Page 183 - Modern physical chemistry

B.2 Useful Equilibria Expressions 175

Putting these into the equilibrium expression yields

(0.0200 mol- 2x) I n
-'--------'--- = 812
(x Int2(x Int2

or
812x + 2x = 0.0200 mol
whence
x = 0.0200 mol = 2.46 x 10-5 mol.
814

Example 8.2

For the ionization of acetic acid

constant Kyc is 1.749 X 10-5 M at 25° C. Calculate the hydrogen ion concentration in a solu-
tion formed by adding 0.100 mol acetic acid and 0.0100 mol sodium acetate to enough water
to give 1 liter of solution. Assume that Qy is 1 as an approximation. Taking Qy equal to 1
leads to the equation
Kc = cCHaCOO-cH+ = 1.749 x 10-5 M.
CCHaCOOH
Let us neglect ionization of the acetic acid and hydrolysis of the acetate ion. Then

CCHaCOOH Kc = 0.100 1.749 x 10-5 M=1.75x10-4 M.
cCHaCOO- 0.0100

Example 8.3

Calculate the hydrogen ion concentration in 0.0100 M acetic acid at 25° C.
The acetic acid ionizes following the chemical equation in example 8.2. If we neglect
the contribution to CH+ from the ionization of water, we have

and
CCHaCOOH = 0.0100 M - x.
Taking Qy, equal to 1, as in example 8.2, leads to

CCH coo-cH+ x 2
Kc = --=..::::>ac::...:..::.-=- ----= 1.749 x 10- 5 M
CCHaCOOH 0.0100 M-x
whence
x 2 = 1.749 x 10- 7 -1.749 x 10- 5 x.