correct statement.Solution to Exercise (p. 28)[H3O

+] = 7.5 × 10−3M[HNO2] = 0.127[OH−] = 1.3 × 10−12M[BrO−] = 4.5 × 10−8M[HBrO] = 0.120 M

Solution to Exercise (p. 28)[OH−] = [NO4

+] = 0.0014 M

[NH3] = 0.144 M

[H3O+] = 6.9 × 10−12M

[C6H5NH3+] = 3.9 × 10−8M

[C6H5NH2] = 0.100 M



Solution to Exercise (p. 28)

(a)[H3O

+][ClO−][HClO] = (x)(x)

(0.0092−x) ≈(x)(x)0.0092 = 2.9 × 10−8

Solving for x gives 1.63 × 10−5M. This value is less than 5% of 0.0092, so the assumption that it can beneglected is valid. Thus, the concentrations of solute species at equilibrium are:[H3O

+] = [ClO] = 5.8 × 10−5M[HClO] = 0.00092 M

[OH−] = 6.1 × 10−10M ;

(b)[C6H5NH3

+][OH−][C6H5NH2]

= (x)(x)(0.0784−x) ≈

(x)(x)0.0784 = 4.3 × 10−10

Solving for x gives 5.81 × 10−6M. This value is less than 5% of 0.0784, so the assumption that it can beneglected is valid. Thus, the concentrations of solute species at equilibrium are:[CH3CO2

−] = [OH−] = 5.8 × 10−6M[C6H5NH2] = 0.00784[H3O

+] = 1.7× 10−9M ;

(c)[H3O

+][CN−][HCN] = (x)(x)

(0.0810−x) ≈(x)(x)0.0810 = 4.9 × 10−10

Solving for x gives 6.30 × 10−6M. This value is less than 5% of 0.0810, so the assumption that it can beneglected is valid. Thus, the concentrations of solute species at equilibrium are:[H3O

+] = [CN−] = 6.3 × 10−6M[HCN] = 0.0810 M

[OH−] = 1.6 × 10−9M ;

(d)[(CH3)3NH

+][OH−][(CH3)3N] = (x)(x)

(0.11−x) ≈(x)(x)0.11 = 6.3 × 10−5

Solving for x gives 2.63 × 10−3M. This value is less than 5% of 0.11, so the assumption that it can beneglected is valid. Thus, the concentrations of solute species at equilibrium are:[(CH3)3NH

+] = [OH−] = 2.6 × 10−3M[(CH3)3N] = 0.11 M

[H3O+] = 3.8 × 10−12M ;

(e)[Fe(H2O)5(OH)+][H3O

+][Fe(H2O)6

2+] = (x)(x)(0.120−x) ≈

(x)(x)0.120 = 1.6 × 10−7

Solving for x gives 1.39 × 10−4M. This value is less than 5% of 0.120, so the assumption that it can beneglected is valid. Thus, the concentrations of solute species at equilibrium are:[Fe(H2O)5(OH)

+] = [H3O+] = 1.4 × 10−4M[

Fe(H2O)62+]= 0.120 M

[OH−] = 7.2 × 10−11MSolution to Exercise (p. 28)pH = 2.41Solution to Exercise (p. 28)[C10H14N2] = 0.049 M

[C10H14N2H+] = 1.9 × 10−4M[

C10H14N2H22+]= 1.4 × 10−11M

[OH−] = 1.9 × 10−4M[H3O

+] = 5.3 × 10−11MSolution to Exercise (p. 28)Ka = 1.2 × 10−2

Solution to Exercise (p. 29)Kb = 1.77 × 10−5



Glossary

De�nition 1: acid ionization constant (Ka)equilibrium constant for the ionization of a weak acid

De�nition 2: base ionization constant (Kb)equilibrium constant for the ionization of a weak base

De�nition 3: leveling e�ect of waterany acid stronger than H3O

+, or any base stronger than OH− will react with water to form H3O+,

or OH−, respectively; water acts as a base to make all strong acids appear equally strong, and itacts as an acid to make all strong bases appear equally strong

De�nition 4: oxyacidcompound containing a nonmetal and one or more hydroxyl groups

De�nition 5: percent ionizationratio of the concentration of the ionized acid to the initial acid concentration, times 100


Relative Strengths of Acids and Bases - cnx.org10.pdf...Water is the acid that reacts with the base,...

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