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.