NEETChemistryIonic EquilibrumMCQ+4 / −1
If pH of a saturated solution of Ba(OH)2 is 12, the value of its Ksp is
- A4.00 106 M3
- B4.00 107 M3
- C5.00 106 M3
- D5.00 107 M3
View written solutionFree
Correct answer: D
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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