Ionic EquilibriumHard
Question
pH of a saturated solution of Ba ( OH )2 is 12. The value of solubility product Ksp of Ba ( OH2 ) is
Options
A.3.3 × 10-7
B.5.0 × 10-7
C.4.0 × 10-6
D.5.0 × 10-6
Solution
Given, pH of Ba ( OH )2 = 12
∴ pOH = 14 - pH
= 14 - 12 = 2
We know that,
pOH = - log [ OH- ]
2 = - log [ OH- ]
[ OH- ] = antilog ( -2 )
[ OH- ] = 1 × 10-2
Ba ( OH )2 dissolves in water as
∴ [ OH- ] = 2s = 1 × 10 -2
[ Ba2+ ]
Ksp = [ Ba2+ ] [ OH- ]2
= 0.5 × 10-6 = 5 × 10-7
∴ pOH = 14 - pH
= 14 - 12 = 2
We know that,
pOH = - log [ OH- ]
2 = - log [ OH- ]
[ OH- ] = antilog ( -2 )
[ OH- ] = 1 × 10-2
Ba ( OH )2 dissolves in water as
∴ [ OH- ] = 2s = 1 × 10 -2
[ Ba2+ ]
Ksp = [ Ba2+ ] [ OH- ]2
= 0.5 × 10-6 = 5 × 10-7
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