Calculate the pH of a 0.2M solution of ammonia. K_{a}=5.62×10^{-10}.

Ammonia is a weak base, so the most convenient approach is to calculate pOH using K_{b}, and then to convert it to pH.

From the Brønsted-Lowry theory we know that K_{a}×K_{b}=K_{w}, that allows us to calculate K_{b}=K_{w}/K_{a}=10^{-14}/5.62×10^{-10}=1.78×10^{-5}.

Calculation of the pOH will be now identical with the calculation of pH of a weak acid.

Let's try first with the simplified formula 8.17:

If we put concentration and K_{b} values into formula we get [OH^{+}]=0.00189 M and **pOH=2.72**. Is it correct? To be sure we have to check if the 5% rule is obeyed, that is if the dissociation fraction is less than 5%.

Dissociated base concentration is the same as [OH^{-}] - if we assume that we can neglect water dissociation. It is obvious that we can, as 0.00189 M is about five orders of magnitude larger than the concentration of OH^{-} from the water. So dissociation fraction is 0.00189/0.2 - below 1%, much less than 5%, so the 5% rule is obeyed.

Now we need the final touch - we know pOH=2.72, but we need pH. Time to use equation 1.6. **pH=14-2.72=11.28**.

Our pH calculator shows 11.27. Difference in the last digit can be probably attributed to different dissociation constant used.