### Calculate the pH of a solution of sulfuric acid

Problem: What is the pH of a 0.025 M H2SO4 solution?

Discussion: Calculating the pH of H2SO4 is a complicated topic. The complication lies in the differing behavior of the two dissociation steps H2SO4 can undergo. What follows is a detailed solution to the problem given above.

Solution:

1) Here are the two dissociation reactions for sulfuric acid, along with their associated Ka values:

 H2SO4 ---> H+ + HSO4¯ Ka1 = large HSO4¯ ---> H+ + SO42¯ Ka2 = 1.2 x 10¯2

By convention, the Ka1 value is simply given as large (sometimes the word strong is used in place of large). This means that, in its first step, sulfuric acid is 100% dissociated. I have seen the actual Ka1 value be given as 1100, but I have never tried to track down if this is actually correct. No matter. Large means 100% dissociation.

The Ka2 value, being less than 1, indicates that sulfuric acid only partially dissociates in its second step.

We must calculate the contribution of hydrogen ion to the solution from both steps. We will use a different technique for each of the two steps.

2) The first dissociation step goes 100% to the right, resulting in this:

[H+] = 0.025 M

[HSO4¯] = 0.025 M

This is the behavior of a strong acid, said behavior being 100% dissociation.

3) For the second dissociation step, we use an ICE chart (I will not write the M for molarity):

 [HSO4¯] ⇌ [H+] [SO42¯] Initial 0.025 0.025 0 Change −x x x Equilibrium 0.025 − x 0.025 + x x

4) Write the equilibrium expression and substitute values ino it:

 [H+] [SO42¯] Ka2 = –––––––––– [HSO4¯]

 (x) (0.025 + x) 1.2 x 10¯2 = ––––––––––– (0.025 − x)

5) The proper technique now is to use the quadratic equation. Often, the 'subtract x' (and in this case, an 'add x') portion can be ignored. However, in this problem, that results in an answer that fails the 5% rule. I will discuss this failure after I finish with the solution which employs the quadratic. Here we go:

(0.012) (0.025 − x) = (x) (0.025 + x)

0.0003 − 0.012x = 0.025x + x2

In standard form, this is:

x2 + 0.037x − 0.003 = 0

6) Using an online quadratic equation solver, we determine that x can be −0.04384 or 0.0068426. Since a negative concentration is impossible, we select the 0.0068426 value to use.

[H+] = 0.025 + 0.0068426 = 0.0318426 M

pH = −log 0.0318426 = 1.49699

The most appropriate pH to report would be 1.50

1) After dropping the '− x' and the '+ x", we have this:

 (x) (0.025) 1.2 x 10¯2 = ––––––––– 0.025

2) Which results in :

x = 0.012

3) Carrying out the 5% rule calculation, we have this:

(0.012 / 0.025) *100 = 48%

Well, that's a big, ole oopsie right there!!