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Ionic Equilibrum question

2012 · Q120
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Ionic Equilibrum question

2012 · Q120

NEETChemistryIonic EquilibrumMCQ+4 / −1
pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
  1. A
    3.3 ×\times× 10−-−7
  2. B
    5.0 ×\times× 10−-−7
  3. C
    4.0 ×\times× 10−-−6
  4. D
    5.0×\times× 10−-−6
View written solutionFree

Correct answer: B

Ba(OH)2⇌Ba2++2OH
At equilibriumx2x


pH = – log[H+]

12 = – log [H+]

$ \Rightarrow $ [H+] = 10–12

As, [H+][OH– ] = 10–14

10–12 [OH– ] = 10–14

$ \Rightarrow $ [OH– ] = 10–2

As [OH– ] = 2x = 10–2 then x = 5.0 × 10–3

Now, Ksp = [Ba2+][OH– ]2

Ksp = (5 × 10–3) (10–2)2 = 5.0 × 10–7
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