Since equal volumes are taken, consider volume = 1 Liter of each solution.
HCl is a strong acid. So it dissociates completely.
HCl + H2O <-------> H3O+ + Cl -
Hence [H3O+] = 2.3 x 10-4 M
Since volume = 1 lilter, number of moles = 2.3 x 10-4 = 0.00023
NH3 is a weak base with Kb value of 1.8 x 10-5
Since volume gets doubled, new concentration =
2.3 
10
-4 / 2 = 1.15 x 10
-4 ammonia reacts with H3O+ and abstracts a proton to form NH4+
NH
3 + H
2O <----------> NH
4 + + OH
-
Initial
1.15 10
-4 0 0
Change -x +x +x
Equilibrium ( 1.15 x 10-4 -x) x x
Kb = x .x / (1.15 x 10-4 -x) = 1.8 x 10-5
Solving for x, we get x = 0.0000373
Hence [OH-] = 0.0000373 M
Number of moles of OH-= 0.0000373 x 2
= 0.0000746 moles
Number of moles of [H3O+] = 0.00023
Since moles of acid are more,
Moles of acid left over after neutralisation = 0.00023 - 0.0000746
= 0.0001554
Molarity of [H3O+] = [H+] = 0.0001554
pH = - log [H+]
= - log [ 0.0001554]
= 3.808