AP Chemistry
Chapter 16. Acid-Base Equilibria
Chapter 16. Acid-Base Equilibria
16.6 Weak Acids
•
Weak acids are only partially ionized in aqueous solution.
•
There is a mixture of ions and un-ionized acid in solution.
•
Therefore, weak acids are in equilibrium:
•
•
HA(aq) + H2O(l)  H3O+(aq) + A–(aq)
Or:
HA(aq)  H+(aq) + A–(aq)
We can write an equilibrium constant expression for this dissociation:
H O A  or K  H A 


Ka
•


3
HA
a

HA
•
Ka is called the acid-dissociation constant.
•
Note that the subscript “a” indicates that this is the equilibrium constant for the dissociation of an acid.
•
Note that [H2O] is omitted from the Ka expression. (H2O is a pure liquid.)
The larger the Ka, the stronger the acid.
•
Ka is larger since there are more ions present at equilibrium relative to un-ionized molecules.
•
If Ka >> 1, then the acid is completely ionized and the acid is a strong acid.
-1-
AP Chemistry
Chapter 16. Acid-Base Equilibria
Calculating Ka from pH
•
In order to find the value of Ka, we need to know all of the equilibrium concentrations.
•
The pH gives the equilibrium concentration of H+.
•
Thus, to find Ka, we use the pH to find the equilibrium concentration of H+ and then the stoichiometric
coefficients of the balanced equation to help us determine the equilibrium concentration of the other
species.
•
We then substitute these equilibrium concentrations into the equilibrium constant
expression and solve for Ka.
Sample Exercise 16.10 (p. 688)
A student prepared a 0.10 M solution of formic acid (HCHO2) and measured its pH. The pH at 25oC was found
to be 2.38. Calculate the Ka for formic acid at this temperature. (1.8 x 10-4)
-2-
AP Chemistry
Chapter 16. Acid-Base Equilibria
Practice Exercise 1 (16.10)
A 0.50 M solution of an acid HA has pH = 2.24. What is the value of Ka for the acid?
a) 1.7 x 10-12
b) 3.3 x 10-5
c) 6.6 x 10-5
d) 5.8 x 10-3
e) 1.2 x 10-2
Practice Exercise 2 (16.10)
Niacin, one of the B vitamins, has the following molecular structure:
A 0.020 M solution of niacin has a pH of 3.26.
What is the acid-dissociation constant, Ka, for niacin?
(1.6 x 10-5)
-3-
AP Chemistry
Chapter 16. Acid-Base Equilibria
•
Weak acids are only partially ionized.
•
Percent ionization is another method to assess acid strength.
•
For the reaction
HA(aq)  H+(aq) + A–(aq)
H 


% ionization
equilibrium
HA initial
100
•
Percent ionization relates the equilibrium H+ concentration, [H+]equilibrium, to the initial HA concentration,
[HA]initial.
•
The higher the percent ionization, the stronger the acid.
•
However, we need to keep in mind that percent ionization of a weak acid decreases as the molarity of
the solution increases.
-4-
AP Chemistry
Chapter 16. Acid-Base Equilibria
Sample Exercise 16.11 (p. 689)
A 0.10 M solution of formic acid (HCOOH) contains 4.2 x 10-3 M H+(aq). Calculate the percentage of the acid
that is ionized.
(4.2%)
Practice Exercise 1 (16.11)
A 0.077 M solution of an acid HA has pH = 2.16. What is the percentage of the acid that is ionized?
a)
0.090%
b) 0.69%
c) 0.90%
d) 3.6%
e) 9.0%
Practice Exercise 2 (16.11)
A 0.020 M solution of niacin has a pH of 3.26. Calculate the percent ionization of the niacin.
(2.7%)
-5-
AP Chemistry
Chapter 16. Acid-Base Equilibria
Using Ka to Calculate pH
•
Using Ka, we can calculate the concentration of H+ (and hence the pH).
1. Write the balanced chemical equation clearly showing the equilibrium.
CH3COOH(aq)  H+(aq) + CH3COO-(aq)
2. Write the equilibrium expression. Look up the value for Ka (in a table).
Ka = [H+][ CH3COO-] = 1.8 x 10-5
[CH3COOH]
3. Write down the initial and equilibrium concentrations for everything except pure water.
• We usually assume that the equilibrium concentration of H+ is x.
4. Substitute into the equilibrium constant expression and solve.
Ka = 1.8 x 10-5 = [H+][C2H3O2-] = (x)(x)
[HC2H3O2]
0.30 - x
Note that Ka is very small (1.8 x 10-5) relative to [HC2H3O2] (0.30 M).
Keep this x

x2 = 1.8 x 10-5
0.30 – x
Neglect x in the denominator since it is extremely small relative to 0.30.
(Use ballpark figure of 103 X difference for neglecting x in the denominator )

x2 = (1.8 x 10-5 )(0.30) = 5.4 x 10-6
  x2 = 5.4 x 10-6)
X = 2.3 x 10-3 M = [H+]
Check: Compare x with original [HC2H3O2] of 0.30 M:
2.3 x 10-3 M x 100% = 0.77%, which is < 5%
0.30 M
5. Convert x ([H+]) to pH.
pH = - log (2.3 x 10-3) = 2.64
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AP Chemistry
Chapter 16. Acid-Base Equilibria
• What do we do if we are faced with having to solve a quadratic equation in order to determine the value of
x?
• Often this cannot be avoided.
•
However, if the Ka value is quite small, we find that we can make a simplifying assumption.
• Assume that x is negligible compared to the initial concentration of the acid.
• This will simplify the calculation.
• It is always necessary to check the validity of any assumption.
• Once we have the value of x, check to see how large it is compared to the initial concentration.
• If x is <5% of the initial concentration, the assumption is probably a good one.
• If x>5% of the initial concentration, then it may be best to solve the quadratic equation or use
successive approximations.
Sample Exercise 16.12 (p. 691)
Calculate the pH of a 0.20 M solution of HCN. Refer to Table 16.2 for Ka. (5.00)
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AP Chemistry
Chapter 16. Acid-Base Equilibria
Practice Exercise 1 (16.12)
What is the pH of a 0.40 M solution of benzoic acid, C6H5COOH? (The Ka value for benzoic acid is given in
Table 16.2.)
a) 2.30
b) 2.10
c) 1.90
d) 4.20
e) 4.60
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AP Chemistry
Chapter 16. Acid-Base Equilibria
Practice Exercise 2 (16.12)
The Ka for niacin is 1.6 x 10-5. What is the pH of a 0.010 M solution of niacin? (3.41)
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AP Chemistry
Chapter 16. Acid-Base Equilibria
Sample Exercise 16.13 (p. 693)
Calculate the percentage of HF molecules ionized in
a) a 0.10 M HF solution (7.9%)
b) a 0.010 M HF solution (23%)
- 10 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Practice Exercise 1 (16.13)
What is the pH of a 0.010 M solution of HF?
a) 1.58
b) 2.10
c) 2.30
d) 2.58
e) 2.64
Practice Exercise 2 (16.13)
In Practice Exercise 16.11, we found that the percent ionization of niacin (Ka = 1.5 x 10-5) in a 0.020 M solution
is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is
a) 0.010 M
(3.9%)
b) a 1.0 x 10-3 M (12%)
- 11 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Polyprotic Acids
•
Polyprotic acids have more than one ionizable proton.
•
The protons are removed in successive steps.
• Consider the weak acid, H2SO3 (sulfurous acid):
H2SO3(aq)  H+(aq) + HSO3–(aq)
Ka1 = 1.7 x 10–2
–
+
2–
HSO3 (aq)  H (aq) + SO3 (aq)
Ka2 = 6.4 x 10–8
• Where Ka1 is the dissociation constant for the first proton released, Ka2 is for the second, etc.
•
It is always easier to remove the first proton in a polyprotic acid than the second.
• Therefore, Ka1 > Ka2 > Ka3, etc.
•
The majority of the H+(aq) at equilibrium usually comes from the first ionization (i.e., the Ka1 equilibrium).
• If the successive Ka values differ by a factor of 03, we can usually get a good approximation of the pH
of a solution of a polyprotic acid by only considering the first ionization.
Weak Polyprotic Acids
Because Ka1 is so much larger than subsequent
dissociation constants for these polyprotic acids,
almost all of the H+ in the solution comes from
the first ionization reaction.
- 12 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
16.7 Weak Bases
•
Weak bases remove protons from substances.
•
There is an equilibrium between the base and the resulting ions:
Weak base + H2O(l)  conjugate acid + OH–(aq)
• Example:
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq).
•
The base-dissociation constant, Kb, is defined as
NH OH 

Kb 
•

4
[ NH 3 ]
The larger Kb, the stronger the base.
- 13 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Sample Problem 16.15 (p. 697)
Calculate the concentration of OH- in a 0.15M solution of NH3.
(1.6 x 10-3 M)
Practice Problem 1 (16.15)
What is the pH of a 0.65 M solution of pyridine, C5H5N? Kb = 1.7 x 10-9
a) 4.48
b) 8.96
c) 9.52
d) 9.62
e) 9.71
Practice Problem 2 (16.15)
Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine,
or nitrous acid?
(methylamine)
- 14 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Types of Weak Bases
•
Weak bases generally fall into one of two categories.
• Neutral substances with a lone pair of electrons that can accept protons.
• Most neutral weak bases contain nitrogen.
• Amines are related to ammonia and have one or more N–H bonds replaced with N–C bonds (e.g.,
CH3NH2 is methylamine).
Like NH3, amines can abstract a proton from a water molecule by forming an additional N-H bond,
as shown in this figure for methylamine.
•
•
Anions of weak acids are also weak bases.
Example: ClO– is the conjugate base of HClO (weak acid):
ClO–(aq) + H2O(l)  HClO(aq) + OH–(aq)
Kb = 3.3 x 10–7
Sample Exercise 16.16 (p. 698)
A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a
pH of 10.50. How many moles of NaClO were added to the water? (See info immediately above.)
(0.60 mol)
- 15 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Practice Exercise 1 (16.16)
-
The benzoate ion, C6H5COO , is a weak base with Kb = 1.6 x 10-10. How many moles of sodium benzoate are
present in 0.50 L of a solution of NaC6H5COO if the pH is 9.04?
a) 0.38
b) 0.66
c) 0.76
d) 1.5
e) 2.9
Practice Exercise 2 (16.16)
What is the molarity of a solution of NH3 in water has a pH of 11.17?
(0.12 M)
- 16 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
16.8 Relationship Between Ka and Kb
•
We can quantify the relationship between strength of an acid and the strength of its conjugate base.
•
Consider the following equilibria:
•
NH4+(aq)  NH3(aq) + H+(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
We can write equilibrium expressions for these reactions:
Ka 
•
•
•


Kb 
[NH 4 ][OH - ]
[ NH3 ]
[ NH 4 ]
If we add these equations together:
NH4+(aq)  NH3(aq) + H+(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)
The net reaction is the autoionization of water.
H2O(l)  H+(aq) + OH–(aq)
Recall that:
•
•
[NH3 ][H + ]
Kw = [H+][OH–]
We can use this information to write an expression that relates the values of Ka, Kb and Kw for a
conjugate acid-base pair:
Ka x Kb = Kw
• Alternatively, we can express this as:
pKa + pKb = pKw = 14.00 (at 25oC)
Thus, the larger Ka (and the smaller pKa), the smaller Kb (and the larger pKb).
• The stronger the acid, the weaker its conjugate base and vice versa.
- 17 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Sample Exercise 16.17 (p. 701)
Calculate
a) the base-dissociation constant, Kb, for the fluoride ion (F-); (1.5 x 10-11)
b) the acid-dissociation constant, Ka, for the ammonium ion (NH4+). (5.6 x 10-10)
Practice Exercise 1 (16.17)
By using information from Appendix D, put the following three substances in order of weakest to strongest
base:
(i) (CH3)3N
(ii) HCOO(iii) BrOa)
b)
c)
d)
e)
(i) < (ii) < (iii)
(ii) < (i) < (iii)
(iii) < (i) < (ii)
(ii) < (iii) < (i)
(iii) < (ii) < (i)
Practice Exercise 2 (16.17)
a) Which of the following anions has the largest base-dissociation constant: NO2-, PO43-, or N3-?
(PO43-, Kb = 2.4 x 10-2)
b) The base quinoline has the following structure:
Its conjugate acid is listed in handbooks as having a pKa of 4.90.
What is the base-dissociation constant for quinoline?
(7.9 x 10-10)
- 18 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
16.9 Acid-Base Properties of Salt Solutions
•
•
Nearly all salts are strong electrolytes.
• Therefore, salts in solution exist entirely of ions.
• Acid-base properties of salts are a consequence of the reactions of their ions in solution.
Many salt ions can react with water to form OH– or H+.
• This process is called hydrolysis.
An Anion’s Ability to React with Water
•
Consider an anion, X–, as the conjugate base of an acid.
• Anions from weak acids are basic.
• They will cause an increase in pH.
• Anions from strong acids are neutral.
• They do not cause a change in pH.
• Anions with ionizable protons (e.g., HSO4– ) are amphiprotic.
• They are capable of acting as an acid or a base.
• If Ka > Kb, the anion tends to decrease the pH.
• If Kb > Ka, the anion tends to increase the pH.
A Cation’s Ability to React with Water
•
•
•
Polyatomic cations that have one or more ionizable protons are conjugate acids of weak bases.
• They tend to decrease pH.
Metal cations of Group 1A and heavy alkaline earth metals are cations of strong bases and do not alter pH.
Other metal ions can cause a decrease in pH.
Combined Effect of Cation and Anion in Solution
•
The pH of a solution may be qualitatively predicted using the following guidelines:
• Salts derived from a strong acid and a strong base are neutral.
• Examples are NaCl and Ca(NO3)2.
• Salts derived from a strong base and a weak acid are basic.
• Examples are NaClO and Ba(C2H3O2)2.
• Salts derived from a weak base and a strong acid are acidic.
• An example is NH4Cl.
• Salts derived from a weak acid and a weak base can be either acidic or basic.
• Equilibrium rules apply!
• We need to compare Ka and Kb for hydrolysis of the anion and the cation.
• For example, consider NH4CN.
• Both ions undergo significant hydrolysis.
• Is the salt solution acidic or basic?
• • The Ka of NH4+ is smaller than the Kb of CN–, so the solution should be basic.
- 19 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Sample Exercise 16.18 (p. 704)
Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral:
a) Ba(CH3COO)2,
b) NH4Cl
c) CH3NH3Br
d) KNO3
e) Al(ClO4)3
Practice Exercise 1 (16.18)
Order the following solutions from lowest to highest pH:
(i) 0.10 M NaClO
(ii) 0.10 M KBr
(iii) 0.10 M NH4ClO4
a)
b)
c)
d)
e)
(i) < (ii) < (iii)
(ii) < (i) < (iii)
(iii) < (i) < (ii)
(ii) < (iii) < (i)
(iii) < (ii) < (i)
Practice Exercise 2 (16.18)
In each of the following, indicate which salt in each of the following pairs will form the more acidic (or less
basic) 0.010 M solution:
a) NaNO3 or Fe(NO3)3
b) KBr, or KBrO
c) CH3NH3Cl or BaCl2
d) NH4NO2 or NH4NO3
- 20 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Sample Exercise 16.19 (p. 705)
Predict whether the salt Na2HPO4 will form an acidic or a basic solution on dissolving in water.
(basic)
Practice Exercise 1 (16.19)
How many of the following salts are expected to produce acidic solutions (See Table 16.3 for data):
NaHSO4, NaHC2O4, NaH2PO4, NaHCO3
a) 0
b) 1
c) 2
d) 3
e) 4
Practice Exercise 2 (16.19)
Predict whether the dipotassium salt of citric acid (K2HC6H5O7) will form an acidic or basic solution in water.
(see Table 16.3 for data)
(acidic)
- 21 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
16.10 Acid-Base Behavior and Chemical Structure
Factors That Affect Acid Strength
•
Consider H–X.
•
For this substance to be an acid:
• The H–X bond must be polar with H+ and X-.
•
In ionic hydrides, the bond polarity is reversed.
• The H–X bond is polar with H- and X+.
• In this case, the substance is a base.
•
Other factors important in determining acid strength include:
• The strength of the bond.
• The H–X bond must be weak enough to be broken.
• The stability of the conjugate base, X–.
• The greater the stability of the conjugate base, the more acidic the molecule.
Binary Acids
•
The H–X bond strength is important in determining relative acid strength in any group in the periodic table.
• The H–X bond strength tends to decrease as the element X increases in size.
• Acid strength increases down a group; base strength decreases down a group.
•
H–X bond polarity is important in determining relative acid strength in any period of the periodic table.
• Acid strength increases and base strength decreases from left to right across a period as the
electronegativity of X increases.
•
For example, consider the molecules HF and CH4.
• HF is a weak acid because the bond energy is high.
•
The electronegativity difference between C and H is so small that the C–H bond is nonpolar, and CH4 is
neither an acid nor a base.
- 22 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Oxyacids
•
Many acids contain one or more O–H bonds.
• Acids that contain OH groups (and often additional oxygen atoms) bound to the central atom are called
oxyacids.
• All oxyacids have the general structure Y–O–H.
e.g. H2SO4
•
The strength of the acid depends on Y and the atoms attached to Y.
• As the electronegativity of Y increases, so does the acidity of the substance.
• The bond polarity increases and the stability of the conjugate base (usually an anion) increases.
•
We can summarize how acid structure relates to the electronegativity of Y and the number of groups
attached to Y:
• For oxyacids with the same number of OH groups and the same number of oxygen atoms:
• Acid strength increases with increasing electronegativity of the central atom, Y.
• Example: HClO > HBrO > HIO
• For oxyacids with the same central atom, Y:
• Acid strength increases as the number of oxygen atoms attached to Y increases.
• Example: HClO4 > HClO3 > HClO2 > HClO
The acidity of oxyacids increases with increasing electronegativity of the central atom.
As the electronegativity of the atom attached to an OH group increases, the ease with which the hydrogen ion is
released increases. The drift of electron density toward the electronegative atom further polarizes the O—H
bond, which favors ionization. In addition, the electronegative atom will help stabilize the conjugate base, which
also leads to a stronger acid. Because Cl is more electronegative than I, HClO is a stronger acid than HIO.
- 23 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Strength of oxyacids
The strength of the oxyacids of chlorine steadily increases from hypochlorous acid (HClO) to perchloric acid
(HClO4) as shown in the figure. Because the oxidation number of the central atom increases as the number of
attached O atoms increases, this correlation can be stated in an equivalent way: In a series of oxyacids, the
acidity increases as the oxidation number of the central atom increases.
Sample Exercise 16.20 (p.708)
Arrange the compounds in each of the following series in order of increasing acid strength:
a) AsH3, HBr, KH, H2Se;
b) H2SO4, H2SeO3, H2SeO4.
Practice Exercise 1 (16.20)
Arrange the following substances in order from weakest to strongest acid:
HClO3, HOI, HBrO2, HClO2, HIO2
a) HIO2 < HOI < HClO3 < HBrO2 < HClO2
b) HOI < HIO2 < HBrO2 < HClO2 < HClO3
c) HBrO2 < HIO2 < HClO2 < HOI < HClO3
d) HClO3 < HClO2 < HBrO2 < HIO2 < HOI
e) HOI < HClO2 < HBrO2 < HIO2 < HClO3
- 24 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Practice Exercise 2 (16.20)
In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution:
a) HBr, HF;
b) PH3, H2S;
c) HNO2, HNO3;
d) H2SO3, H2SeO3.
Carboxylic Acids
•
There is a large class of acids that contain a –COOH group (a carboxyl group).
•
Acids that contain this group are called carboxylic acids.
• Examples: acetic acid, benzoic acid, formic acid.
The portion of the structure shown in blue is called the carboxyl group, which is often written as COOH. Thus,
the chemical formula of acetic acid is written as CH3COOH, where only the hydrogen atom in the carboxyl
group can be ionized. Acids that contain a carboxyl group are called carboxylic acids, and they form the largest
category of organic acids.
•
Why are these molecules acidic?
1. The additional oxygen atom on the carboxyl group increases the polarity of the O–H bond and stabilizes
the conjugate base.
2. The conjugate base exhibits resonance.
• This gives it the ability to delocalize the negative charge over the carboxylate group, further
increasing the stability of the conjugate base.
- 25 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Resonance. The conjugate base of a carboxylic acid
(a carboxylate anion) can exhibit resonance, which
Contributes further to the stability of the anion by
spreading the negative charge over several atoms.
spreading
•
The acid strength also increases as the number of electronegative groups in the acid increases.
• For example, acetic acid is much weaker than trichloroacetic acid.
Comparison of Different Types of Acids and Bases
Arrhenius (traditional) acids and bases (C19th)
•
Acid: compound containing H that ionizes to yield H+ in solution
•
Base: compound containing OH that ionizes to yield OH- in solution
(Note: does not describe acid/base behavior in solvents other than water)
Note: Every Arrhenius acid/base is also a Brønsted-Lowry acid/base.
Brønsted-Lowry Acids and Bases (1923)
•
•
Acid: H+ (proton) donor
Base: H+ (proton) acceptor
NH3
+
ammonia
(B-L base)
H2O
water
(B-L acid)

NH4+
+
ammonium ion
(B-L acid)
OHhydroxide ion
(B-L base)
Note: Every Brønsted-Lowry acid/base is also a Lewis acid/base.
Lewis Acids and Bases (1920’s)
•
Acid: accepts pair of e-‘s
•
Base: donates pair of e-‘s
H+
Lewis acid
+
[:O:H]-  H:O:H
Lewis base
- 26 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
16.11 Lewis Acids and Bases
•
A Brønsted-Lowry acid is a proton donor.
•
Focusing on electrons: A Brønsted-Lowry acid can be considered as an electron pair acceptor.
•
Lewis proposed a new definition of acids and bases that emphasizes the shared electron pair.
• A Lewis acid is an electron pair acceptor.
• A Lewis base is an electron pair donor.
• Note: Lewis acids and bases do not need to contain protons.
• Therefore, the Lewis definition is the most general definition of acids and bases.
For a substance to be a proton acceptor (a BrønstedLowry base), it must have an unshared pair of electrons
for binding the proton. NH3, for example, acts as a
proton acceptor. Using Lewis structures, we can write
the reaction between H+ and NH3 as shown in the firugre.
•
What types of compounds can act as Lewis acids?
• Lewis acids generally have an incomplete octet (e.g., BF3).
Consider the reaction between NH3 and BF3.
This reaction occurs because BF3 has a
vacant orbital in its valence shell. It therefore
acts as an electron-pair acceptor (a Lewis
acid) toward NH3, which donates the electron
pair. The curved arrow shows the donation of
a pair of electrons from N to B to form a
covalent bond.
•
•
•
Transition-metal ions are generally Lewis acids.
Lewis acids must have a vacant orbital (into which the electron pairs can be donated).
Compounds with multiple bonds can act as Lewis acids.
• For example, consider the reaction:
H2O(l) + CO2(g)  H2CO3(aq)
• Water acts as the electron pair donor and carbon
dioxide as the electron pair acceptor in this reaction.
• Overall, the water (Lewis base) has donated a pair of
electrons to the CO2 (Lewis acid).
- 27 -
AP Chemistry
Chapter 16. Acid-Base Equilibria
Hydrolysis of Metal Ions
•
The Lewis concept may be used to explain the acid properties of many metal ions.
•
Metal ions are positively charged and attract water molecules (via the lone pairs on the oxygen atom of
water).
•
Hydrated metal ions act as acids.
• For example:
Fe(H2O)63+(aq)  Fe(H2O)5(OH)2+(aq) + H+(aq)
Ka = 2 x 10–3.
In general:
• The higher the charge, the stronger the M–OH2 interaction.
• Ka values generally increase with increasing charge
• The smaller the metal ion, the more acidic the ion.
• Ka values generally decrease with decreasing ionic radius
• Thus the pH of an aqueous solution increases as the size of the ion increases (e.g., Ca2+ vs. Zn2+)
and as the charge increases (e.g., Na+ vs. Ca2+ and Zn2+ vs. Al3+).
•
The acidity of a hydrated cation depends
on cation charge and size.
The interaction between a water molecule and a
cation is much stronger when the cation is a
smaller ion of higher charge. The pull of electron
density toward the cation weakens the polar O—
H bond of the water molecule and allows the
transfer of a H+ ion to a nearby water molecule.
As a result, hydrated cations tend to be acidic,
with their acidities increasing with increasing
charge and decreasing size.
The Amphoteric Behavior of Amino Acids
•
•
•
Amino acids: building blocks of proteins.
Each contains a carboxyl group AND an amine group.
Thus amino acids have both acidic and basic groups.
• They undergo a proton transfer in which the proton of the carboxyl is transferred to the basic nitrogen
atom of the amine group.
• A zwitterion or dipolar ion results.
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