JEE Advanced levelOne or more correct optionsDifficulty 3 of 3g2036
Use pKa values (in water): acetic acid 4.76, phenol 10.0, ethanol 15.9, pentane-2,4-dione 9.0, propanone 19.3. For which of the following reactions is the equilibrium constant much greater than 1 (more than 103)?
K equals 10pKa(acid on right)−pKa(acid on left). Option 1: 1010.0−4.76=1.7×105. Option 2: 1010.0−15.9=1.3×10−6. Option 3: 1015.9−9.0=7.9×106. Option 4: 1015.9−19.3=4.0×10−4. So options 1 and 3.