Transcript Acid-Base Equilibria Part III

Acid - Base Equilibria
Part III: Ka and Kb Calculations
Jespersen Chap. 17 Sec 4 & 5
Dr. C. Yau
Spring 2014
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Calculating Ka and pKa from pH
Example 17.1 p. 785
Lactic acid (HC3H5O3), which is
present in sour milk, also gives
sauerkraut its tartness. It is a
monoprotic acid.
In a 0.100 M soln of lactic acid, the pH
is 2.44 at 25oC. Calc the Ka and pKa
for lactic acid at that temperature.
This calls for an ICE table!
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Calculating Kb and pKb from % Ionization
Example 17.2 p. 786
Methylamine, CH3NH2, is a weak base and
one of several substances that give herring
brine its pungent odor.
In 0.100 M CH3NH2 only 6.4% of the base is
ionized. what are Kb and pKb of
methylamine?
What does "6.5% ionized" mean?
% ionization

# mol/L ionized
x 100
# mol/L available
6 . 4 % of 0.100 M  0 . 06 4 x 0.100 M  0.0064
M ionized
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In 0.100 M CH3NH2 only 6.4% of the base is
ionized. what are Kb and pKb of methylamine?
CH3NH2 + H2O
CH3NH3+ + OH–
I
0.100
0
0
-x
+x
+x
C
-0.0064
+0.0064 +0.0064
E (0.100-0.0064)
0.0064 0.0064
6.4 % of 0.100 M  0.064 x 0.100 M  0.0064 M ionized
T his m eans in 1 L, starting w ith 0.100 m oles C H 3 N H 2
0.0064 m oles have been con verted in t o 0.0064 m o les C H 3 N H 3
Now, simply calculate Kb and pKb.

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Nicotinic acid (niacin) is a monoprotic acid with the
formula HC6H4NO2. A solution that is 0.012 M in
nicotinic acid has a pH of 3.39 at 25 °C. What are
the acid-ionization constant, Ka, and pKa for this acid
at 25 °C? What is the degree of ionization of
nicotinic acid in this solution?
Let HNic = nicotinic acid and Nic– = anion.
HNic(aq) + H2O(l)
Nic–(aq) + H3O+(aq)
0.012
0
0
I
-x
+x
+x
C
x
x
E (0.012 –x)
What does pH = 3.39 tell us?
[H+] = ?
Ka = ?
pKa = ?
Degree of ionization = ?
Do Pract Exer 17, 18, 19 p. 787
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Calculating Equilibrium Concentrations
from Ka (or Kb) and Initial Concentrations
Almost any problem where you are given Ka
or Kb falls into one of 3 categories:
1. Only solute is weak acid
2. Only solute is weak base
3. Two solutes, one is weak acid, other is
conjugate base (buffer problem-section 17.7)
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HA + H2O
H3O+ + A-
Jespersen/Brady/Hyslop
HA + H2O
H3O+ + A-
Chemistry: The Molecular Nature of Matter, 6E
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Determining the pH Of
Aqueous Weak Acid Solutions
Previous examples (17.1 & 17.2) were ones where
you had to calculate Ka and Kb. Here you are
GIVEN Ka and Kb and have to calculate pH.
• Dominant equilibrium is Ka reaction
– Write the net ionic equation.
– Look up the Ka value for the acid.
– Set up ICE table.
– Solve for x.
– Calculate pH from the hydronium
concentration at equilibrium.
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Example 17.3 p. 790
A student planned an experiment that would
use 0.10 M propionic acid, HC3H5O2.
Calculate the value of [H+] and the pH for this
soln.
For propionic acid, Ka = 1.34x10-5
Simplification General Rule:
X is negligible if [HA] >>100xKa
For bases....... If [B] >> 100xKb
Rule is only for Ka and
Kb where coefficient of
reactant is always one.
Why 100xKa? See explanation on p. 793
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Example 17.3 p. 790
A student planned an experiment that would use 0.10 M propionic acid,
HC3H5O2. Calculate the value of [H+] and the pH for this soln.
For propionic acid, Ka = 1.34x10-5
Do Pract Exer 20, 21 p. 791
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Now let’s see how this works with a weak base.
Example 17.4 p. 791
A solution of hydrazine, N2H4 has a conc of
0.25 M. what is the pH of the soln, and
what is the percentage ionization of the
hydrazine? Hydrazine has Kb = 1.3x10-6
Check: 0.25  100 x1.3x10-6
What does this mean??
To save lecture time, you should finish this
yourself before checking your answer.
Ans. 0.23% ionized.
Do Pract Exer 22, 23, 24 p. 793
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Simplifications in Acid-Base Equilibrium Calc’s
Ex.Morphine, C17H19NO3, is administered
medically to relieve pain. It is a naturally
occurring base, or alkaloid.
What is the pH of a 0.075 M solution of morphine
at 25 °C? The base-ionization constant, Kb, is
1.6x10–6 at 25 °C. Let Mor = base morphine
and HMor+ = conjugate acid.
Mor(aq) + H2O(l)  OH–(aq) + HMor+(aq)
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(cont’d)
[Mor] (M)
I
C
E
[HMor+] (M)
[OH–] (M)
0.075
–x
0
+x
~0
+x
0.075 – x
x
x

Kb 

[O H ][H M or ]
[M or]

x
2
0.075  x
= 1.6x10
6
 Assume x << 0.075, [Mor]=0.075>>100xKb
 then
0.075 - x  0.075
 Plugging in for Kb and x gives
x
2
0 . 075
 1 . 6  10
6
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(cont’d)
x
2
0 . 075
2
 1 . 6  10
x  (1.6  10
x 
6
1 . 2  10
6
)(0.075)  1.2  10
7
7
= 3.5 x 104 M
• x  [OH] = 3.5 x 104 M
– check assumption: 0.075 - 0.00035 = 0.075
• pOH = log (3.5 x 104) = 3.46
• pH = 14.00 – pOH
•
= 14.00 – 3.46 = 10.54
Make sure you re-do this problem yourself
without looking at the solution.
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