Question

Find the pH during the titration of 20.00 mL of 0.2300 M nitrous acid, HNO2 (Ka = 7.1 ✕ 10-4), with 0.2300 M NaOH solution after the following additions of titrant. a) 0mL b)10.00mL c) 15.00mL d) 19.00mL e)19.95 mL f) 20.00mL g) 20.05mL h) 25.00mL

Answer 1

a)

pka of HNO2 = -LOGKa = -log(7.1*10^(-4)) = 3.15

pH = 1/2(pka-logC)

    = 1/2(3.15-log0.23) = 1.9

b)

No of mol of HNO2 = 20/1000*0.23 = 0.0046 mol

No of mol NaOH added = 10/1000*0.23 = 0.0023 mol

it is at half equivalence point

pH = pka = 3.15

c)

No of mol of HNO2 = 20/1000*0.23 = 0.0046 mol

No of mol NaOH added = 15/1000*0.23 = 0.00345 mol

pH = pka+log(salt/acid)

   = 3.15+log(0.00345/(0.0046-0.00345))

   = 3.627

d) 19 ml

No of mol of HNO2 = 20/1000*0.23 = 0.0046 mol

No of mol NaOH added = 19/1000*0.23 = 0.00437 mol

pH = pka+log(salt/acid)

   = 3.15+log(0.00437/(0.0046-0.00437))

   = 4.428

e) 19.95 ml

No of mol of HNO2 = 20/1000*0.23 = 0.0046 mol

No of mol NaOH added = 19.95/1000*0.23 = 0.00459 mol

pH = pka+log(salt/acid)

   = 3.15+log(0.00459/(0.0046-0.00459))

   = 5.81

f) 20 ml

Concentration of salt = 20*0.23/40 = 0.115 M

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

   = 7+1/2(3.15+log0.115)

= 8.1

g)

excess NaOH concentration = (20.05-20)/40.05*0.23

   = 0.000287

pOH = -log0.000287 = 3.54

pH = 14-3.54 = 10.46

h)
excess NaOH concentration = (25-20)/45*0.23

   = 0.0255

pOH = -log0.0255 = 1.6

pH = 14-1.6 = 12.4