If 200 mL of a 0.031 molar solution of H2SO4 is added to 84 mL of a 0.150 M KOH solution, The value of pH of the resulting solution is

mmol of H+(initial) =200×0.031×2=12.4 mmol of OH−(initial) =84×0.15=12.6 mmol of OH−(left) after neutralisation =12.6−12.4=0.2 OH−final =0.2284=7×10−4M pOH=−log⁡OH−=−log⁡7×10−4 pOH=3.15 and pH=10.85 ∴pH=14−pOH=14−3.15=10.85≈10.9