Question

Estimate the pH of a 50 mM solution of acetic acid made up in 0.1 M K₂SO_4?

Answer 1

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 K_a 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