Acid/Base Definition
Acid/Base Reactions
• some covalent compounds have weakly bound
H atoms and can lose them to water (acids)
• some compounds produce OH- in water
solutions when they dissolve (bases)
• Arrhenius definition:
– anything that produces H3O+ in solution (or H+)
– anything that produces OH- in solution
• Bronsted-Lowry definition:
– an acid is a proton (H+) donor (HCN, HCl, etc.)
– a base is a proton (H+) acceptor (CN-, OH-, etc.)
• acid/base reaction are very important to
biochemistry and environmental chemistry
Terms/Items you Need to Know
you need to memorize these !!
•
•
•
Major vs. Minor Species
• For a strong acid, a large Ka means no reactant
is left, so products are the major species.
HNO3 (aq) + H2O (l) 

x NO3- (aq) + H3O+ (aq)
• For a weak acid, a small Ka means only a small
amount of product forms, so reactants are the
major species and products are the minor species
HNO2 (aq) + H2O (l)  NO2- (aq) + H3O+ (aq)
Ka = 4.4710-4
Monoprotic, diprotic, polyprotic acids
Amphoteric
The difference between a strong
acid/base and a weak acid/base
The pH Scale
• What’s the pH of neutral water?
=  log[10
pH<7 : solution is acidic
pH>7 : solution is basic
Same thing for bases….
• Define pOH = log[OH]
104  less
Equilibrium Calculations
• Since Kw = [H3O+][OH] = 10
-log[H3O+] - log[OH] = - log(10)
or
pH + pOH = 14
Equilibrium Calculations
• Calculate the pH of a solution of 2.5102 M
HClO4. (perchloric acid)
• Calculate the pH of a solution of
2.5102 M HClO? (hypochlorous acid)
• Step #1: identify the acid
If strong acid = complete dissociation
HClO4 (aq) 
H+ (aq) + ClO4 (aq)
2
2.510 M
0
0
init
2
2
0
2.510 2.510
equil
• Step #1: identify the acid
if weak acid = look up Ka = 3.5  10-8
HClO (aq)  H+ (aq) + ClO- (aq)
2.510-2 M
0
0
init
-x
x
x
change
-2
(2.510 -x)
x
x
equil
• Step #2: find x, then pH = - log [H+]
- log (3.010-5) = 4.53
• Step #2: pH =  log [H+]
 log (2.5102) = 1.60
Weak Base Example
• Calculate the pH of a solution of 2.5102
M trimethylamine (CH3)3N ?
•Step #1: identify it as a weak base (pKb=4.19)
(CH3)3N(aq) H2O  OH(aq)+(CH3)3NH+(aq)
2.5102 M
0
0
init
x
+x
+x change
(2.5102x)
x
x
equil
• Step #2: find x, pOH =  log [OH]
 log (1.3103) = 2.89
• Step #3: 14  pOH = pH = 11.11
Buffers
• Buffering demo
Watch pH of water + acid/base
Watch pH of buffer + acid/base
Definition:
A buffer is usually a mixture of conjugate
acid-base pairs in solution. A buffer resists
strong changes in pH.
How do you make a buffer?
add a weak acid and its conjugate base to a
solution
A weak acid only partially breaks up:
CH3COOH + H2O  CH3COO- + H3O+
A weak base only pulls some H off H2O:
CH3COO- + H2O  CH3COOH + OH-
How do you make a buffer?
add some NaOH to a weak acid
CH3COOH + OH-  CH3COO- + H2O
don’t add so much that you react all of the weak acid
get as close to 50/50 mixture as possible
add some HCl to a weak base
CH3COO- + HCl  CH3COOH + Cldon’t add so much that you react all of the weak base
You end up with mostly this in solution
get as close to 50/50 mixture as possible
Buffers
 Buffer capacity refers to how much of the
buffer is there and how much acid or base it
can absorb before it no longer works
 The more buffer you add to the solution, the
higher the buffer capacity
 I could add:
 1 mole of HOAc and 1 mole of NaOAc
 OR 10 mol of HOAc and 10 mol of NaOAc
 Which has the largest buffer capacity?
 10 mol of HOAc and 10 mol of NaOAc ( 10 !)
Acid/Base Titrations
• titration : a way of determining the
concentration of one solution by using another
solution of known concentration
• The molar mass of the chemical can be
determined from the equivalence point
• The identity of the acid/base can be determined
from the midpoint
tries to maintain this pH
usually ± 1 pH unit
Terms
• equivalence point (or stoichiometric point) 
the calculated point when enough titrant has
been added to react with all of the unknown
• physical change = color, turbidity,
temperature, conductivity, voltage
• mid-point  the point in a weak acid titration
where pH = pKa
weak acid
•
•
•
•
Buffer region
Stoichiometric (or
equivalence) point
mid-point, pH = pKa
weak acid & weak
base, so buffer
1. Starting pH
2. Buffer region calc
3. Equivalence point
4. Post-equivalence point
• see AcidBaseTitrations worksheet
of OH added
What type of Keq problem is it?
Well, what is in the solution?
weak acid,
so ICE
Regions in a Titration
weak base,
so ICE
What is being titrated by what?
starting with
acid
or
base?
Are you titrating
with an
aan
weak
acidor
orstrong
base? base?
base
weak acid
strong base, so
pH=14-pOH
acid
Dissolved Ionic Compounds
• some ionic compounds do not dissolve in
water (at least not very much)
PbSO4 (s)  Pb2+(aq) + SO42-(aq)
Reaction Types - Precipitation
• when solutions of ions mix, sometimes an
insoluble salt forms
Pb(NO3)2 (aq) + 2 NaI (aq)  PbI2(s) + 2 NaNO3(aq)
soluble
soluble
insoluble
• the net ionic reaction is:
Pb2+(aq) + 2 I- (aq)  PbI2(s)
• spectator ions: NO3- and Na+
soluble
Barium Sulfate
• Ba2+ is a mildly toxic heavy metal when it is soluble in
water
Ksp Example
What mass of AgCl will dissolve in a 500 mL sol’n
of 5.010-4 M NaCl?
AgCl(s)  Ag+(aq) + Cl-(aq)
– linked to elevated blood
pressure
– ingestion can cause nausea,
vomiting, diarrhea, and
crampy abdominal pain
within minutes of consuming
the meal
initial(M)
change(M)
eq. (M)
0
x
x
pKsp=9.74
5.010-4
+x
5.010-4 + x
10-9.74 = 1.82 10-10 = (x)(5.010-4 + x)
assume: x<< 5.010-4 
x = 1.82 10-10 / 5.010-4 = 3.6410-7M
• BaSO4 is so insoluble
(pKsp = 9.96) that it is
ingested as a contrast agent
in X-rays and CAT scans
find the mass of AgCl: (3.6410-7M)(0.5L)(143g/mol)
= 2.6 10-5g
Olmsted and Williams 3rd ed pg. 161
Another Ksp problem
Another Ksp problem (cont.)
Mix 500 mL of 0.200 M magnesium chloride
with 200 mL of 0.400 M NaOH, what mass
of precipitate Mg(OH)2 forms?
(0.500L)(0.200 mol/L) = 0.100 mol MgCl2
(0.200L)(0.400 mol/L) = 0.0800 mol NaOH
0.500 L + 0.200 L = 0.700 L
0.100/0.700 = 0.142 M MgCl2
0.0800/0.700 = 0.114 M NaOH
What is the reaction of interest?
Mg2+ Cl Na+ OH
Mg(OH)2(s)  Mg2+ + 2OH
Another Ksp problem (cont.)
Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq) pKsp=11.25
I(M)
0
0.143
0.114 Ksp says?
NI(M) 0.0570
0.0859
0
C(M)
-x
+x
+2x
E(M) 0.0570-x
0.0859+x
2x
Ksp = 10-11.25 =(2x)2(0.0859+x)
assume: x<<0.0859check
-11.25
x=[(10
)/(0.0859*4)]½=4.0410-6 M
[Mg2+] = 0.0859+x = 0.086 M
[OH-] = 2x
= 8.1E-6 M
g Mg(OH)2 = (0.057M)(.7L)(58.3g/mol)=2.3 g