In: Chemistry
Estimate the pH of a 50 mM solution of acetic acid made up in 0.1 M K2SO4?
Determining the pH of a weak acid solution is more complicated than for a strong acid. This is because a weak acid does not completely dissociate.
CH3COOH 
H+ + CH3COO-
To simplify matters, it is helpful to construct a grid:
| Reaction | CH3COOH | H+ | CH3COO- | 
|---|---|---|---|
| Initial | 
 0.05 M  | 
 0 M  | 
 0 M  | 
| Change | 
 (-x) M  | 
 + x M  | 
 +x M  | 
| Equilibrium | 
 (0.05-x) M  | 
 x M  | 
 x M  | 
To determine the [H+], we must employ the Ka of acetic acid, which is 1.8 x 10-5.
| 
 Ka =  | 
 [H+][CH3COO-]/[CH3COOH]  | 
= 1.8 x 10-5 | 
Substituting the equilibrium values in the table above, we get:
| 
 1.8 x 10 -5 =  | 
 (x M)(x M)/(0.05 - x) M  | 
Since acetic acid is a weak acid, we can assume that very little of it will dissociate. This simplifies the above equation to:
| 
 1.8 x 10 -5 =  | 
 (x M)(x M)/0.05 M  | 
This can be easily solved:
9 x 10-7 M = x2 M
x = 9.5 x 10-4 M
x = 0.00095 M
Now that we have the [H+], we can solve for pH:
pH = -log[0.00095]
pH = 3