# Finding pH and molarities

1. Mar 13, 2005

### Clari

Assuming the first ionization of sulphuric acid is 100% and the $$K_{a2}$$ of sulphuric acid is 0.10M at 25 degree celsius.(Ka is the acid disocciation constant)
a. Find the pH of a 0.1M $$NaHSO_4$$
b. Calculate the molarities of $$H^+$$, $$HSO_4^-$$, $$SO_4^{2-}$$ respectively in 0.1M $$H_2SO_4$$

Here are my steps:
a. Let a be the degree of dissociation.
$$K_{a2} = \frac{a^2}{0.1-a} = 0.1$$
a = 0.0618
pH = - log 0.0618 = 1.21

I have no idea of how to do part b. Please help me with it and tell me whether I have done right in part a.

2. Mar 13, 2005

### GCT

I've already explained this problem in multiple threads, nevertheless,

a)

$$Ka_2= \frac{[SO_4^{-2}][H_3O^+]}{[HSO_4^-]}$$

$$Ka_2= \frac{[x][x]}{[.1M-x]}$$

b) The first acid will dissociate completely, thus the initial concentration of both the HSO4- and H3O+ will be .1M. Thus

$$Ka_2= \frac{[SO_4^{-2}][H_3O^+]}{[HSO_4^-]}$$

$$Ka_2= \frac{[x][.1M+x]}{[.1M-x]}$$