Question

Consider a weak base-strong acid titration in which 25 mL of 0.160 M ammonia is titrated with 0.160 M HCl.

a) What is the pH of the ammonia solution before the addition of HCl? pKb of ammonia= 4.75

b) Calculate the pH after the addtion of 3 mL of HCl.

c) Calculate the pH after the addition of 12.5mL of HCl. This is the half-neutralization point: the partial neutralization of ammonia converts some of the NH3 molecules to NH4+ ions and produces a buffer whose pH depends on the acid/base ratio (NH4+/NH3).

d)Calculate the pH after the addition of 25mL of HCl (equivalence point).

e) Calculate the pH after the addition of 35mL of HCl.

f) Suggest an indicator, other then methyl red, that would be suitable for this titration, and explain why it would be effective.

Thanks alot

Answer 1

a) before addition of HCl :

NH3 + H2O -----------------------> NH4+   + OH-

0.160                                             0            0

0.160-x                                         x            x

Kb = x^2 / 0.160-x

1.8 x 10^-5 = x^2 / 0.160-x

x^2 + 1.8 x 10^-5 x - 2.88 x 10^-6 = 0

x = 1.69 x 10^-3

x = [OH-] = 1.69 x 10^-3 M

pOH = -log [OH-] = -log (1.69 x 10^-3) = 2.77

pH + pOH = 14

pH = 11.23

b) after addtion of 3 mL HCl

millimoles of NH3 = 25 x 0.160 = 4

millimoles of HCl = 0.48

NH3 + H+ ----------------> NH4+

4           0.48                        0

3.52        0                         0.48

pOH = pKb + log [NH4+/NH3]

pOH = 4.75 + log (0.48 /3.52)

pOH = 3.88

pH + pOH = 14

pH = 10.12

c) after addtion of 12.5 mL HCl

at half-neutralisation point . pOH = pKb

pOH = 4.75

pH + pOH = 14

pH = 9.25

d) after addition of 25 mL HCl

at equivalence point only salt is remains

salt NH4+ concentration = millimoles / total volume

                                        = 4 / (25 +25)

                                        = 0.08 M

NH4+ salt is the weak base and strong acid. so pH < 7

pH = 7 - 1/2[pKb + logC]

pH = 7 - 1/2 [4.75 +log 0.08]

pH = 5.17

e) after addtion of 35 mL HCl

millimoles of NH3 = 25 x 0.160 = 4

millimoles of HCl = 35 x 0.160 = 5.6

NH3 + H+ ----------------> NH4+

4          5.6                       0

0       1.6                        4

strong acid HCl remains . so [H+] = 1.6 / (25 +35) = 0.0267 M

pH = -log (0.0267)

pH = 1.57