What is the pH of a #1*10^-5# #M# solution of sulfuric acid?
I make it 4.73
Sulphuric acid is diprotic, and it dissociates in two steps:
This has equilibrium constant that is very large and can be assumed to be "total dissociation".
The second step is the dissociation of the bisulphate ion
Bisulphate dissociates as follows:
If we let the hydrogen ion concentration be X, then:
X = [H+] =
The concentration of bisulphate remaining in solution is going to be the original concentration of bisulphate less the amount that dissociated, which is 0.00001 - X.
We know that
0.0120 = [X]² / (0.00001-X)
So [X]² = 0.0120*( 0.00001 - X)
X² = 0.0000001 - 0.0120X
Now rearrange to quadratic equation::
X² +0.0120X - 0.0000001 = 0
X = 0.00000833 (the other root is negative and can be ignored)
So now we know the total [H+] is 0.00001 (from initial dissociation) and 0.00000833 (from dissociation of bisulphate dissociation) = 0.00001833
pH = -log [H+]
pH = -log 0.00001833
pH = 4.73