Ch. 17 Packet: Acids and Bases
Definitions for acids and bases: There are three definitions we’ll learn eventually. (the
first two at the beginning of Ch. 17, the last at the end of the chapter. These are 1) the
Arrhenius definition, the Brønsted-Lowry definition, and the Lewis definition.
Arrhenius acids produce
Arrhenius bases produce
when in water.
when in water.
The Bronsted-Lowry definition is most definitely the most important of the three.
Bronsted acid-base reactions occur as a proton (H+ ion) is transferred.
Bronsted acids
Bronsted bases
a proton in the acid-base reaction.
a proton in the acid-base reaction.
A second idea involved with the B-L is the concept of conjugate acid-base pairs. These
conjugate pairs sorta look like each other: They differ only by one proton (H+). One of
the pair will be on the left side of the equation; one will be on the right. Each B-L
acid/base reaction will have two conjugate pairs. Look at this example:
C 2H 5COOH (aq) + F – (aq)  C 2H 5COO– (aq) + HF(aq)
C2H5COOH and C2H5COO– are one conjugate acid-base pair. In the reaction,

C2H5COOH
a proton to become C2H5COO–. Therefore:
C2H5COOH is the
, and C2H5COO– is the conjugate
.
a proton to become HF; they’re the other conjugate pair.
F-
The fluoride ion is thus the
, and HF is the conjugate
.
Polyprotic acids have more than one acidic proton. Examples: H2CO3, H3PO4, H2C2O4
–
H 2CO3 (aq) + H 2O (aq)  HCO 3 (aq) + H 3O + (aq) [1st ionization]
–
–2
HCO 3 (aq) + H 2O (aq)  CO3 (aq) + H 3O + (aq)

[2 nd ionization]
In this series of ionizations, we see that H2CO3 is an acid; CO3-2 is a base, but HCO3– is the
“interesting” species. In the first ionization, it is the CB of the acid H2CO3 . In the 2nd
ionizaiton, it is the CA of the base CO3-2.
Substances, like HCO3–, that can be either a Bronsted base or a Bronsted acid are called
amphiprotic substances. Many of these are the “middle guys” in polyprotic acid
dissociation series, but water is a very important amphiprotic substance:
H2O’s conjugate acid is
. H2O’s conjugate base is
.
***Futureview: At the end of the chapter we’ll discuss amphoteric substances. Similar
names, different idea. Be prepared to discern between them!***
Strong acids are strong because they dissociate 100% in water. Strong bases are the
soluble hydroxides.
There are strong bases that aren’t hydroxides, but we don’t often see them. The chart on p. 799 shows a
few examples.
Using the Bronsted concept, we can think of the “job” of acids as donating a proton. The
“job” of a base is to accept a proton. The stronger the acid (or base), the better it is at its
respective “job.” SA’s dissociate 100% (All molecules donate a proton); WA’s tend to
dissociate <5% in water (most molecules remain protonated). We saw in chapter 5 how
this impacts how we write acids in net ionic equations.
Since a SA is so darn good at its “job,” it makes sense that its CB is awful at the job of
receiving a proton. There is a ‘spectrum’ of sorts at work, where the stronger an acid, the
weaker its CB. The weaker an acid is, the stronger its CB. I personally like to think of the
conjugates of “the strongs” as worthless. HNO3, HClO4, HCl, HBr, and HI are strong
acids; NO3-1, ClO4-1, Cl-1, Br-1, and I-1 are ‘worthless’ bases. [I’ve intentionally left H2SO4 out of
this example. Since sulfuric acid is polyprotic, its CB is amphiprotic. As it happens, HSO 4-1 is quite acidic in water.]
CB’s of weak acids, on the other hand, are ‘decent’ bases. Acetic acid is a WA, acetate is
a decent base. Neither is “strong,” but both are OK at their respective jobs. This idea will
be very important when we consider hydrolysis reactions in section 17.8.
***It is true that the stronger acid and the stronger base will always be on the same side
of the equation. (Can you explain why? It’ll take some serious thought!)***
We can predict whether an acid-base reaction will be reactant-favored or product-favored
by comparing the strengths of the two acids and the two bases. Since the “job” of acids is
to donate a proton, the stronger acid will be the one that does this more effectively,
pushing the reaction toward the side with the weaker acid. Likewise, since bases are
acceptors of electrons, the stronger base will tend to become protonated, forming its
conjugate acid.
CH3CO2H(aq) + NH3 (aq)  NH+1
4 (aq) + CH3CO2 (aq)
-1
Acid

base
acid
base
In this example, we can see from acid/base strength chart (p. 794 or p. 799) that acetic
acid is a stronger acid than is the ammonium ion, while ammonia is a stronger base than
is the acetate ion. As a result, we can predict that this reaction is product favored, and that
equilibrium will lie to the right. A general rule of acid/base problems is that the “best
acid” will react with the “best base” in a system.
Towards which side does equilibrium lie for each of the following acid-base equations?
1
1
H2SO4 (aq) + HCO3 (aq)  H2CO3 (aq) + HSO 4 (aq)
1
H2O(l) + HCO 3 (aq)  H2CO3 (aq) + OH1(aq)



H 2O(l) + H 2S(aq)  HS -1 (aq) + H 3O 1(aq)
I can’t overstate the importance of being comfortable with writing equations to show the
equilibrium that exists when acids or bases are mixed in water. You must be able to write
the Ka (or Kb) reaction, and the resulting equilibrium expression Ka (or Kb), as if you
were on autopilot. So be sure you can easily do the skills shown on this page for any
given acid or base!
When an acid is mixed in water, they will establish an equilibrium whose equilibrium
constant is called Ka. For the generic acid HA,
HA(aq) + H 2O(l)  A-1(aq) + H 3O 1(aq)

and
[H 3O + ][A - ]
Ka =
[HA]
We know that the stronger the acid, the more it dissociates. Therefore, the stronger the
acid, the larger its Ka value. If you’re given Ka values for two acids, the stronger acid is
the one with the larger value of Ka.

Bases are similar:
B(aq) + H 2O(l)  BH + (aq) + OH1(aq)

and
[OH - ][BH + ]
Kb =
[B]
like acids, the stronger the base, the more it dissociates in water, so a base with larger Kb
is stronger than a base with a smaller value of Kb.

Water/pH scale
We’ve seen that water is capable of acting as an acid or as a base. An important
consequence of this is shown in this reaction, the autoionization of water***:
2 H 2O(l)  H 3O + (aq) + OH1(aq)
The equilibrium expression for this reaction is called Kw, the dissociation
constant for water.

K w = [H 3O + ][OH - ]
***Sometimes we see H2O(l)  H+ (aq) + OH-(aq) in this case Kw = [H+][OH-]***

At 25°C, Kw = 1.0 x 10-14, a very small number. This shows us that equilibrium lies far
to the left for this reaction. Additionally, the stoichiometry of the reaction shows us that
for each molecule of water that dissociates, one hydronium ion and one hydroxide ion
will be formed. For a neutral water solution, at 25°C, [H3O+] = [OH-] = 1.0 x 10-7 M.
***Multiple choice question: What makes a solution neutral?
a) its pH is 7 b) [OH-] = 1.0 x 10-7 M c) [H3O+] = 1.0 x 10-7 M
d) [H3O+] = [OH-]
In an acid solution, [H3O+] > [OH-]
In a base solution [H3O+] < [OH-]
Regardless, at 25°C, [H3O+] x [OH-] = 1.0 x 10-14
Let’s look at a conjugate acid/base pair: HCN and CN- . Let’s start by writing the Ka and
Kb equations and expressions for both the acid and base (I strongly recommend
practicing this on your own for random acids and bases!):
HCN + H2O  H3O+ + CN-
CN- + H2O  HCN + OH-

Ka =
[H 3O + ][CN - ]
[HCN]
Kb =
[HCN][OH - ]
[CN - ]

If we add the two equations above, we get the following equation:
2 H 2O(l)  H 3O + (aq) + OH1(aq) and as we learned in Ch. 16, when we add two
equations to get a third, K3 = K1 • K2 . In this case, Knew = Ka • Kb = Kw . (see how [CN] and
[HCN] cancel?)

This relationship is always true:
Ka • Kb = Kw
The pH scale
pH = -log[H3O+]
Furthermore, at 25°C,
and
pOH = -log[OH-]
pH + pOH = 14.00
Easier acid/base pH problems have strong acids or strong bases: we know that these acids
and bases dissociate completely, therefore [strong acid] = [H3O+] and we can then
compute pH directly. Also, [strong base] = [OH-] for IA hydroxides, or in the case of the
IIA strong bases where each mole of strong base (Sr(OH)2 or Ba(OH)2) produces 2 moles
of OH-, so [strong IIA base] = 0.5[OH].
Example 17.3 in your book: An aqueous solution contains 0.700g NaOH and has volume
485 mL. What is the pH?
• 0.700g NaOH is 0.0175 moles of NaOH (will give 0.0175 moles OH-)
• 485 mL is 0.485 L of solution
• [OH-] = 0.0175 moles OH-  0.485L = 0.0361 M
• pOH = -log(.0361) = 1.442
• pH = 14 – 1.442 = 12.558
We could have also determined [H3O+] from = Kw / [OH-] and then taken pH = -log[H3O+].
Both should give the identical answer.
••• See the yellow tips box on page 808. Shows how to use sig figs with logarithms, and
show the important “other” equation relating pH and [H3O+] or [OH-]:
[H3O+] = 10-pH
and
[OH-] = 10-pOH
This has been shown to you as part of “The Box.” You must be able to convert to/from any
of these, given one of the others: pH, pOH, [H3O+], or [OH–].
17.7: Equilibrium Problems using Ka and Kb
In many ways, equilibrium is equilibrium: ICE tables, products and reactants, etc. Just like
in Chapter 16, there are two major types of equilibrium problems: Ones where you are
given (or can ‘easily’ find) x in the ICE table; and ones where you know K and need to
solve for x.
So how are Acid/Base equilibrium problems unique?
• There are things that are special for A/B problems: “x” is related to either [H3O+]
or [OH–], so we can find x if we’re given pH (or pOH). OR, if we’re solving for x, we’ll
likely have turn it into pH after the equilibrium problem is done. Another way of saying
this is “sometimes we do an acid/base problem before the equilibrium problem; sometimes
its equilibrium problem then an acid/base problem.”
Example: If we’re given pH at equilibrium, we can use [H3O+] = 10-pH to solve for
the equilibrium concentration of H3O+, which is x in most problems. Since we know x, we
can plug that same value into all places in the ICE table, and then solve for Ka using K
expression.
Example 2: We’re given Ka. Using our ICE skills, we solve for x. “x” is now our
equilibrium concentration of [H3O+], so we can now find pH using it.
• We need to combine what we know about acids/bases with what we know about
equilibrium. Small K means reactant-favored, so we expect small K’s for weak acids and
weak bases. We also know that strong acids dissociate fully; therefore we expect  zero
reactants to remain when a SA dissociates in water. This is why there aren’t number values
for Ka for SA’s: The K expression products/reactants approaches infinity.
• Here’s the good news: Since we know that weak acids have small K’s, and we
know that they dissociate very little (<5%). Therefore in our ICE table, we can ASSUME
that x is small compared to the initial concentration of the weak acid, and make the
assumption that [Acid]equilb  [Acid]initial. That is, “[Acid]initial – x”  [Acid]initial. Why Is this
good news? This allows us to avoid the quadratic formula; a beautiful thing.
To Be Continued…