# Buffers & Titrations

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```					Buffers & Titrations
Chapter 18

1
Buffers
• A soln that resists change in pH when
strong acid or strong base is added
• Or
• Strong acids and strong bases don’t make
buffers!
• So how does it work?
2
How buffers work

• Acetic acid/acetate buffer system:
C2H3O2H(aq) + H2O(l)  C2H3O2-(aq) + H3O+(l)
C2H3O2H(aq) + OH-(aq)  C2H3O2-(aq) + H2O(l)
H3O+(aq) + C2H3O2-(aq)  H2O(l) +C2H3O2H(aq)

3
Common Ion Effect: CIE
• The ionization of an acid or a base is limited by
the presence of its conjugate base or acid.
HAc(aq) + H2O(l)  Ac-(aq) + H3O+(aq)
•   Acetate ion is added in form of NaAc
– Which way will this shift the rxn?
– Would you expect a greater or lesser acidity if the CIE
was lacking?
• Let’s look at the next problem

4
Calculating pH of a buffer soln

• You have an acetic acid/acetate buffer
with a 0.700 M conc of acetic acid and a
0.600 M conc of acetate ion. What’s the
pH of the buffer? (Ka = 1.8 x 10-5)

5
Solution
HAc(aq) + H 2 O(l)  Ac-(aq) + H3O + (aq)
I 0.700M --               0.600   0
C -x        --             +x    +x
E 0.700 - x --            0.600+x x
[0.600+x][x]
K a = 1.8 x 10-5 
[0.700  x]
CIE: addition or subtraction of "x" from original concentrations very small
[0.600][x]

[0.700]
x=2.110-5 M
pH=-log(2.110-5 )=4.68

6
Let’s work on this

• Consider 100.0 mL of a buffer solution
that is 1.00M in HAc and 1.00M in NaAc.
What is the pH after addition of 25.0 mL
of 1.00M NaOH?

7
Solution
HAc(aq) + H 2 O(l)  Ac-(aq) + H3O + (aq)
I 1.00M --            1.00      0
C -x       --         +x       +x
E 1.00 - x --      1.00 + x x
[1.00+x][x] [1.00][x]
K a = 1.8 x 10-5               
[1.00  x]    [1.00]
x=1.8 x 10-5 M
Thus, [HAc]  1.00M  1.8 x 10-5 M =1.00M
And [Ac- ]=1.00M +1.8 x 10-5 M=1.00M
(By the way, the pH=4.74)
HAc(aq)  OH -(aq)  Ac-(aq) + H 2 O(l)
mol                            1
acid: 1.00         0.1000L acid                     0.800M
L acid                    0.1250L total
mol                  1
base: 1.00       0.0250L              0.200M
L               0.1250L
mol                    1
acetate ion: 1.00      0.1000L                 0.800M
L                 0.1250L
I 0.800M 0.200 0.800              --
C -0.200 -0.200 +0.200 --
E 0.600       0        1.000
[1.000][x]
So K a = 1.8 x 10-5             , x = [H 3O + ]  1.1 x 10-5
[0.600]
pH = -log(1.1 x 10-5 )  4.96
8
Henderson-Hasselbalch equation

• Useful for previous problem
– Let’s take a look
[H 3O + ][A - ]
Ka =
[HA]
[HA]
Re-arrange: [H 3O ]= -  K a
+

[A ]
Take the negative log
[A - ]
pH = pKa + log(         )
[HA]
9
Titrations
• Used to determine quantity of acid or base in
unknown (analyte)

10
• When sample is neutralized (H3O+ = OH-)
– Equivalence point
• Determined by use of pH indicator

11
Different titration types

• Strong acid-strong base: equivalence pt =
7.0 (contains neutral salt)

12
Different titration types

• Weak acid-strong base: equivalence pt is
greater than 7 (basic salt)
• pH @ half-equivalence pt (halfway pt) =
pKa

13
Problem

25.0 mL of 0.100 M NaOH is then added.
What is the pH of the resulting solution?

14
Solution
HAc(aq)  OH - (aq)  Ac- (aq) + H 2O (l)
mol                  1
acid: 0.100      0.0500L           0.0667M
L              0.0750L
mol                  1
base: 0.100      0.0250L           0.0333M
L              0.0750L
I 0.0667 0.0333      0        --
C -0.0333 -0.0333 +0.0333 --
E 0.0334       0         0.0333     --
[A - ]                           [0.0333]
pH = pKa + log(         )= -log(1.8 x 10 )  log(
-5
)  4.74
[HA]                              [0.0334]
15

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