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pH of
Weak Acids
Calculate the pH and percent dissociation of
0.10 M CH
3
COOH
CH
3
COOH +
H
2
O
⇄
H
3
O
+
+ CH
3
COO

K
a
= 1.8 x 10
5
Start
0.10
0
0
Change
x
+x
+x
equil
0.10 x
x
x
x
10
.
0
x
2

= 1.8 x 10
5
Using the approximation that x is small compared with 0.10, we should be able to solve quickly:
x = [H
3
O
+
]
= 1.34 x 10
3
M
pH
=  log[H
3
O
+
] = log (1.34 x 10
3
) = 2.87
Percent ionization =
10
.
0
10
x
34
.
1
3

x 100 = 1.34%
Since the percent ionization is less than 5%, our approximation is considered valid.
If we
solved exactly, we would get x = 1.33 x 10
3
M, very close to our approximate solution.
Calculate the pH and percent dissociation of
0.0010 M CH
3
COOH
CH
3
COOH +
H
2
O
⇄
H
3
O
+
+ CH
3
COO

K
a
= 1.8 x 10
5
Start
0.0010
0
0
Change
x
+x
+x
equil
0.0010 x
x
x
x
0010
.
0
x
2

= 1.8 x 10
5
Using the approximation that x is small compared with 0.0010, we should be able to solve quickly:
x = [H
3
O
+
]
= 1.34 x 10
4
M
Percent ionization =
0010
.
0
10
x
34
.
1
4

x 100 = 13.4%
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View Full Document Since the percent ionization is more than 5%, our approximation is suspect.
Using successive
approximations, or solving the quadratic equation exactly, we get
x = [H
3
O
+
] =
1.26 x 10
4
pH
=  log[H
3
O
+
] = log (1.26 x 10
4
) = 3.90
Percent ionization =
0010
.
0
10
x
26
.
1
4

x 100 = 12.6%
Comparing 0.10 M CH
3
COOH and 0.0010 M CH
3
COOH
[H
3
O
+
]
pH
% ionization
0.10 M CH
3
COOH
1.34 x 10
3
2.87
1.34%
0.0010 M CH
3
COOH
1.26 x 10
4
3.90
12.6%
The more dilute acid has lower [H
3
O
+
] and consequently a higher pH, but it also has a higher percent
ionization.
Calculate the pH and percent dissociation of
0.10 M HCN
HCN + H
2
O
⇄
H
3
O
+
+ CN

K
a
= 6.2 x 10
10
0.10  x
x
x
We have set up our equilibrium calculation in one line. We start with 0.10M HCN and x reacts.
This forms x in both H
3
O
+
and CN

and leaves over 0.10x in HCN.
To save time and space, further
equilibrium calculations shown here will be done in one line.
x
10
.
0
x
2

= 6.2 x 10
10
Solving, using the usual approximation:
x = [H
3
O
+
] =
7.9 x 10
6
M
pH =  log[H
3
O
+
] = log(7.9 x 10
6
) = 5.10
Percent ionization =
10
.
0
10
x
9
.
7
6

x 100 = 0.0079 %
Compare the answers for 0.10 M CH
3
COOH and 0.10 M HCN.
These are both weak acids, both HCN is a weaker acid (smaller K
a
) than CH
3
COOH.
The HCN
solution has a lower [H
3
O
+
], a higher pH and a smaller percent ionization—all consistent with a smaller
value of K
a
.
A 0.020 M solution of acid HX is measured
to have pH = 2.90.
Calculate the K
a
and percent
ionization of HX.
In the previous examples, we were given the concentration of a weak acid and its K
a
value and
were asked to calculate the pH and percent ionization.
In this example, we are given the concentration and pH of a weak acid, and asked to calculate its
K
a
and percent ionization.
Let’s first set up a general approach to this type of problem.
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This note was uploaded on 04/04/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.
 Spring '08
 siegal

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