Unit #3: Acids and Bases
Andrea Houg
Emily Lichko
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Definitions of Acids and
Bases
•Arrhenius Acid =
releases an H+
•Arrehenius Base =
releases an OH•Brönsted-Lowry Acid =
proton donor
•Brönsted-Lowry Base =
proton acceptor
•Lewis Acid = electron
acceptor
•Lewis Base = electron
donor
Lewis
Brönsted - Lowry
Arrhenius
Dissociation
•The strength of an acid or a base depends on
how much it dissociates in water
•Kw = Ka•Kb= 1E-14= [H+][OH-]
•EX: Acetic Acid’s Ka = 1.8 E-5. Kb = Kw/Ka = (1E14)/(1.8E-5) = 5.6 E -10
•Strong Acids dissociate 100%
•Weak Acids typically dissociate less than 5%
H 

• % dissociation = HA  100%
•The stronger the acid or base, the weaker its
conjugate
Strong Acids
Strong Bases
eq
0
HCl
Alkaline metals with OH-
HBr
Ca(OH)2
HI
Sr(OH)2
HNO3
Ba(OH)2
H2SO4
HClO4
http://myphlip.pearsoncmg.com/phproducts/student/
ab2page.cfm?vbcid=9029&vid=10007#oa231470
Neutralization
•Combining an acid with a base produces water and a salt
•For strong acids and bases:
•NH+CaVa = NOH-CbVb
•EX: How much 0.3000M NaOH is needed to neutralize 23.98mL of
0.8000M HNO3?
Answer:
(1)(0.8000M)(23.98mL) = (1)(0.3000M)(x)
X= (0.8000M)(23.98mL)/(0.3000M) =63.95mL
HCl
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679899/2/istockphoto_2679899_strong_man.jpg
Calculating pH
14-pH = pOH
14-pOH= pH 
[H+]
pOH
 -log [OH-] =pOH
-log [H+] =pH
pH
1 E-14 / [H+] = [OH-]
1 E-14 / [OH-] = [H+]
[OH-]
Henderson-Hasselbalch equation

base
pH  pK a  log
acid 
pOH
http://www.nku.edu/~russellk/TV/simon.gif

acid 
 pKb  log
base
Titrations
Strong acid / strong base
Weak acid / strong base
Polyprotic acid / strong base
Back Titration: when base is not
very soluble,
dissolve in excess acid and titrate
back with base until excess acid is
used up (endpoint)
http://www.sparknotes.com/chemistry/acidsbases/titrations/section1.html
ICE chart
EX: pH of original 0.100 weak acid solution, Ka given
HA +
I
0.100M
C
-x
E
0.100M

H2 O
H3O+
0

H O A 
K

+
A0
+x
+x
x
x

3
HA
Ka = x2/0.100 x= [H3O+]
-log[H3O+] =pH
Use to find new pH, at any point in titration, by finding initial amount
of A- from base added.
EX: (5mL)(0.250M NaOH) = 1.25mmol
Amphiprotic
Amphiprotic: Can act as either an acid or a base
If Ka > Kb then it acts as an acid at the equivalence point
If Kb > Ka then it acts as a base at the equivalence point
(for diprotic acids, always compare Ka2 to Kb1 since the will follow
one of these equilibria at the equivalence point
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Buffers
•Buffer
-weak acid base and its conjugate
-minimizes change in pH when acid or base is added
•Buffer capacity -the quantity of acid or base that a buffer can
neutralize before pH begins to change significantly
•Common ion effect -a slightly soluble salt’s solubility decreases
with the addition of a solute with a common ion
EX1:
Make a buffer with pH 3.50
Choose an acid with a Ka close to 1E -3.5, like HOCN
HOCN + H2O ↔ H3O+ + OCNKa = [H3O+][OCN-] / [HOCN]
With pH 3.50, [H3O+] = 3.16 E -4
3.5 E -4 = [3.16 E-4][OCN-] / [HOCN]
[HOCN] = 0.9035[OCN-]
This is the ratio. The amounts depend on desired buffer capacity….
Buffers
EX 2: Make it so the pH of the buffer changes by 0.50 pH unit when 0.50
mol acid is added.
Answer:
pH is allowed to change to 3.00, so [H3O+] will change to 1 E-3. [OCN-] will
decrease and [HOCN] will increase.
3.5 E -4 = ([1 E-3]•([OCN-]-0.50)) /(0.9035[OCN-]+0.5)
1 E-3[OCN-] - 5 E-4 = 3.16 E-4[OCN-] + 1.75 E-4
6.837722 E-4[OCN-] =6.75 E-4
[OCN-] = 0.987 M
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