# Homework Help: Find pH of 0.1 MNaHCO3

1. Feb 13, 2009

### atavistic

1. The problem statement, all variables and given/known data

Find pH of $$0.1M NaHCO_3$$

$$Ka_1 (H_2CO_3) = 4.3 * 10^{ - 7} , Ka_2 (H_2CO_3) = 5.61 * 10^{ - 11}$$

2. Relevant equations

$$pH = \frac {1}{2}(14 + pKa_1 + logC)$$

3. The attempt at a solution

Now $$K_h = \frac {Kw}{Ka_1} = 2.3 * 10^{ - 8} > > Ka2$$

So I neglect dissociation and use salt hydrolysis formula.