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Question
The solubility product of BaCl2 is 4.0 × 10-8. What will be its molar solubility in mol dm-3?
Solution
Given: Solubility product (Ksp) = 4.0 × 10-8
To find: Molar solubility, S
Formula: Ksp = xx yy Sx+y
Calculation: Solubility equilibrium of BaCl2 is:
\[\ce{BaCl_{2(s)} <=> Ba^{2+}_{ (aq)} + 2Cl^{-}_{ (aq)}}\]
x = 1, y = 2
Ksp = xx yy Sx+y = (1)1 (2)2 S1+2 = 4S3
The molar solubility (S) of BaCl2 is
S = `root(3)("K"_"sp"/4) = root(3)((4.0 xx 10^-8)/4) = root(3)(1.0 xx 10^-8)` = 1 × 10-2 = 2.154 × 10-2 mol dm-3
Molar solubility (S) of BaCl2 is 2.154 × 10-2 mol dm-3
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