| ⇦ | 
| ⇨ |
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: - Helium atom is two times heavier than a hydrogen molecule, the average kinetic energy
- Which is insoluble in water
- NaOCl is used as a bleaching agent and sterilising agent. It can be synthesized
- If the value of an equilibrium constant for a particular reaction is 1.6 ˣ 10¹²,
- When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M
Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Helium atom is two times heavier than a hydrogen molecule, the average kinetic energy
- Which is insoluble in water
- NaOCl is used as a bleaching agent and sterilising agent. It can be synthesized
- If the value of an equilibrium constant for a particular reaction is 1.6 ˣ 10¹²,
- When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M
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