JEE Advanced levelSingle correct optionDifficulty 3 of 3g2035
The pKa values in water are ethanol 15.9 and ethyne 25. The equilibrium constant for HC≡CH+CX2HX5OX−HC≡CX−+CX2HX5OH (treating the pKa values as valid for the medium) is closest to
A7.9×10−10
B1.3×109
C1.0×10−15
D7.9×10−1
Answer: A
K=Ka(ethyne)/Ka(ethanol)=10−25/10−15.9=10−9.1=7.9e−10. Ethoxide is too weak a base to deprotonate ethyne to any useful extent. The value 1.3×109 is the inverse.