Question

Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.00 mL sample of the cleaner, the equivalence point is reached after 12.12 mL of 0.152 M HCl has been added. What is the initial concentration of ammonia in the solution? What is the pH of the solution at the equivalence point?

Answer 1

initial pH:

[NH3] = M = mmol/ml =

Kb = 1.8*10^-5

find M

mmol = MV = 12.12*0.152 = 1.84224 mmol

so

[NH3] = M = mmol/ml = 1.84224/25 = 0.0736896 M

then

Kb = x*x/(M-x)

1.8*10^-5 = x*x/(0.0736896-x)

x = OH = 0.00114 M

pO H= -log(0.00114) = 2.943

pH = 14-2.943 = 11.057

in the equivalence point:

only NH4+ left

then

NH4+ + H2O <-> NH3 + H3O+

Ka = [NH3][H3O+]/[NH4+]

Ka = Kw/Kb = 5.55*10^-10

[NH3]= x = [H3O+]

[NH4+] = M-x = M -x

then

M = mmol of NH3 /VT = 1.84224/(25+12.12) =0.07368

Ka = [NH3][H3O+]/[NH4+]

5.55*10^-10 = x*x/(0.07368-x)

x = H+ = 6.36*10^-6

pH = -log(6.36*10^-6

pH = 5.1965