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