PH sulphuric acid??

Question

What us the PH of sulphuric acid in the following concentrations:
- 0.05Molar
- 0.025 Molar
- 0.25 Molar
- 0.5 Molar
Thanks alot fot your help!!!

Answer 1

Pay no attention to the guy who says there are 2 moles of H+ per mole of H2SO4. There is 1 mole of H+ and 1 mole of HSO4-, then there is a smaller contribution from partial dissociation of HSO4- into H+ and SO4--. To a first approximation, the answers would be 1.3, 1.6, 0.6, and 0.3. If the second were fully dissociated, they would be 1.0, 1.3, 0.3, and 0.0, respectively. The actual answers fall somewhere in between, and from the best I can seem to find on the web, I believe they are just about halfway between, i.e, 1.15, 1.45, 0.45, and 0.15.

Don't take these as the perfectly correct answers, but if your answers don't at least come close to these, you should rethink your calculations. I haven't done any chemistry since 1981, and my specialty was organic synthesis, so I'm rusty in general and physical chemistry.

Answer 2

In terms of an exact answer, this may not be so easy to figure out. Sulfuric acid is a dibasic acid, and the second equilibrium HSO4-> SO4= + H+ has a Ka of about 0.01. Thus for the given H2SO4 concentrations, partial dissocaiation may have to be taken into account, particularly at the lower molarities of your question.

To find the pH, you start with the above equation.
HSO4- initially = molarity of added H2SO4; the first H+ is totally dissocated, and little if any H2SO4 is left.
If x moles/liter dissociates, at equilibrium,
[HSO4-] = [(H2SO4)o - x]
[SO4=] = x
[H+] = [(H2SO4)o +x]
and K2 = [H+][SO4]/[HSO4-]
so if (H2SO4)o= .25 M
0.01 = (0.25+x)(x)/(0.25-x)
x= 0.01 appx, and [H+]=0.26 M.....
So pH = 0.6 appx

Answer 3

For each mole of H2SO4, there are 2 moles of H+ ions.

so double your concentrations to get the amount of H+ ions.

pH is the negative log of the concentration of the H+ ions.

Answer 4

it will range from 4-1 i guess. I don't know the exact pH levels though.

Answer 5

indeed it does react!!!