Protolysis of sulfuric acid

Question

Hi!

I have the following question: 0.05 moles of sulfuric acid was mixed with water to the volume of 1 cubic decimetres. What is the pH-value of the solution?

My answer: First I wrote the reaction: H2SO4 + H2O -> HSO4- + H3O+

So reaction shows that 0.05 moles of sulfuric acid gives 0.05 moles of H3O+.

The concentration of it is 0.05/1 = 0.05 M

I recalculate it to pH by -lg0.05 = 1.3

Answer= 1.3

But my compendium answers differently. They say that 1 mole of H2SO4 gives 2 moles of HSO4-

But the further reaction of HSO4- is not complete, because HSO4- is a weak acid: HSO4- + H20 <-> SO42- + H3O+. Surely this reaction would work, but not all 0.05 moles of HSO4- from the first reaction should react here since we have a weak acid and I don't know how the book thinks that both reaction one and two reacts competely and gives us a total of 0.1 moles of H3O+

Thankful for help!

Answer 1

You are correct in stating that the 2nd reaction (second ionization) isn't complete because HSO4- is a weak acid. In this case, you'd have to use the Ka value for HSO4- ==> H+ + SO42-. Of course, this will add relatively little to the [H+] relative to the first ionization, but if pH = 1.3 isn't correct, perhaps they want to to include the 2nd ionization as well.

The value I have for Ka HSO₄^- ==> H⁺ + SO₄^2- is 1.2x10^-2

From the first ionization you will have 0.05 M H+ and 0.05 M HSO4-. Then you have...

HSO₄^- ==> H⁺ + SO₄^2- ...... Ka = 1.2x10^-2

0.05..........0..........0......Initial

-x...........+x..........+x.....Change

0.05-x......x...........x.......Equilibrium

Ka = [H⁺][SO₄^2-]/[HSO₄^-] = 1.2x10^-2

1.2x10^-2 = (x)(x) / 0.05-x

Solve for x, and add this value to 0.05 M from the first ionization to get the TOTAL H⁺ concentration. Take the negative log to get the final pH.

I hope this helps.