Question

Show the calculation of the pH of a solution obtained by adding 0.0100 mole of H+ to a buffer of pH 9.26 which originally contains 0.100 M NH4Cl and 0.100 M NH3.

NH3+H2O --> NH4+ + OH-
kb= 1.8*10^-5

Answer 1

You have not specified any indication of volume, but I will assume this is in a vessel of 1.00 L.
If 0.0100 mole of H+ is added to this buffer, then in the equilbrium
NH3+H2O ⇄ NH4+ + OH-
the H+ reacts with OH- and decreases [OH-] and shifts the equilibrium to the products. [NH4+] will increase by 0.0100 M and [NH3] will decrease by 0.0100M. So the new concentrations are [NH4+] = 0.110 M and [NH3] = 0.09 M. Then
[OH-] = [NH3]/[NH4+] * Kb
= 0.09/0.110 *1.8*10^-5
= 1.47 x 10^-5
and pOH = -log(1.47 x 10^-5) = 4.8318.
Since pH + pOH = 14,
pH = 14 - 4.8318 = 9.168
The answer to this question is therefore 9.168.