Themes > Science > Chemistry > Inorganic Chemistry > Acids and Bases > Acids and Bases Index > Equilibrium Constants for Weak acids

When a weak acid is added to water, an equilibrium is quickly reached:

HA(aq) + H2O(l) < = > A-(aq) + H3O+(aq)
More simply, this can be expressed as
HA(aq) < = > H+(aq) + A-(aq)
The equilibrium constant for a weak acid is thus
Ka = [H+][A-]/[HA]

A common way to express the strength of an acid is the pKa, which is similar in form to the pH

pKa = -log10Ka
The lower the Ka and pKa values for an acid, the stronger the acid.

Example 1: Acetic acid (CH3COOH) the acid in vinegar, has a pKa value of 4.74. What is Ka for acetic acid?

Solution 1: The pKa is the negative log of Ka, so just set up the equation.

pKa = -log10Ka
4.74 = -log10Ka
-4.74 = log10Ka
Ka = 1.8*10-5

Example 2: Acetic acid has a Ka value of 1.8*10-5. What is the pH of a 0.100 M solution of acetic acid?

Solution 2: To compute the pH, you need to find the hydrogen ion concentration. This is an equilibrium problem at heart- you need to find the concentrations of the species in an equilibrium reaction.

First, write down the reaction, then write the equilibrium constant expression for the reaction:

CH3COOH < = > H+ + CH3COO-
Ka = [H+][CH3COO-]/[CH3COOH]
Next, write out the table of initial concentrations and their change when a reaction occurs. Here, for every molecule of acetic acid that dissociates, we form one H+ ion and one CH3COO- ion. (Check the stoichiometry in the problem above.) If we represent the amount of acetic acid used up as x, then
 [CH3COOH] (M) [H+] (M) [CH3COO-] (M) Initial 0.100 0.000 0.000 Change -x +x +x Equilibrium 0.100-x x x
Now we can write out the equilibrium expression and solve for x
Ka = [H+][CH3COO-]/[CH3COOH]
1.8*10-5 = (x)(x)/(0.100-x)
This looks like we might have to use the quadratic equation. Luckily, if we consider Ka, we realize that it is very small. Thus, very little acetic acid will dissociate, and x will be very small compared to 0.100M. Thus, we can assume that 0.100-x ~ 0.100 without much loss of accuracy.
1.8*10-5 = (x)(x)/(0.100)
x = 0.00134
x is indeed only about 1% of 0.100, so this is not a bad approximation.

Now, we wanted the pH. Our equilibrium H+ concentration is just equal to x, so [H+] = 0.00134 M. Thus, the pH is

pH = -log10([H+])
pH = 2.87

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