What is the pH of a #1*10^-5# #M# solution of sulfuric acid?

1 Answer
Jul 8, 2016


I make it 4.73


Sulphuric acid is diprotic, and it dissociates in two steps:

#H_2SO_4 --> H^+ + HSO_4^- #

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 #HSO_4^-#, but this only partially dissociates and the equilibrium constant of this reaction is much lower, in fact the #K_a# value for bisulphate is only 0.0120.

Bisulphate dissociates as follows: #HSO_4^-# = #H^+ + SO_4^ 2-#

#K_a = ([H+]*[SO_4^2-]) / [[HSO4-]#

If we let the hydrogen ion concentration be X, then:

X = [H+] = #[SO_4^ 2-]# , therefore [H+].#[SO_4^2-]# = [X]²

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 #K_a# is 0.0120, so now we can write::

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