Question

Consider the titration of 30.0ml of 0.030 M NH3 (Kb=1.8 x 10^-5) with 0.025M HCl. What is the pH after the addition of 36.0mL of HCl?

Answer 1

millimoles of base = 30 x 0.03 = 0.9

millimoles of acid = 36 x 0.025 = 0.9

Kb = 1.8 x 10^-5

pKb = 4.74

NH3 + HCl --------------------> NH4Cl

0.9         0.9                                  0

0             0                                  0.9

here salt only remained. so

[salt] = salt millimoles / total volume

         = 0.9 / (30 + 36)

        = 0.0136 M

pH = 7 - 1/2 (pKb + log C)

      = 7 - 1/2 (4.74 + log 0.0136)

      = 5.56

pH = 5.56