For carbonic acid H2C03, for example, we write HC03 aq +.==' H+aq + CO~ - aq Note that the conjugate base in the first ionization is the acid in the second ionization, Table 16,8 sh
Trang 1656 CHAPTER 16 Acids and Bases
Remember t h at the hy d ron iu m io n can be
expressed as either H + or H 30 + These are two
e qu iv a lent ways to represent the autoionization
of water
Think About It Because the conjugates of weak acids and bases 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 bases to calculate pH for solutions containing dissolved salts
-' -' -' ' -~-~ -
-•
•
A s for any c hemical equations, we ca n add the s e two equilibria and cancel identical terms:
_ CH3COO II (a q ) :;:::, ====:' H + (a q ) + _ CH3COO (aq)
DA "'1(-;;'a q;;) ) + H )O(l ) , ' S:: H 3 COOII(aq) + OH - Caq)
The s um i s the autoionization of wate r In fact , this i s the case for any weak acid and its conj u gate
ba s e:
1M :;:::, ====:' H + + Pi"
+ K + H20 , ' 1M + OH ,
H20 :;::, =::z:' H + + OH
-ar for any weak ba se and it s conjugate acid
.B + H20 :;::, ====:' wr + 0 H
-+ ~ + H20 ' J1 + H 3 0 +
2H20 :;:: ====:' H 3 0 + + OH
-R eca ll that when we add t wo equilibria, the equilibrium constant for the net reaction i s the
product o f the eq uilibrium con s tant s for the indi v idual equation s [ ~~ Section 15 3 ] Thus, for any
conjugate acid - base pair,
Equation 16.7
Equation 16.7 gives th e quantitative ba s i s for the reciprocal re lation s hip between the strength of an acid and that of it s c onjugate ba se Co r between the s tr e ngth of a base and that of it s conjugate acid) Becau se K w is a con sta nt , K b mu st decrease if K a increa s e s, and vice versa
Sample Problem 16.16 s h ows how to d e termine ionization constant s for conjugates
Sample Problem 16.16
Determine (a) Kb of the acetate ion (CH3COO-), (b) Ka of the methylammonium ion (CH 3 NH ~ ) ,
(c) Kb of the fluoride ion (F-), and (d) Ka of the anunonium ion (NHt)
Strategy Each species listed is either a conjugate base or a conjugate acid Determine the identity
of the acid corresponding to each conjugate base and the identity of the base corresponding to each conjugate acid; then, consult Tables 16.6 and 16.7 for their ionization constants Use the tabulated ionization constants and Equation 16.7 to calculate each indicated K value
Setup (a) A Kb value is requested, indicating that the acetate ion is a conjugate base To identify the corresponding BrSilnsted acid, add a proton to the formula to get CH3COOH (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 methylammonium ion is a conjugate acid Determine the identity of the corresponding BrSilnsted base by removing a proton from the formula to get
CH3NH2 (methylamine) The Kb of methylamine (from Table 16.7) is 4.4 X 10- 4
(c) F- is the conjugate base of HF; K a = 7.1 X 10-4
(d) NH t is the conjugate acid ofNH3; Kb = 1.8 X 10- 5 Solving Equation 16.7 separately for Ka and
Kb gives, respectively,
1.8 X 10 - )
Trang 2SECTION 16.8 Diprotic and Polyprotic Acids 657
,
I Practice Problem A Determine (a) Kb of the benzoate ion (C 6 H sCOO - ), (b) Kb of the ascorbate ion
I (HC6H60 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
16 7 1 Calculate the Kb of the cyanide ion 16 7 2 Which of the anions listed is the
a) 4.9 X 10 - 10 a) Ascorbate ion (HC6H60(j)
c) 4.9 X 10-24 c) Nitrite ion (N02 ) d) 1.0 X 10-7 d) Phenolate ion (C 6 H s O- )
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 (H2C03), for example, we write
HC03 (aq) +.==' H+(aq) + CO~ - (aq)
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 specie s 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 (H2C20 4) is a poisonous s ubstance 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 corres ponding ionization constants are
K a = 6,1 X 10 - 5
2
Construct an equilibrium table for each ionization, using x a s the unknown in the fir s t ionization and
y as the unknown in the second ionization
(Continued)
,
A triprotic acid h as K a Ka , a n d K a
I ' 1 3
Trang 3658 CHAPTER 16 Acids and Bases
* Th e second ionization constant of H25 is very low and difficult to measure The va lue in this table is an est im ate
H 2 C 2 0iaq) :;:.=::!:' H +( aq) + HC 2 0 4 (aq)
the starting concentration for the second ionization Additionally, the equilibrium concentration of
Trang 4SECTION 16.8 Diprotic and Polyprotic Acids 659
Testing the approximation,
This time, because the ionization constant is much smaller, the approximation is valid At
equilibrium, the concentrations of all species are
[H2C20 4l = 0.046 M
[HC20 4 l = (0.0 54 - 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 HZC20 4 , HC20 4 , C20 ~- , and H+ ions in a
0.20 M oxalic acid solution at 2S°C
Practice Problem B Calculate the concentrations of H2S04, HS04 , 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
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
Trang 5660 CHAPTER 16 Ac ids and Bases
The polarity of the H - X b o nd actually
decreases from H - F to H -I , large ly be c ause
F i s the most electronegative element T h i s
would suggest tha t H F would be the str o nge s t
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 mor e
important f actor in determining the stre n gths
o f these acids
Figure 16.2 Lewis structures of
some common oxoacids
Molecular Structure and Acid Strength
The st ren g th of an acid i s mea sure d b y it s tendenc y t o ionize:
HX • H + + X
-Two factors influence the ex tent to which the acid und e rgoes ionization One is the strength of the H- X bond The stro nger the bond, the more difficult it i s for the HX molecule to break up and hence the weaker the acid The o ther fa cto r i s the polarity of the H - X bond The difference in the
electronegativities between H and X re s ult s in a polar bond like
0+ H-X
0-If the bond i s highly p o lari ze d ( i.e , if there i s a large accumulation of po s itive and negative charges
o n the H a nd X atoms, re s pecti ve ly ), HX will tend to break up into H + and X - ions A high degree
of polarity , therefore , gives ri se to a s tronger acid In thi s s ection , we consider the role s of bond
s tren gt h and b o nd polarity in determining the s trength of an acid
Hydrohalic Acids
The halogen s form a se rie s of binary acids called the hydrohalic acids (HF, HCI, HBr, and HI) Tabl e 16 9 s h ows that o f thi s se rie s 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 s trength of the hydrohalic acids i s
0- H bond s If the c entral a tom i s an electronegative element, or i s in a high oxidation state, it will
attract el ec tron s, causing the O - H bond to be more p o lar Thi s makes it easier for the hydrogen to
be l ost a s H + , making the acid s tron ge r
Trang 6SECTION 16.9 Molecular Structure and Acid Strength 661
•• • • H-O-CI: H-O-C1-0: •• •• ••
To compare their strengths, it is convenient to divide the oxoacids into two groups:
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
• 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, HCI04 is the strongest
acid because it has the largest number of oxygen atoms attached to Cl The acid strength
decreases as follows:
HCIQ4 > HCI03 > HCI02 > 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 RN02
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 electronegat ivitie 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) RN03 is a stronger acid than HN02
Practice Problem Indicate which is the stronger acid: (a) HBr03 or HBr04; (b) H 2 Se04 or H• 2S04,
.' , , ,
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 thes e
two is to remember that HN03 is one of the
seven strong acids, HN02 is not
Think About It Four of the strong acids are oxoacids: RN03, HCI04, HCI03, and H2S04,
Trang 7662 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 HC2H302 or as CH3COOH
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:
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 it s
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 ple, acetic acid and chloroacetic acid:
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:
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 wh i 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 so l ution will be neutral, basic, or acidic Note in the preceding example that sodium ions (Na +) do not hydrolyze and thus h ave no impact on the pH
of the solution
Basic Salt Solutions
Sodium fluoride is a salt t h at dissolves to give a bas i c 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 (CH3COO - ), the nitrite ion (N02 ), t h e sulfite ion (SO~ - ), and the hydrogen carbona t e
Trang 8SECTION 16.10 Acid-Base Properties of Salt Solutions 663
ion (HCO)) Each of these anions undergoes hydrolysis to produce the corresponding weak acid
A - (aq) + H 20(l) :;:: =::=:" HA(aq) + OH - (aq)
We can therefore make the qualitative prediction that a solution of a salt in which the anion is
same way we calculate the pH of any weak ba se solution, u s ing the Kb value for the anion The
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 so lution of sodium fluoride (NaF) at 25 ° C
Strategy A solution of NaF contains Na + ion s and F - ions The F - ion is the conjugate base of
the weak acid, HF Use the K a 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
, Then, solve this pH problem like any equilibrium problem, u s in g 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) + H z O(l) ::;: ==" HF(aq) + OH - (aq)
Solution Substituting the equilibrium concentrations into the eq uilibrium expression and u sing the
shortcut to solve for x, we get
z
1.4 X 10 - 11 = X =
x = ~( 1.4 X 10 - 11)(0.10) = 1.2 X 10 - 6 M
According to our equilibrium table, x = [OH - j In thi s 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 i s 8.05
Practice Problem A Determine the pH of a 0.15 M so lution of sod ium acetate (CH3COONa) at
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 s ure that your calculated pH
agrees with your prediction In this
case, we would predict a ba s ic pH becau se the anion in the salt (F - ) is the conjugate base of a weak acid
indeed basic
Trang 9664 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
(NH3) 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 ions 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 K a for NH t 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
Calculate the pH of a 0.10 M solution of ammonium chloride (NH4Cl) at 25°e
Strategy A solution ofNH4Cl contains NHt cations and Cl- anions The NH t ion is the conjugate acid of the weak base NH3 Use the K b value for NH3 (1.8 X lO - s from Table 16.7) and Equation
Solution Substituting the equilibrium concentrations into the equilibrium expression and using the
shortcut to solve for x, we get
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 p y ridinium (see Table 16.7), and nitrate ions (N0 3 ).]
Practice Problem B Determine the concentration of a solution of ammonium chloride (NH4Cl) 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 yations such as AI3+, Cr3+,
Fe3+, Bi3+, and Be2+ For example, when aluminum chloride dissolves in water, each AI3+ ion becomes associated with six water molecules (Figure 16.4)
Trang 10SECTION 16.10 Acid-Base Properties of Salt Solutions 665
Consider one of the bonds that forms between the metal ion and an oxygen atom from one
of the s ix water molecules in Al(H20)~+:
•
H
AI ' I cf l
~ H
The positively charged A13+ ion draw s 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
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 m e tal cations of Groups lA and
·· 2 :.f · · ···
2A (the cations of the strong bases), do not undergo s ignificant hydroly sis (Be i s an exception)
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, undergoe s 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 so lution basic is
COO-A cation that will not affect the pH of a so lution is
• A Group lA or heavy Group 2A cation (except Be2+)
An anion that will not affect the pH of a solution is
•
Figure 16.4 The six H20 molecule s s urround the AIH ion in an
octahedral arrangement The attraction
of the s mall AIH ion for the lone pairs
on the oxygen atoms is so great that the 0- H bonds in an H20 molecule
attached to the metal cation are
weakened, allowing the loss of a proton
(51"'+ and Ba 2+)
Trang 11666 CHAPTER 16 Acids and Bases
Think About It It's very important
that you be able to identify the ions
in solution correctly If necessary,
review the formulas and charges
of the common polyatomic ions
[ ~ Section 2.7, Table 2.8]
Sample Problem 16.21 let s yo u practice predicting the pH of salt solutions
Predict whether a 0.10 M solution of each of the following salts will be basic, acidic, or neutral:
(a) LiI, (b) NH4N03, (c) Sr(N03h , (d) KN02, (e) NaCN
Strategy Identify the ions present in each solution, and determine which, if any, will impact the pH
of the solution
Setup (a) Ions in solution: Li+ and r Li+ is a Group lA cation; 1 - is the conjugate base of the
strong acid HI Therefore, neither ion hydrolyzes to any significant degree
(b) Ions in solution: NH 1 and NO) NH 1 is the conjugate acid of the weak base NH3; N03" is the
conjugate base of the strong acid HN03 In this case, the cation will hydrolyze, making the pH
N02 (aq) + H 2 0(l) :;:: ==' HN0 2 ( aq) + OW(aq)
(e) Ions in solution: Na + and CN- Na + is a Group 1A cation; CN- is the conjugate base of the weak
acid HCN In this case, too, the anion hydrolyzes, thus making the pH basic:
CW(aq) + H 2 0(l) :;::: ==:!:' H CN(aq) + OH - (aq)
Practice Problem A Predict whether a 0.10 M solution of each of the following salts will be basic,
acidic, or neutral: (a) CH3COOLi, (b) C s HsNHCI , (c) KF, (d) KN03, (e) KCl04
Practice Problem B In addition to those given in Sample Problem 16.21 and Practice Problem
A, identify two salts that will dissolve to give (a) an acidic solution, (b) a basic solution, and (c) a neutral solution
•
Salts in Which Both the Cation and the Anion Hydrolyze
So far we have considered sa lts in which only one ion undergoes hydrolysis In some salts, both the cation and the anion hydroly ze Whether a so lution of s uch a salt is basic, acidic, or neutral depend s on the relativ e strengths of the weak acid and the weak ba se Although the process of calculating the pH in these cases i s more complex than in cases where only one ion hydrolyzes ,
we can make qualitative predictions regarding pH u s ing the values of Kb (of the salt's anion) and
Ka (of the salt's cation):
• When Kb > Ka, the so lution is ba s ic
• When Kb < Ka, the solution is acidic
• When Kb = Ka, the solution i s neutral or nearly neutral
The salt NH4NOz, for example, dissociates in so lution to give NHt (Ka = 5.6 X 10- 1°) and
NO z (Kb = 2.2 X 10 - 11) Because K a for the ammonium ion i s larger than Kb for the nitrite ion ,
we would expect the pH of an ammonium nitrite solution to be s lightly acidic
Trang 12SECTION 16.11 Acid-Base Properties of Oxides and Hydroxides 667
16.10.1 Calculate the pH of a 0.075 M solution
16.10.3 Which of the following salts will
produce a basic solution when dissolved
in water? (Select all that apply.)
16.10.4 Which of the following salts will produce
water? (Select all that apply.)
Acid-Base Properties of Oxides and Hydroxides
As we saw in Chapter 8, oxides can be classified as acidic, basic, or amphoteric Thus, our discussion of
acid - base reactions would be incomplete if we did not examine the properties of these compounds
Oxides of Metals and Nonmetals
Figure 16.5 shows the formulas of a number of oxides of the main group elements in their highest
oxi-dation states All alkali metal oxides and all alkaline earth metal oxides except BeO are ba s ic
Beryl-lium oxide and several metallic oxides in Groups 3A and 4A are amp h oteric Nonmetallic oxides in
which the oxida ti on number of the main group e l ement is high are acidic (e.g., N10 s, S03, and CI20 7),
but those in which the oxidation number of the main group element is low (e.g , CO and NO) show no
measurable acidic properties No nonmetallic oxide s are known to have basic properties
Trang 13668 CHAPTER 16 Acids and Bases
The basic metallic oxides react with water to form metal hydroxides:
BaO(s) + H ? O(I) • Ba(OHMaq)
The reactions between acidic oxides and water are as follows:
The reaction between CO2 and H20 explains why pure water gradually becomes acidic when it is
5.5 The reaction between S0 3 and H ? O is largely responsible for acid rain
Reactions between acidic oxides and bases and those between basic oxides and acids ble normal acid -b ase reactions in that the products are a salt and water:
resem-CO2(g) + 2NaOH(aq) - _ Na2C0 3 (aq) + H20(l) BaO (s) + 2HN0 3 (aq) • Ba(N0 3Ma q) + H 2 0(l)
Aluminum oxide ( Al20 3 ) is amphoteric Depending on the reaction conditions, it can behave either
as an acidic oxide or as a basic oxide For example, Ah0 3 acts as a base with hydrochloric acid to produce a salt (A ICI 3) and water:
and acts as an acid with sodium hydroxide:
Only a salt, sodium aluminum hydroxide [NaAI(OH)4, which contains the Na + and AI(OH)4 ions]
is formed in the reaction with so dium hydroxide no water is produced Nevertheless, the reaction
i s still classified as an acid-base reaction because Al20 3 neutralizes NaOH
Some transition metal oxides in which the metal has a high oxidation number act as acidic
of which react with water to produce acids:
p e rman gan ic a c id
Cr0 3 (S) + H 2 0(l) - _ H2Cr04(aq)
c hromi c acid
Basic and Amphoteric Hydroxides
All the alkali and alkaline earth metal hydroxides , except Be(OHh, are basic Be(OH)2, as well as AI(0H) 3, Sn ( OH) 2> Pb(OH )z, Cr(OH) 3 , CU(OH)2, Zn(OHh, and Cd(OH)2> is amphoteric All ampho- teric hydroxides are insoluble, but beryllium hydroxide reacts with both acids and bases as follows:
Be(OHMs) + 2H + (aq) - _ Be 2+( aq) + 2H20(l)
Be(OH)is) + 20H - (aq) • Be(OH)~- (aq) Aluminum hydroxide reacts with both acids and bases in a similar fashion:
AI(OHMs) + 3H + (aq) - _ AI 3+ (aq) + 3H20(l)
AI(OH Ms) + OH -(a q) • AI(OH)4 (aq)
So far we have discussed acid-base properties in terms of the Br0nsted theory For example, a Br0n s ted ba se is a substance that must be able to accept protons By this definition, both the hydroxide ion and ammonia are bases: