JEE Advanced levelSingle correct optionDifficulty 3 of 3g2159
For the equilibrium 2-OX2N−CX6HX4−NHX2+CX6HX5NHX3X+2-OX2N−CX6HX4−NHX3X++CX6HX5NHX2 in water, the conjugate acids have pKa of −0.3 (2-nitroanilinium) and 4.6 (anilinium). The equilibrium constant is closest to
A7.9×104
B2.0×104
C1.3×10−5
D5.0×10−5
Answer: C
logK=pKa(conjugate acid of the base)−pKa(acid)=(−0.3)−4.6=−4.9, so K=1.26e−05≈1.3×10−5. The weaker base 2-nitroaniline cannot take the proton from anilinium.