NTA Abhyas JEE Main2020ChemistryIonic EquilibriumPractice
The K sp of FeS = 4 × 10 − 19 at 298 K . The minimum concentration of H + ions required to prevent the precipitation of FeS from a 0 .01 M solution Fe 2 + salt by passing H 2 S (Given H 2 S k a 1 × k b 1 = 10 − 21 )
Options
- A1.6 × 10 – 3 M
- B2.5 × 10 – 4 M
- C2.0 × 10 – 2 M
- D1.2 × 10 – 4 M
Correct answer
A. 1.6 × 10 – 3 M
Step-by-step solution
F e 2 + S – 2 = 4 × 10 – 19 ⇒ S – 2 = 4 × 1 0 – 19 1 × 1 0 – 2 = 4 × 10 – 17 M In order to precipitate F e S , S – 2 required is 4 × 10 – 17 M from 0.01 M F e 2 + salts. Now H + 2 4 × 1 0 – 17 0.1 = 1 × 10 – 21 ⇒ H + 2 = 2.5 × 10 – 6 ⇒ H + = 1.6 × 10 – 3