Please guide me for finding pH value of sulphuric acid?
That depends on the concentration of the acid. I.E: The amount of water in there.
I'd say around Ph 2 or if it is even more concentrated then its Ph 1
Just to confuse you even more:
Super-concentrated sulphuric acid. (Also known as: Oleum) would have a value of Ph 7 (the reason for this is because it has NO water and therefore cannot really be called an acid as such; but BELIEVE ME !! get a drop of that stuff on ya and you'll be carbon in seconds !!!)
Hope this has helped.
You need the concentration of the acid. Sulfuric acid is an incredibly industrially and synthetically useful strong acid, that is also diprotic (and while the Ka of the first H is very high, the second H is still on the high end of the scale for weak acids and its contribution to pH will most usually not show up in pH values).
With that said, a 1 molar solution of sulfuric acid has a pH value slightly less than zero, which shows that it is very strongly acidic.
It's simple to calculate for a strong acid. All you have to do is take the -log of the molar concentration.
100% pure oleum wouldn't have a pH value at all since it wouldn't dissociate at all. However, 100% pure oleum would probably be damn hard to get anyway, I'd guess it's pretty intensely hygroscopic (absorbs water from the air) - hence the name fuming sulfuric acid.Originally Posted by leohopkins
Anyway, oleum would make a nice aqueous acid if it didn't spontaneously react with water to make sulfuric acid (since it's really just a solution of sulfur trioxide in sulfuric acid). The definition of an acid is also not limited to aqueous solutions either - for example, someone started a thread about carboranes recently, and compounds like p-toluenesulfonic acid are important reagents for systems using organic solvents.
Anyway, talking pH -
pH = -log[H<sup>+</sup>], i.e., negative log of the H<sup>+</sup> concentration.
For a strong acid, it's assumed that the acid HA completely dissociates into H<sup>+</sup> and A<sup>-</sup> (in this case HSO<sub>4</sub><sup>-</sup> is A<sup>-</sup>.
For 1 M acid,
pH = -log(1), pH = 0
You can work out conc from pH too, [H<sup>+</sup>] = 10<sup>-pH</sup>]
For pH 1, 10^-1 = 0.1 M
Rough guess - but the assumption (total dissociation) is close enough for it to not really make much difference.
look up the sulfuric acid article on Wikipedia and go down to "Forms of Sulfuric acid". I think that'll cover pretty much all the info you'd want on it. Also, Matt, I'm the one who started the carboranes thread. I'd like to mention that carborane acid is 1,000,000 times stronger than pure sulfuric acid, classifying it as a "Super acid". A super acid, i believe, is any acid that ionizes to a greater extent than sulfuric.
If you can find the concentration of the acid working out the pH shouldn't be hard. Remeber: pH=-log[H+] (some people also use [H3O+] instead of [H+]).
So what is the pH value for carborane acid then ?
And what is the chemical formula ?
I presume it consits of Carbon, Boron, Hydrogen & Oxygen ?
And what will the practical uses for carborane acid be ?
Why the reluctance to to type 'carborance acid wiki' instead of a whole post - much easier... :-DOriginally Posted by leohopkins
pH is concentration dependent, the convention of comparing acidity is to use pKa - see http://en.wikipedia.org/wiki/Acid_Dissociation_ConstantOriginally Posted by leohopkins
I'd estimate the pKa of the strongest carborane acids to be in the range -10 to -20 (big range, but compare that with -3 for sulfuric acid and -25 for HSbF<sub>6</sub>, the strongest acid of them all.
H2SO4 + H2O ---> H3O+ + HSO4-
H2O + HSO4- ----> H3O+ + SO4-- K = 0.01
If we assume that the c ( H3O+ )= 0.1 mol/dm3, that we will have for the first stage of reaction:
H2SO4 + H2O ---> H3O+ + HSO4-
---------------------( 0.1 )------- ( 0.1 )--------------------------
H2O + HSO4- ----> H3O+ + SO4--
------( 0.1-x )-------- ( x )------- ( x )---------------------------
So, the c ( H3O+ ) = 0.1 + x, if we add first and second equasion. (1)
K = ( [H3O+]*[SO4--] )/[HSO4-] = ( (0.1+x)*x )/(0.1-x) = 0.01
Solving the quadrant equasion, we have:
x2 + 0.11*x - 0.001 = 0
x = ( -0.11 +- squareroot( square(0.11)+0.004) ) / 2 =( -0.11+-0.1269 ) / 2
x = 8.45 * 10-3 mol-dm3 (2)
If we change the value (2) in the equasion (1), we get:
[ H3O+ ] = 0.10845 mol/dm3, approximately 0.109 mol/dm3
pH = -log [ H3O+ ] = 0.96
Can also be determined by using activation.
Now, the answer is in a afact that the 2nd dissolving process is effecting the first one by lovering the pH value.
|« Mendeleev | Chem - salt problem AJ »|