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The solubility product of BaCl2 is 4.0 × 10-8. What will be its molar solubility in mol dm-3? - Chemistry

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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?

बेरीज
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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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पाठ 3: Ionic Equilibria - Short answer questions (Type- I)
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