**Problem:** What is the pH of a 0.025 M H_{2}SO_{4} solution?

**Discussion:** Calculating the pH of H_{2}SO_{4} is a complicated topic. The complication lies in the differing behavior of the two dissociation steps H_{2}SO_{4} 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 K_{a} values:

H _{2}SO_{4}---> H^{+}+ HSO_{4}¯K _{a1}= largeHSO _{4}¯ ---> H^{+}+ SO_{4}^{2}¯K _{a2}= 1.2 x 10¯^{2}

By convention, the K_{a1} 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 K_{a1} 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 K_{a2} 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[HSO

_{4}¯] = 0.025 MThis 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):

[HSO _{4}¯]⇌ [H ^{+}][SO _{4}^{2}¯]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 ^{+}] [SO_{4}^{2}¯]K _{a2}=–––––––––– [HSO _{4}¯]

(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 + x

^{2}In standard form, this is:

x

^{2}+ 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 MpH = −log 0.0318426 = 1.49699

The most appropriate pH to report would be 1.50

**Comment about the 5% rule:**

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!!

**Comment about an alternate solution:**

In a big over-simplification of the problem, the writer of the problem will ask you to consider that BOTH dissociations are strong (this is, 100% dissociation). So, let's do that.

0.025 M H_{2}SO_{4}is 0.050 M in H^{+}pH = −log 0.050 = 1.30

This approach is generally used early in an acid base unit, before the concept of K_{a} is introduced. It is possible that your teacher may address sulfuring acid in this manner and never discuss the more complex situation detailed above.

Here are several calculations like the one I did above. You will notice that a number of them stop at the answer which assumes 100% dissociation in both hydrogens.