Question

Calculate the pH at the equivalence point for the following titration:

0.1 M HCl and 0.1 M NH3

The finale answer should be 5.28

And there is no volume

Answer 1

0.1 M of HCl and NH3 will react to form 0.1 M of NH4+

find Ka of NH4+

use:

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

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

Ka = 5.556*10^-10

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

0.1 0 0

0.1-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)*0.1) = 7.454*10^-6

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

so, x = 7.454*10^-6 M

so.[H+] = x = 7.454*10^-6 M

use:

pH = -log [H+]

= -log (7.454*10^-6)

= 5.13

Answer: 5.13