Question

Determine the pH during the titration of 35.9 mL of 0.280 M ammonia (NH₃, K_b = 1.8×10^-5) by 0.280M HCl at the following points. (Assume the titration is done at 25 °C.)
Note that state symbols are not shown for species in this problem.
(a) Before the addition of any HCl

(b) After the addition of 13.9 mL of HCl

(c) At the titration midpoint

(d) At the equivalence point

(e) After adding 51.7 mL of HCl

Answer 1

a) before addition of HCl :

pKb = 4.74

pOH = 1/2 (pKb - log C)

       = 1/2 (4.74 - log 0.280)

pOH = 2.65

pH = 11.35

b) after addtion of 13.9 mL HCl

millimoles of NH3 = 35.9 x 0.280 =10.052

millimoles of HCl = 3.892

NH3    + H+ ----------------> NH4+

10.052           3.892                      0

6.16         0                       3.892

pOH = pKb + log [NH4+/NH3]

pOH = 4.74 + log (3.892 /6.16)

pOH = 4.54

pH = 9.46

c) At the titration midpoint

at half-neutralisation point . pOH = pKb

pOH = 4.74

pH = 9.26

d) At the equivalence point l

at equivalence point only salt is remains

salt NH4+ concentration = millimoles / total volume

                                        = 10.052 / (35.9 +35.9)

                                        = 0.14 M

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

pH = 7 - 1/2 [4.75 +log 0.14]

pH = 5.06

e)

pH = 1.30