5 mL of 0.4 N NaOH is mixed with 20 mL of 0.1 N HCl. The pH of the resulting solution will be

Milliequivalents of NaOH=N1V1=0.4×5=2  Milliequivalents of HCl=N2V2=0.1×20=2 ∴ Milliequivalents of NaOH (base) = milliequivalents of HCI (acid) ∴ Resulting solution wiII be neutral. ∴ Its PH = 7