| ⇦ |
| ⇨ |
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 volume of water to be added to 100 cm³ of 0.5 N H₂SO₄ to get decinormal
- The interatomic distances in H₂ and Cl molecules are 74 and 198 pm respectively
- The equivalent weight of H₃PO₂, when it disproportionates into PH₃ and H₃PO₃ is
- Which one of the following is not a condensation polymer
- The crystal structure of solid Mn(II) oxide is
Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The volume of water to be added to 100 cm³ of 0.5 N H₂SO₄ to get decinormal
- The interatomic distances in H₂ and Cl molecules are 74 and 198 pm respectively
- The equivalent weight of H₃PO₂, when it disproportionates into PH₃ and H₃PO₃ is
- Which one of the following is not a condensation polymer
- The crystal structure of solid Mn(II) oxide is
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