[kissoftalons]

Hi. I'm doing a package for AP chem and I'm having some problems with calculating pH. There are 3 questions. Step by step answer would be GREATLY appreciated!!

1. Calculate the pH of 0.025 mol/L of HCL (aq)
2. Give the Hydronium ion concentration in a solution with a pH of 3.
3. What is the concentration of sulfate ions in a 2 mol/L solution of Aluminum Sulfate.

If anyone has any good sites to help I'm kind of lost with this section. Anything to clearify would be great!

Thanks!

[Borek]

In general:

http://www.chembuddy.com/?left=pH-calculation&right=toc

For questions 1 and 2 check equation 7.6:
http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base#7.6
and example question:
http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-strong-acid-q1

3^rd question has nothing to do with pH calculation (well, to be precise - it can have something to do, but I doubt as the first two questions are pretty easy, while the last is a real hard one).

[Blueshawk]

3. What is the concentration of sulfate ions in a 2 mol/L solution of Aluminum Sulfate.

assuming that (Al)₂(SO₄)₃ completely dissociates.

(Al)₂(SO₄)₃ -> 2Al⁺³ + 3(SO₄^-2)

so assume you have 1 liter solution, so you have 2 mols of aluminum sulfate

So by the reaction equation..for every 2 mols of starting material, you get 2 mols of Al ion and 3 mols of sulfate ion.  

From there you can find what you are looking for.

What this has to do with pH i dont know...but maybe something...my guess is the solution is acidic.

[kissoftalons]

so assume you have 1 liter solution, so you have 2 mols of aluminum sulfate

So by the reaction equation..for every 2 mols of starting material, you get 2 mols of Al ion and 3 mols of sulfate ion.  

Do you mean for every ONE mol of starting material you get 2 mols of Al and 3 mols of sulfate....and since you have 2 mols of aluminum sulfate..you'd get 4 mols of Al and 6 mols of sulfate?

And since concentration is mols/L ...wouldn't the concentration of sulfate ions be 6 mols/L..? I don't think this seems right.

[Blueshawk]

Do you mean for every ONE mol of starting material you get 2 mols of Al and 3 mols of sulfate....and since you have 2 mols of aluminum sulfate..you'd get 4 mols of Al and 6 mols of sulfate?

And since concentration is mols/L ...wouldn't the concentration of sulfate ions be 6 mols/L..? I don't think this seems right.

I am very sorry for a typo.  for every one mol of starting material you get 2 mols Al and 3 mols of sulfate.  Again I am sorry if I confused you.

[Blueshawk]

When I am calculating the [] of sulfate, I am assuming that aluminum sulfate completely dissociates.

If it does.....by the reaction equation, one mol Al sulfate = 2 mol Al and 3 mol sulfate.

So for 2 mols Al sulfate, yes 6 mols of sulfate ion is produced.

If it does not completely dissociate, then less then 6 mols of sulfate is produced.

[kissoftalons]

So it is possible that 6 mols of Sulfate ions are present in a 2 mol/L aluminum sulfate compound? I'm just not quite sure that it means by "concentration of sulfate ions" so there are 6 mols of sulfate ions for every L...but the original compound (aluminum sulfate)'s concentration is only 2 mol/L.

[Blueshawk]

There is nothing wrong with getting 6 mols from 2 mols.

look at these examples.

(1)CaCl2 -> (1)Ca+2 + (2)Cl-    from one mol of calcium chloride, you get one mol calcium II ion and 2 mols of chloride ion.

2(H2O) + NRG  ->  2H2 + O2   2 mols water after electrolysis gives 2 mols hydrogen gas and 1 mol oxygen gas.