By adding 20 ml 0.1 N HCl to 20 ml 0.001 N KOH, the pH of the obtained solution will be

20 ml of 0.1 NHCl = 0.11000 × 20 g  eq = 2 × 10-3  g eq.20 ml of 0.001 KOH =  0.0011000 × 20 g  eq = 2 × 10-5  g eq.∴    HCl  left  unneutralised = 210-3 - 10-5 = 2 × 10-3  1-0.01 = 2 × 0.99 × 10-3  =1.98 × 10-3 g eq.Volume of solution  = 40 ml∴   HCl = 1.98 × 10-340  ×  1000 M  =  4.95 ×  10-2 ∴    pH = 2-log 4.95 = 2 -0.7 = 1.3