The first and second ionization constants of H2X are 2.5 × 10 –8 and 1.0 × 10–13 respectively. The c — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

H2X H + + HX –, 0.1–x x +y x–y 2.5 × 10 –8 = (x y) (x y) 0.1 x    HX – H + + X 2–, x–y x +y y 1 × 10 –13 = (x y) (y) x y   Approximate : 2 a a K K  So x > > y. x + y  x, x – y  x –13 = x.y x y = 10 –13 [X 2–] = 10 –13 [X 2–] = 100 × 10 –15