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

##### 1 Answer

I make it 4.73

#### Explanation:

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