### Calculate the Ka of a weak acid given the pH and molarity

This is a favorite problem for teachers to test

Problem #1: A 0.120 M solution of a generic weak acid (HA) has a pH of 3.26. Determine the Ka.

Notice that a generic weak acid is used, symbolized by the formula HA. No one cares what the specific acid is because the technique to be explained works for all weak acids. What happens is that some teachers will use the name of a specific weak acid while others go the generic route.

Solution

1) Write the dissociation equation for the acid:

HA ⇌ H+ + A¯

2) Write the equilibrium expression:

Ka = ( [H+] [A¯] ) / [HA]

3) Our task now is to determine the three concentrations on the right-hand side of the equilibrium expression since the Ka is our unknown.

a) We will use the pH to calculate the [H+]. We know pH = -log [H+], therefore [H+] = 10¯pH

[H+] = 10¯3.26 = 5.4954 x 10¯4 M

I've kept a couple guard digits; I'll round off the final answer to the proper number of significant figures.

b) From the dissociation equation, we know there is a 1:1 molar ratio between [H+] and [A¯]. Therefore:

[A¯] = 5.4954 x 10¯4 M

c) the final value, [HA] is given in the problem. In the example being discussed, 0.120 M is the value we want. Some teachers will use 0.120, while others would say to subtract the 5.4954 x 10¯4 value from 0.120 first. Let's do both.
c1) Ka = [(5.4954 x 10¯4) (5.4954 x 10¯4)] / 0.120

Ka = 2.52 x 10¯6

c2) Ka = [(5.4954 x 10¯4) (5.4954 x 10¯4)] / (0.120 minus 5.4954 x 10¯4)

Ka = 2.53 x 10¯6

In reality, it makes very little difference if we use the unmodified concentration of the acid (the 0.120 value) or if we do the subtration. In the end, you do what your teacher recommends. So, ask your teacher if you're not sure.

In this example, I'll use a real acid and numbers that lead to the actual Ka for the acid.

Before that, a comment: one reason teachers might tend to avoid real substances in this type of question is that you can just look up the answers on the Internet. For example, I used this page to get the Ka for the following problem.

Problem #2: A 0.128 M solution of uric acid (HC5H3N4O3) has a pH of 2.39. Calculate the Ka of uric acid.

Another reason for using the generic acid formula of HA is that this avoids the need to constantly write a somewhat complex formula for the anion portion of the weak acid. In fact, as I do this problem, I will write Ur¯ for the anion portion of uric acid (the C5H3N4O3¯).

Solution

a) [H+] = 10¯pH = 10¯2.39 = 4.0738 x 10¯3 M

b) [Ur¯] = [H+] = 4.0738 x 10¯3 M

c) I will use 0.128 M

Ka = [(4.0738 x 10¯3) (4.0738 x 10¯3)] / 0.128 = 1.30 x 10¯4

Problem #3: HC9H7O4 (MW = 180. g/mol) is prepared by dissolving 3.60 g into a 1.00 L solution. The pH of this solution was determined to be 2.60. What is the Ka?

Solution

3.60 g / 180. g/mol = 0.0200 mol

0.0200 mol / 1.00 L = 0.0200 mol/L

2) Then apply the usual technique:

Ka = [(2.5 x 10¯3) (2.5 x 10¯3)] / 0.0200 = 3.2 x 10¯4

Problem #4: A student prepares a 0.45 M solution of a monoprotic weak acid and determines the pH to be 3.68. What is the Ka of this weak acid? The answer.

Comment: the type of problem discussed in this tutorial covers only monoprotic acids. The techniques for di- and triprotic are more complex and will not be covered.

More comment: if you don't know the formula of the monoprotic weak acid, that's OK. Simply use HA as the formula. It does not matter what the anion portion is, it only matters that the acid is weak and monoprotic.