Ionic EquilibriumHard
Question
If pH of a saturated solution of Ba(OH)2 is 12, the value of its Ksp is
Options
A.4.00 × 10-6 M3
B.4.00 × 10-7 M3
C.4.00 × 10-6 M3
D.4.00 × 10-7 M3
Solution
Given, pH of Ba(OH)2 = 12
∴ [H+]= [1 × 10-12]
and [OH-] =
mol
[∵[H+][OH-] = 1×10-14]
= 1×10-2 mol/L
Ba(OH)2 →
Ksp = [Ba2+][OH-]2
= [s][2s]2
=
(1 × 10-2)2
= 0.5 × 10-6
= 5 × 10-7 (M/lit)3
∴ [H+]= [1 × 10-12]
and [OH-] =
mol[∵[H+][OH-] = 1×10-14]
= 1×10-2 mol/L
Ba(OH)2 →
Ksp = [Ba2+][OH-]2
= [s][2s]2
=
(1 × 10-2)2 = 0.5 × 10-6
= 5 × 10-7 (M/lit)3
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