If the solubility product of AB₂ is 3.20 × 10⁻¹¹ M³, then the solubility of AB₂ in pure water is _____ × 10⁻⁴ mol L⁻¹. [Assuming that neither kind of ion reacts with water]
If the solubility product of AB₂ is 3.20 × 10⁻¹¹ M³, then the solubility of AB₂ in pure water is _____ × 10⁻⁴ mol L⁻¹. [Assuming that neither kind of ion reacts with water]
-
1 Answer
-
(x/m) = k (P)¹/?
log (x/m) = log k + (1/n) log P
Slope = 1/n = 2 So n = ½
Intercept ⇒ log k = 0.477 So k = Antilog (0.477) = 3
So (x/m) = k (P)¹/? = 3 [0.04]² = 48 × 10?
Similar Questions for you
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 681k Reviews
- 1800k Answers