NTA Abhyas JEE Main2020ChemistryElectrochemistryPractice
For sparingly soluble salt M NH 3 4 Br 2 H 2 PO 2 , what will be the solubility product constant in ( mol litre ) 2 ? Given, λ 0 M NH 3 4 Br 2 + = 100 S m 2 mol − 1 , λ 0 H 2 PO 2 − = 50 S m 2 mol − 1 Take specific resistance as 2 0 0 Ω cm
Options
- A1.11 × 10 − 11
- B1.11 × 10 − 3
- C3.33 × 10 − 6
- DNone of these
Correct answer
A. 1.11 × 10 − 11
Step-by-step solution
Saturated solution corresponding to infinite dilution ∧ m o for salt = λ ° M NH 3 4 Br 2 + + λ ° H 2 PO 2 − = 100 S m 2 mole − 1 + 50 S m 2 mol − 1 = 150 S m 2 mol − 1 κ = 1 ρ = 1 2 0 0 Ω -1 cm -1 = 1 2 Ω -1 m -1 ∧ m o salt = κ s = 1 2 Ω - 1 m - 1 s moles m - 3 s moles m - 3 = 1 2 × 1 5 0 = 1 3 0 0 = 3 · 3 3 × 1 0 - 3 s moles L - 1 = 3 · 3 3 × 1 0 - 6 K sp = 3 · 3 3 × 1 0 - 6 2 = 1.11 × 10 − 11 M 2