Question
The first and second ionization constants of H _ 2 X are 2.5 10^ -8 and 1.0 10^ -13 respectively. The concentration of X ^ 2- in 0.1 M H _ 2 X solution is \_\_\_\_ 10^ -15 M . (Nearest Integer)
JEE Main
Chemistry
Ionic Equilibrium
2026
JEE Main 2026 (21 January Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. A
- B. B
- C. C
- D. D
Answer
A. A
Step-by-step solution
For a weak diprotic acid H_2X, the ionization steps are: H_2X H^+ + HX^- with K_ a1 = 2.5 10^ -8 HX^- H^+ + X^ 2- with K_ a2 = 1.0 10^ -13 Since K_ a1 K_ a2 , the concentration of H^+ is primarily determined by the first ionization. [H^+] K_ a1 C = 2.5 10^ -8 0.1 = 2.5 10^ -9 = 5 10^ -5 M Also, from the first ionization, [HX^-] [H^+] = 5 10^ -5 M For the second ionization: K_ a2 = [H^+][X^ 2- ] [HX^-] Substituting the values: 1.0 10^ -13 = (5 10^ -5 )[X^ 2- ] 5 10^ -5 Thus, [X^ 2- ] = K_ a2 = 1.0 10^ -13 M To express this in the form x 10^ -15 M: 1.0 10^ -13 = 100 10^ -15 M The value is 100.
Related: Chemistry — Ionic Equilibrium · All PYQ Banks