CHAPTER 16 (MOORE)
ACIDS, BASES AND ACID-BASE EQUILIBRIA
This chapter deals with acids and bases and their equilibria. We will treat acids as proton (H+)
donors in solution and bases as proton acceptors. According to this approach, acid-base reactions
are merely proton transfer reactions from A (an acid) to B (a base).
We will examine how the structure of an acid influences its strength or ability to give up its proton
in aqueous solution.
Recall that, in Chapter 5, acids and bases were classified as strong or weak, based upon their degree
of ionization. See Moore, pp. 171-179 to review what you should already know about acids and
bases as you begin this chapter.
Key Concepts:
electrolytes – strong and weak
strong acids and weak acids
acid-base reactions
Brønsted–Lowry Theory
Arrhenius theory: an acid forms H+ in water; a base forms OH1– in water.
NaOH (+ H2O) Na+(aq) + OH1-(aq)
base
hydroxide ion
HCl (+ H2O) H+(aq) + Cl1-(aq)
acid
proton
But not all acid–base reactions involve water, and many bases (NH3, CO32-) don’t contain any
OH1–, so Brønsted–Lowry theory defines acids and bases in terms of a proton (H+) donor –acceptor
model.
A Brønsted–Lowry acid is a proton donor:
HCl(aq) + H2O H3O+(aq) + Cl1-(aq)
donor
acceptor
A Brønsted–Lowry base is a proton acceptor.
NH3(aq) + H2O NH41+ + OH1-(aq)
acceptor donor
Ionization of Acetic Acid in Water
HC2H3O2(aq) + H2O(l) ' H3O+(aq) + C2H3O21-(aq)
Ionization of Ammonia in Water
NH3(aq) + H2O(l) NH41+(aq) + OH1-(aq)
Ammonia is a weak base in water.
Strength of Conjugate Acid–Base Pairs
Stronger acids donate H+ more readily than a weaker acid. The stronger an acid, the weaker is its
conjugate base. The stronger a base, the weaker is its conjugate acid.
Conjugate Acids and Conjugate Bases
conjugate acid: a base plus an attached H+
conjugate base: an acid minus an attached H+
HCl(aq) + H2O(l) H3O+(aq) + Cl1-(aq)
conjugate acid
conjugate base
What is the conjugate base of HCl? We say that Cl1- is the conjugate base of HCl. HCl and Cl1- are
an example of a conjugate acid-conjugate “base pair.”
Example 16.1. Identify the Brønsted–Lowry acids and bases and their conjugates in:
(a)
H2S + NH3 ' NH4+ + HS1–
(b)
OH1– + H2PO41– ' H2O + HPO42–
Acid Strengths. See Table 15.1 (Hill, p. 620)
Ka and Kb
The equilibrium constant for a Brønsted acid (HA) is represented by Ka, and that for a Brønsted
base (B) is represented by Kb.
Weak Acid
HA(aq) + H2O(l) ' H3O+(aq) + A1–(aq)
Ka = ([H3O+][A1-]) / [HA]
Weak Base
B(aq) + H2O(l) ' BH+(aq) + OH1–(aq)
Kb = ([BH+][OH1-]) / [B]
Acid/Base Strength and Direction of Equilibrium
Consider the following equilibrium. Since HBr is a stronger acid than CH3COOH, the equilibrium
for the reaction lies to the left.
weaker base
stronger base
CH3COOH + Br1- ' HBr + CH3COO1weaker acid
stronger acid
C equilibrium lies in this direction
Strong Acids. Strong acids ionize completely in water.
The strong acids are HCl, HBr, HI, HNO3, H2SO4 and HClO4. Note that the “strong” acids all
appear above (i.e. are stronger acids than) H3O+ in the table. Strong acids are all “leveled” to the
same strength—that of H3O+—when they are in water. They are all essentially 100% dissociated
(ionized) in water. (HI> HBr> HCl > H2SO4 > HNO3 > H3O+)
Periodic Trends in Acid Strength
The greater the tendency for HX (general acid) to transfer a proton to H2O, the more the forward
reaction is favored and the stronger the acid. Any factor that makes it easier for the H+ to “leave”
will increase the strength of the acid. Acid strength is inversely proportional to H—X bonddissociation energy. Weaker H—X bond means a stronger acid.
Periodic Trends in Acid Strength (see pp.790-794, Moore)
Strength of Oxoacids
Acid strength increases with the electronegativity of the central atom and with the number of
terminal oxygen atoms.
Strength of Carboxylic Acids
Carboxylic acids all have the –COOH group in common. Differences in acid strength come from
differences in the R group attached to the carboxyl group. In general, the more electronegative
atoms there are in the R group, the stronger the acid will be.
Example 16.2 Select the stronger acid in each pair:
(a) nitrous acid, HNO2, and nitric acid, HNO3 [HNO3 is the stronger acid (more O atoms on N)]
(b) Cl3CCOOH and BrCH2COOH [Cl3CCOOH = stronger acid - Cl more electronegative than Br)]
Self-Ionization of Water (“autoprotolysis”)
Fact: Pure water conducts some electricity, since water “self-ionizes:”
H2O(l) + H2O(l) ' H3O+(aq) + OH1-(aq)
The equilibrium constant for this process is called the ion product of water (Kw).
At 25 °C, Kw = 1.0 x 10–14 = [H3O+][OH1–]
This equilibrium applies to all aqueous solutions—acids, bases, salts, and nonelectrolytes and not
just to pure water.
The pH Scale
Concentration of H3O+ can vary over a wide range in aqueous solution, from about 10 M to about 1
x 10–14 M. A more convenient expression for H3O+ is pH.
pH = –log [H3O+]
and [H3O+] = 10–pH
p-functions: p( ) means take the – log( )
pOH = –log [OH1–]
and
[OH1–] = 10–pOH
pKw = –log Kw = –log (1.0 x 10-14) = 14.00
Recall that Kw = 1.0 x 10–14 = [H3O+][OH1–] at 25 °C
[H3O+][OH1–] = 1.0 x 10–14
Taking –log of each side:
–log { [H3O+][OH1–] } = –log (1.0 x 10–14)
–log [H3O+] –log [OH1–] = 14.00 = pKw
pH = –log [H3O+]
pOH = –log [OH1–]
pKw = pH + pOH = 14.00
The pH Scale (see Figure 16.2, Moore, p. 782)
Example 16.3 A Conceptual Example
Is the solution 1.0 x 10–8 M HCl acidic, basic, or neutral?
pH = - log(1.0 x 10-8) = 8.00 Basic! (but only slightly basic)
Equilibria in Solutions of Weak Acids and Weak Bases
1. Write the equation for the reversible reaction.
2. Set up the ICE table.
3. Find the changes that occur at equilibrium
4. Use “simplifying assumptions” (“5% rule”).
5. Calculate equilibrium concentrations / constant.
Weak Acid and Weak Base Ionization Constants. See Table 16.2 (Moore, p.788)
Example 16.4 Ordinary vinegar is approximately 1 M CH3COOH and as shown in Figure 16.2, it
has a pH of about 2.4. Calculate the expected pH of 1.00 M CH3COOH(aq), and show that the
calculated and measured pH values are in good agreement.
CH3COOH(aq) + H2O ' H3O+(aq) + CH3COO1-(aq)
Ka = 1.8 x 10-5 = [H3O+][CH3COO1-] / [CH3COOH]
Calculate the expected pH of 1.00 M CH3COOH(aq):
I
C
E
[CH3COOH]
1.00
−x
1.00 – x
[H3O+]
0
+x
x
Substituting equilibrium values into the Ka expression,
Ka = 1.8 x 10-5 = [H3O+][CH3COO1-] / [CH3COOH]
Ka = 1.8 x 10-5 = [(x)(x) / (1.00 – x), so, x2 / (1.00 – x) = 1.8 x 10-5
Assume that x << 1.00 D 1.00 – x ≈ 1.00, so,
(x2 / 1.00) ≈ 1.8 x 10-5
2
-5
+
-5 1/2
x ≈ 1.8 x 10 and [H3O ] = x ≈ (1.8 x 10 ) = 4.2 x 10-3 M
[CH3COO1-]
0
+x
x
Check assumption: (1.00 - 4.2 x 10-3 = 0.9958 = 1.00 (so assumption is valid)
pH = – log (4.2 x 10-3) = 2.38
(very close to 2.4)
Example 16.5 What is the pH of 0.00200 M ClCH2COOH(aq)? Ka = 1.4 x 10-3
Ka = 1.4 x 10-3 = [H3O+][ClCH2COO1-] / [ClCH2COOH]
Now set up the ICE table:
I
C
E
[H3O+]
0
+x
x
[ClCH2COOH]
0.00200
−x
0.00200 – x
[ClCH2COO1-]
0
+x
x
Substituting equilibrium values into the Ka expression,
Ka = 1.4 x 10-3 = [H3O+][ClCH2COO1-] / [ClCH2COOH] = (x)(x) / (0.0200 – x)
so, x2 / (0.0200 – x) = 1.4 x 10-3 or, x2 = (0.0200 – x)(1.4 x 10-3) = 2.8 x 10-6 – 1.4 x 10-3 x
Collecting terms on the left,
x2 + 1.4 x 10-3 x – 2.8 x 10-6 = 0
ax2 + bx + c = 0
(quadratic expression)
-3
a = 1; b = 1.4 x 10 ; c = – 2.8 x 10-6
Quadratic Solution: x = [(–b ± SQRT(b2 – 4ac)] /2a
Roots: x(1) = 1.1 x 10-3 and x(2) = – 0.0025 (negative concentration not possible)
[H3O+] = x = 1.1 x 10-3 M
pH = – log(1.1 x 10-3) = 2.96
Example 16.6 What is the pH of 0.500 M NH3(aq)?
Example 16.7 The pH of a 0.164 M aqueous solution of dimethylamine is 11.98. What are the
values of Kb and pKb? The ionization equation is
(CH3)2NH + H2O ' (CH3)2NH2+ + OH1–
dimethylamine dimethylammonium ion
Kb = ?
Polyprotic Acids
monoprotic acid: one ionizable H+ per molecule
polyprotic acid: more than one ionizable H+ per molecule
Examples:
sulfuric acid, H2SO4
carbonic acid, H2CO3
phosphoric acid, H3PO4
diprotic acid
diprotic acid
triprotic acid
The protons of a polyprotic acid dissociate in steps. Each step has its own Ka.
Ka1 > Ka2 > Ka3, etc. for steps (1), (2) and (3) and so on.
Polyprotic Acids (see Appendix p. A31)
Example of stepwise ionization for a diprotic acid
(1) H2CO3 + H2O ' H3O+ + HCO31Ka1 = 4.4 x 10-7
1+
2(2) HCO3 + H2O ' H3O + CO3
Ka2 = 4.7 x 10-11
H2CO3 + 2 H2O ' 2 H3O+ + CO32- (2-step, overall ionization)
Koverall = Ka1 x Ka2 = (4.4 x 10-7)(4.7 x 10-11) = 2.1 x 10-17
Example 16.8 What is the approximate pH of 0.71 M H2SO4?
Ions as Acids and Bases
HCl is a strong acid: therefore Cl– is so weakly basic in water that a solution of chloride ions (e.g. in
NaCl (aq)) is virtually neutral. Acetic acid, HC2H3O2, is a weak acid, so C2H3O21-, is significantly
basic in water. A solution of sodium acetate (NaC2H3O2) dissociates completely (into sodium and
acetate ions in water) is therefore slightly basic:
C2H3O21-, + H2O ' HC2H3O2 + OH1–
1. Salts of strong acids and strong bases form neutral solutions: NaCl, KNO3
2. Salts of weak acids and strong bases form basic solutions: KNO2, NaClO
3. Salts of strong acids and weak bases form acidic solutions: NH4NO3
4. Salts of weak acids and weak bases form solutions that may be acidic, neutral, or basic; it
depends on the relative strengths of the cations and the anions: NH4NO2, CH3COONH4.
Example 16.9 A Conceptual Example: Is NH4I(aq) acidic, basic, or neutral?
HI(aq) + NH3(aq) NH4I(aq)
strong
weak
acid
base
Recall: Salts of strong acids and weak bases form acidic solutions, e.g. NH4NO3
Thus, we predict that this solution should be … ACIDIC!!
In order to make quantitative predictions of pH for a salt solution, we need an equilibrium constant
for the hydrolysis reaction. The relationship between Ka and Kb of a conjugate acid–base pair is:
Ka x Kb = Kw
or, Kb = Kw / Ka
If, instead, we have values of pKa or pKb: pKa + pKb = pKw = 14.00
Example 16.10 Calculate the pH of a solution that is 0.25 M NaC2H3O2aq).
Example 16.11 What is the molarity of an NH4NO3(aq) solution that has a pH = 4.80?
The Common Ion Effect
Consider a solution of a weak monoprotic acid, HA, like acetic acid below:
CH3COOH (aq) + H2O (l) ' H3O+ (aq) + CH3COO1- (aq)
If we add a sodium salt containing the conjugate base (A1-) ion as a second solute (i.e., NaA), the
pH of the solution increases. Let’s see why: When a salt supplies A1-, the equilibrium shifts to the
left. Then apply LeChâtelier’s principle: what happens to [H3O+] when the equilibrium shifts to the
left? (it decreases!)
common ion effect: the inhibition of the ionization of a weak acid or a weak base by the presence
of a “common ion” from a strong electrolyte The equilibrium is shifted back to the left − this is just
another case of LeChatelier’s principle!
Buffer Solutions
Many biochemical reactions—especially enzyme-catalyzed reactions—are sensitive to pH.
For working with such reactions we often need a solution that maintains a nearly-constant pH, i.e. a
“pH stat.” We use a buffer solution! a solution that changes pH only slightly when small amounts of
a strong acid or a strong base are added. Buffers resist changes in pH when we add small amounts
of strong acid or strong base.
A buffer contains significant concentrations (like 0.1 or even 1.0 M) of both a weak acid and its
conjugate base (an weak acid buffer), or of both a weak base and its conjugate acid (a weak base
buffer).
Weak acid + conjugate base: HA and A1or,
Weak base + conjugate acid: B and BH+
The acid component of the buffer neutralizes small added amounts of OH1–, forming the weaker
conjugate base, which does not affect pH much: HA + OH1– H2O + A1–
Pure water does not buffer at all …
More Buffer Solutions
HA(aq) + H2O H3O+ (aq) + A1– (aq)
Ka = [H3O+] x {[A1-] / [HA]}
pKa = pH −log {[A1-] / [HA]}
Example. Consider 1000 A1- and 1000 HA in a solution:
Initial: {[A1-] / [HA]} = 1000/1000 = 1.000
Final: {[A1-] / [HA]} = 1010/990 = 1.020 Only a 2.0% change in the ratio
Comparison with a Buffer and an Unbuffered Solution.
Pure water increases in pH by about 5 pH units when the OH– is added, and decreases by about 5
pH units when the H3O+ is added. In contrast, the same amounts of OH– and H3O+ added to a
buffer solution barely change the pH.
An Equation for Buffer Solutions. The Henderson–Hasselbalch equation
pH = pKa + log {[A1-] / [HA] }
In order to use this equation, the following must be true:
(1) 0.10 < [A1-]/[HA] < 10 and
(2) [conjugate base] / Ka >100 and [weak acid] / Ka > 100
Example 16.12
Calculate the pH of an aqueous solution that is both 1.00 M HC2H3O2 and 1.00 M NaC2H3O2.
This is an acetic acid-acetate buffer, so the equilibrium we need is
HC2H3O2(aq) + H2O ' H3O+(aq) + C2H3O21- (aq), where
Ka = {[H3O+][C2H3O21-]}/[HC2H3O2] = 1.8 x 10-5
For acid and conjugate base concentrations > 0.1 M, we can “neglect” x, so let x = [H3O+]:
Ka = {(x)(1.00)} / (1.00) = 1.8 x 10-5
x = 1.8 x 10-5 M = [H3O+]
pH = - log(1.8 x 10-5) = 4.74
Alternate solution using Henderson-Hasselbalch: pH = pKa + log{[A1-]/[HA]}
pH = - log(1.8 x 10-5) + log{(1.00)/(1.00)} = 4.74 + 0 = 4.74
Example 16.13 What concentration of acetate ion in 0.500 M HC2H3O2 (aq) produces a buffer
solution with pH = 5.00? (pH 5.00 @ [H3O+] = 1.00 x 10-5 M
HC2H3O2(aq) + H2O ' H3O+(aq) + C2H3O21- (aq)
1.00 x 10-5 M x
0.500 M
+
1Ka = {[H3O ][C2H3O2 ]}/[HC2H3O2] = 1.8 x 10-5
x = [(1.8 x 10-5)(0.500)]/(1.00 x 10-5) = 0.900 M
Buffer Capacity and Buffer Range
Buffer - most effective if the [A1-] and [HA] are equal or nearly equal.
This is when pH ≈ pKa of HA in the buffer.
buffer capacity (β): the number of mol of strong acid or strong base that must be added to change
the pH of 1000 mol of a buffer 1.00 pH unit
Acid–Base Indicators
acid–base indicator: a weak organic acid or base − the acid form (HA) of the indicator has one
color, the conjugate base (A–) form has a different color. Note: one of the forms may be colorless.
In acid solutions, [H3O+] is large. Because H3O+ is a common ion, it “suppresses” the ionization of
the indicator acid, shifting the equilibrium to the left, and we see the color of HA.
? ?
HA + H2O ' H3O+ + A–
Different indicators have different values of Ka and different color transition ranges (different
values of pH across which they change) …
Titration Curve, Strong Acid with Strong Base
A plot of pH vs. mL titrant added, e.g. from a buret. When the titrant is NaOH, the pH increases
since we’re adding a base …
Neutralization Reactions: terminology
titration curve: a graph of pH versus volume of titrant a neutralization titration
equivalence point (in an acid–base titration): precise stoichiometric proportions of acid and base
endpoint: volume of added titrant at the point at which the indicator changes color
Ideally, the indicator is selected so that V(endpoint) = V(equivalence point)
Strong acid titrated with strong base: pH = 7.00 at equivalence point
HCl (aq) + NaOH (aq) H2O (l) + NaCl (aq)