Question

For the titration of 70.0 mL of 0.400 M NH3 with 0.500 M HCl at 25 °C, determine the relative pH at each of these points: (a) before the addition of any HCl; (b) after 56.0 mL of HCl has been added;(c) after 66.0 mL of HCl has been added

Answer 1

a)

NH3 + H2O <-> NH4+ + OH-

Kb = [NH4+][OH-]/[NH3]

[NH4+]= [OH-]= x

[NH3] = M-x = 0.4-x

1.8*10^-5 = x*x/(0.4-x)

x = 0.00267

[OH-] =0.00267

pOH = -log(0.00267) = 2.5734

pH = 14-2.5734= 11.4266

b)

after 56 mL of acid

mmol acid = MV = 0.5*56 = 28

mmol base = MV = 70*0.4 = 28

this ineutralization so..

[NH4+] = M1V1/(V1+V2) = 28/(70+56) = 0.22222

Ka = [NH4´+][H+]/[NH3]

Ka = Kw/Kb = (10^-14)/(1.8*10^-5) = 5.55*10^-10

Ka = 1.8*10^-5

5.55*10^-10 = x*x/(0.22222-x)

x =1.11*10^-5

pH = -log(1.11*10^-5)

pH = 4.9546

finally...

after addition of extra HCl

mmol of acid = MV = 66*0.5 = 33 mmol

mmol of base = MV = 28

33-28 = 5 mmol of acid left

V = V1+V2 = 70+66 = 136

[H+] = 5/136 = 0.036764

pH = -log(0.036764 = 1.4345