Question

Calculate the pH for the titration 20.00mL of 0.500M HF with 0.500M KOH at the following volumes of added titrant:

a) 0.00 mL (before titration begins)

b)10.00mL (at 50% equivalence)

c) 19.00 mL (at 95% equivalence)

d) 20.00 mL (at equivalence point)

e) 21 mL (at 105% equivalence)

Answer 1

no of mol of HF = 20*0.5 = 10 mmol

pka of HF = 3.17

a) before titration begins

pH of weakacid = 1/2(pka-logC)

     C = con of HF = 0.5 M

    pH = 1/2(3.17-log0.5) = 1.73

b) at half equivalence point (50%)

pH = pka of HF = 3.17

c) at 95% equivalence point

no of mol of KOH = 19*0.5 = 9.5 mmole

no of mol of HF = 20*0.5 = 10 mmole

pH = pka + log(base/acid)

    = 3.17 + log(9.5/(10-9.5))

   = 4.45

d) at equivalence point

no of mol of HF = no of mol of KOH

concentration of salt= 10/40 = 0.25 M

pH of salt of weakacid,strongbase = 7+1/2(pka+logC)

c = con of salt = 0.25 M

   = 7+1/2(3.17+log0.25)

   = 8.3

e) 21 mL (at 105% equivalence)

excess KOH = 21-20 = 1 ml

excess No of mol of KOH = 21*0.5 - 20*0.5 = 0.5 mmol

concentration of excess KOH = 0.5/41 = 0.0122 M

pH = 14 - ( -log(oH-))

    = 14 - (-log0.0122)

   = 12.09