Question
The solubility product constants of Ag_2CrO_4 and AgBr are 32x and 4y respectively at 298 K. The value of ( molarity of Ag_2CrO_4 molarity of AgBr ) can be expressed as :
JEE Main
Chemistry
Ionic Equilibrium
2026
JEE Main 2026 (02 April Shift 1)
JEE Main Chemistry Question (2026) — Solution
Options
- A. 2 [3] x y
- B. 2 x y
- C. x y
- D. [3] x y
Answer
D. [3] x y
Step-by-step solution
Let the solubility of Ag_2CrO_4 be S_1. For Ag_2CrO_4 2Ag^+ + CrO_4^ 2- K_ sp = (2S_1)^2(S_1) = 4S_1^3 4S_1^3 = 32x S_1^3 = 8x S_1 = 2 [3] x Let the solubility of AgBr be S_2. For AgBr Ag^+ + Br^- K_ sp = S_2^2 S_2^2 = 4y S_2 = 2 y The ratio of their molarities is: S_1 S_2 = 2 [3] x 2 y = [3] x y Answer: [3] x y
Related: Chemistry — Ionic Equilibrium · All PYQ Banks