Chapter 16 Text

16.6 Weak Acids 635
[TABLE 16.3 Acid-Dissociation Constants of Some Common Polyprotic Acids
Name
Ascorbic
Carbonic
Citric
Oxalic
Phosphoric
Sulfurous
Sulfuric
Tartaric
Formula
H2C6H606
H2C03
HsQ^sO?
H2C2O4
H3P04
H2S03
H2SO4
H2C4H4O6
•K«i
8.0 x 10~5
4.3 x 10~7
7.4 X 10~4
5.9 X 10~2
7.5 X 10~3
1.7 X 10~2
Large
1.0 X 10~3
Ka2
1.6 X 10"12
5.6 X 10"11
1.7 X 10~5
6.4 X 10~5
6.2 X 10~8
6.4 x 10~8
1.2 X 10~2
4.6 X 10"5
-K«3
4.0 X 10"7
4.2 x 10~13
SAMPLE EXERCISE 16.13
The solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037M. The common practice is to assume that all of the dissolved
CO2 is in the form of carbonic acid (H2CO3), which is produced by reaction between the CO2 and H2O:
What is the pH of a 0.0037 M solution of H2CO3?
CO2(«^) + H2O(I) v
H2CO3(ag)
Solution
Analyze: We are asked to determine the pH of a 0.0037 M solution of a polyprotic acid.
Plan: H2CO3 is a diprotic acid; the two acid-dissociation constants, Kal and Ka2 (Table 16.3), differ by more than a factor of 103. Consequently,
the pH can be determined by considering only Kfll/ thereby treating the acid as if it were a monoprotic acid.
Solve: Proceeding as in Sample Exercises
16.11 and 16.12, we can write the
equilibrium reaction and equilibrium
concentrations as follows:
Initial
Change
Equilibrium
0.0037 M
-xM
(0.0037 - x) M
H>y) 4- HC03-(aj)
0
+xM
xM
0
+xM
xM
The equilibrium-constant expression
is as follows:
Solving this equation using an equation-solving
calculator, we get
Alternatively, because Knl is small, we
can make the simplifying approximation
that x is small, so that
= [H+][HC03-] = (*)(*)
[H2CO3] 0.0037 - x
x = 4.0 x 10~5 M
0.0037 ~ * « 0.0037
= 4.3 X 10~7
Thus,
Solving for x, we have
(*)(*) - 4.3 X 10
~7
0.0037
x2 = (0.0037) (4.3 x 10"7) = 1.6 X 10"
The small value of x indicates that our
simplifying assumption was justified.
The pH is therefore
x = [H+] - [HCO3~] - Vl.6 x 10~9 = 4.0 X
pH = -log[H+] = -log(4.0 x 10^5) = 4.40
M
Comment: If we were asked to solve
for [CO32~], we would need to use
Ka2. Let's illustrate that calculation.
Using the values of [HCO3~] and [H+]
calculated above, and setting
[CO3 ] = y, we have the following
initial and equilibrium concentration
values:
Initial
Change
Equilibrium
4.0 x 10"5 M
-yM
(4.0 X 10~5 -.y)M
4.0 X 10~5 M
+yM
(4.0 X 10"5 + y) M
+
0
+i/M
yM
Assuming that y is small compared
to 4.0 X 10~5, we have
[H+][C03 2-1 (4.0 X
~5 - 5.6 X 10~u
[HC03-] 4.0 x 10
y = 5.6 X KT11 M = [CO32~]