JEE Advanced | 2013Ionic EquilibriumMedium
Question
The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2 CrO4 in a 0.1 M AgNO3 solution is
Options
A.1.1 × 10-11
B.1.1 × 10-10
C.1.1 × 10-12
D.1.1 × 10-9
Solution
Ag2CrO4 = 2Ag+(aq) + CrO42- (aq)
(2s + 0.1) s
Ksp = [Ag+]2 [CrO42-]
Ksp = (2s + 0.1)2 (s)
2s < < 0.1
⇒1.1 × 10-12 = (0.1)2 × s
s = 1.1 × 10-10
(2s + 0.1) s
Ksp = [Ag+]2 [CrO42-]
Ksp = (2s + 0.1)2 (s)
2s < < 0.1
⇒1.1 × 10-12 = (0.1)2 × s
s = 1.1 × 10-10
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
The first and second ionization constants of a weak dibasic acid $H_{2}\text{ }A$ are $8.1 \times 10^{- 8}$ and $1.0 \ti...Liquid ammonia ionizes to slight extent. At −50°C, its self-ionization constant, K = [NH4+] [NH2−] = 10–30 M2. How many ...The buffer capacity (β) for a weak acid (A) – conjugate base (B) buffer is defined as the number of moles of strong acid...Which of the following result(s) is/ are correct for the equilibrium state in a solution originally having 0.1 M-CH3COOH...The following five solutions of KOH were prepared as first, 0.1 mole in 1 L; second, 0.2 mole in 2 L; third, 0.3 mole in...