Question

In a titration of 49.51 mL of 0.3605 M ammonia with 0.3605 M aqueous nitric acid, what is the pH of the solution when 49.51 mL of the acid have been added?

Answer 1

NH3 + HCl ----> NH4Cl

   no of mol of NH3 taken = 49.51*0.3605 = 17.85 mmol

   no of mol of NH4Cl formed = HCl added = 49.51*0.3605 = 17.85 mmol

concentration of NH4Cl formed = n/v

                                = 17.85/(49.51+49.51)

                                = 0.18 M

pH of NH4Cl = 7-1/2(pkb+logC)

   pkb of NH3 = 4.75

C = concentration of NH4Cl = 0.18 M

pH = 7-1/2(4.75+log0.18)

    = 5