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

Jul 8, 2016

I make it 4.73

#### Explanation:

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

${H}_{2} S {O}_{4} - \to {H}^{+} + H S {O}_{4}^{-}$

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 $H S {O}_{4}^{-}$, 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: $H S {O}_{4}^{-}$ = ${H}^{+} + S {O}_{4}^{2} -$

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

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

X = [H+] = $\left[S {O}_{4}^{2} -\right]$ , therefore [H+].$\left[S {O}_{4}^{2} -\right]$ = [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