Question

In the Kjeldahl nitrogen analysis, the final product is NH4+ in HCI solution. It is necessary to titrate the HCI without titrating the NH4+ ion. Calculate the pH of pure 0.050M NH4CI. Could someone show me this step by step. I'm not sure how to do this problem.

Answer 1

Kb of NH3 is 1.8*10^-5

First find Ka of NH4+

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

NH4+ + H2O -----> NH3 + H+

5*10^-2 0 0

5*10^-2-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*5*10^-2) = 5.27*10^-6

since c is much greater than x, our assumption is correct

so, x = 5.27*10^-6 M

So, [H+] = x = 5.27*10^-6 M

use:

pH = -log [H+]

= -log (5.27*10^-6)

= 5.2782

Answer: 5.28