Question

Suppose 20.0 mL of a 0.100 M NH3 solution is titrated with 0.100 M HI. Determine the pH of the solution after the addition of the following volumes of titrant.

a. 5.00 mL

b. 10.00 mL

c. 15.00 mL

d. 20.00 mL

e. 25.00 mL

Answer 1

NH3 + HI = H2O + NH4I

mmol of base = MV = 20*.1 = 2 mmol of NH3

mmol of acid = MV = 0.1*V = ?

then

a)

this is a buffer, pKb for nh3 = 4.75

mmol of acid = MV = 0.1*5 = 0.5 mmol of acid

mmol of base left = 2-0.5 = 1.5 mmol of base

pOH = pKb + log(conjugate/base)

pOH = 4.75 + log(0.5/1.5) = 4.2728

pH = 14-pOH = 14-4.2728 = 9.7272

b)

this is a buffer, pKb for nh3 = 4.75

mmol of acid = MV = 0.1*10 = 1 mmol of acid

mmol of base left = 2-1 = 1 mmol of base

pOH = pKb + log(conjugate/base)

pOH = 4.75 + log(1/1) = 4.75

pH = 14-4.75 = 9.25

c)

this is a buffer, pKb for nh3 = 4.75

mmol of acid = MV = 0.1*15= 1.5 mmol of acid

mmol of base left = 2-1 = 1 mmol of base

pOH = pKb + log(conjugate/base)

pOH = 4.75 + log(1.5/0.5) = 5.2271

pH = 14-5.2271= 8.7729

d)

neutralization

VT = V1+V2 = 20+20 = 40

NH4+ in solution... only

then

NH4+ + H2O <--> NH3 + H3O+

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

Ka = Kw/Kb = (10^-14)/(1.8*10^-5) =5.5555*10^-10

then

[H3O+] = [NH3] = x

[NH4+] = 2/(40) - x= 0.05 -x

then

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

5.55*10^-10 = (x+x)/(0.05 -x)

x = [H3o+] = 5.267*10^-6

pH = -log(H) = -log(5.267*10^-6) = 5.2784

e)

extra acid

mmol of acid left = 0.1*25 - 20*0.1 = 2.5-2 = 5 mmol

V T= V1+V2 = 20+25 = 45

[HI] = 0.5/45 = 0.01111

pH = -log(H) = -log(0.01111) =1.954285