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प्रश्न
The solubility product of BaCl2 is 4.0 × 10-8. What will be its molar solubility in mol dm-3?
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उत्तर
Given: Solubility product (Ksp) = 4.0 × 10-8
To find: Molar solubility, S
Formula: Ksp = xx yy Sx+y
Calculation:
Ksp = [Ba2+][Cl−]2
Ksp = (s)(2s)2 = s ⋅ 4s2 = 4s3
4s3 = 4.0 × 10−8
`s^3 = (4.0 xx 10^-8)/4 = 1.0 xx 10^-8`
`s = root3(1.0 xx 10^-8)`
= 2.15 × 10-3 mol dm−3
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