| ⇦ |
| ⇨ |
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: - The C – C bond length in ethane is
- Polarisation power of a cation increases,when
- Pick out the stronger reducing agent among the following oxyacids of phosphorus
- The reaction of an organic compound with ammonia followed by nitration
- Permanent hardness of water is due to the presence of
Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The C – C bond length in ethane is
- Polarisation power of a cation increases,when
- Pick out the stronger reducing agent among the following oxyacids of phosphorus
- The reaction of an organic compound with ammonia followed by nitration
- Permanent hardness of water is due to the presence of
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