Solubility product of BaCl 2 is 4 × 10 –9 . Its solubility in moles/litre would be
Chemistry · JEE Main · NTA Exams — Equilibrium
Solubility product of BaCl2 is 4 × 10–9. Its solubility in moles/litre would be
- 1 × 10–3
- 1 × 10–9
- 4 × 10–27
- 1 × 10–27
Answer
(A) 1 × 10 –3
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- G ^ ( HI ; g ) +1.7 ~kJ . What is the equilibrium constant at 25ºC for 2HI(g) ⇌ H 2 (g) + I 2 (g)?
- In the reaction A(g) + 2B(g) ⇌ C(g) + Q kJ, greater product will be obtained or the forward reaction is…
- Which of the following information can be obtained on the basis of Le-chatelier’s principle?
- Given below are two statements. One is labelled as Assertion and the other is labelled as Reason. STATEMENT-1…
- A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of […
- In chemical reaction A ⇌ B, the system will be known in equilibrium when
- An acid solution of pH = 6 is diluted hundred times. The pH of the solution becomes :
- 4 moles of A are mixed with 4 moles of B. At equilibrium for the reaction A + B ⇌ C + D, 2 moles of C and D…