| ⇦ | 
| ⇨ |
The solubility product of a sparingly soluble salt AX₂ is 3.2 ˣ 10⁻¹¹.Its solubility (in mol/L)is
Options
(a) 5.6 ˣ 10⁻⁶
(b) 3.1 ˣ 10⁻⁴
(c) 2 ˣ 10⁻⁴
(d) 4 ˣ 10⁻⁴
Correct Answer:
2 ˣ 10⁻⁴
Explanation:
K(sp) = 3.2×10⁻¹¹.
AX₂ ⇌ A² + 2X⁻
K(sp) = sx(2s)² = 4s³; i.e) 3.2×10⁻¹¹=4s³. (or) s³ = 0.8×10⁻¹¹
= 8×10⁻¹².
Therefore s = 2×10⁻⁴.
Related Questions: - For the reaction C₂H₅OH + 3O₂(g) → 2CO₂(g) + 3H₂O which one is true
- When 0.1 mol MnO₄²⁻ is oxidised the quantity of electricity required to completely
- The inert gas abundantly found in atmosphere is
- The amount of energy,which is released due to the addition of extra electron
- During a redox titration involving a solution containing Fe²⁺ ions against MnO₄⁻
Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- For the reaction C₂H₅OH + 3O₂(g) → 2CO₂(g) + 3H₂O which one is true
- When 0.1 mol MnO₄²⁻ is oxidised the quantity of electricity required to completely
- The inert gas abundantly found in atmosphere is
- The amount of energy,which is released due to the addition of extra electron
- During a redox titration involving a solution containing Fe²⁺ ions against MnO₄⁻
Topics: Equilibrium (104)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply