Question

If exactly 50 ml of a 0.050m solution of hydrochloric acid is added to exactly 50 ml of 0.050m ammonia, what is the ph of the resulting solution?

Answer 1

NH3(aq) + HCl(aq) ⇒ NH4Cl(aq) >>> (1)
∵ C = n/V; C= concentration, n= No. of moles, and V= volume (L)
∴ n = C*V, n(HCl) = 0.050*(50/1000) = 0.0025 moles
n(NH3) = 0.050*(50/1000) = 0.0025 moles
So, the limiting no. of moles is 0.0025 moles >>> (2)
∵ NH3 is weak base, and HCl is strong Acid (and have the same number of moles) >>> So, without any calculation we can notice that the formed salt (NH4Cl) is acidic salt and the pH is less than 7.
From (1) and (2), The no. of moles of NH4Cl is 0.0025 moles >>> (3)
∴ the concentration of [NH4Cl] = 0.0025 / (total volume per L) 
                                                    = 0.0025 / ((50 + 50) / 1000) = 0.025 M
NH4+(aq) ⇔ NH3(aq) + H+(aq)   >>> (4)
(0.025 - x)          (x)              (x)       >>> (5)
∵ Ka = [NH3] [H+] / [NH4+]    >>>> (6)
Ka = Kw / Kb,    Kb = 1.8 * 10^-5 >>> (7)
∴ Ka = 10^-14 / 1.8*10^-5 = 5.56*10^-10  >>> (8)
From (4), (5), (7) and (8) 
 Ka = 5.56*10^-10 = (x * x) / (0.025-x) , we will assume that (0.025 - x) = 0.025
∴ x^2 = (5.56*10^-10)(0.025) = 139*10^-13 
∴ x = 3.73*10^-6 = [H+]
∵ pH = - log [H+]
∴ pH = - log 3.73*10^-6 = 5.43