Question

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. (Assume that the temperature is 25°C.)

 (a) 101.5 mL of 0.19 M HCO2H (Ka= 1.8 ✕ 10^-4) titrated with 0.19 M KOH 

 (b) 104.1 mL of 0.18 M (C2H5)3N (Kb = 4.0 ✕ 10^-4) titrated with 0.36 M HClO4 

 (c) 100.9 mL of 0.47 M HClO4 titrated with 0.24 M NaOH 

Answer 1

a) it is weak acid vs strong base titration

   pka of HCO2H = -logKa

                = -log(1.8*10^-4)

       = 3.74

at halfway point = pH = pka = 3.74

at euivalence point ,

no of mol of HCO2H = 101.5*0.19 = 19.3 mmol

no of mol of KOH required = 19.3 mmol

volume of KOH required = 19.3/0.19 = 101.5 ml

HCO2H + KOH ---> HCO2K + H2O

concentration of salt(HCO2K) formed(C) = 19.3/(101.5+101.5)

                                       = 0.095 M

pH = 7+1/2(pka+logC)

    = 7+1/2(3.74+log0.095)

    = 8.36

b)

it is weak base vs strong acid titration

pkb of (C2H5)3N = -logKb

                = -log(4*10^-4)

                = 3.4

at halfway point = pOH = pkb = 3.4

          pH = 14-3.4 = 10.6

at euivalence point ,

no of mol of (C2H5)3N = 104.1*0.18 = 18.74 mmol

no of mol of HClO4 required = 18.74 mmol

volume of HClO4 required = 18.74/0.36 = 52.05 ml

(C2H5)3N + HClO4 ---> (C2H5)3NHClO4

concentration of salt((C2H5)3NHClO4) formed(C) = 18.74/(104.1+52.05)

                                               = 0.12 M

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

    = 7-1/2(3.4+log0.12)

    = 5.76

c)

it is strong base vs strong acid titration

no of mol of HClO4 taken = 100.9*0.47 = 47.423 mmol

at halfway point , no of mol of HClO4 remaining = 47.423/2 = 23.711 mmol

volume of NaOH required to reach halfway point = 23.711/0.24 = 98.8 ml

concentration of HClO4 remaining = 23.711/(98.8+100.9)

                                 = 0.119 M
1 mol HCLo4 = 1 mol H+

pH = -log(H3O+)

    = -log0.119

    = 0.924

b) at equivalence point , pH = 7