To a one litre solution containing 4 grams of sodium hydroxide, 9.8 grams of sulphuric acid is added pH of the solution at 298 K is

Number of gram equivalents of base = 4/40 = 0.1 Number of gram equivalents of acid = 9.8/49 = 0.2Nature of resultant solution is acidicNumber of gram equivalents of acid left = (0.2 - 0.1)Number of gram equivalents of acid left = 0.1∴H+=0.1×11H+=0.1MpH=1