determine the molar solubility of BaF2 in a solution containing 0.0870 m LiF. Ksp (BaF2)=1.6x10^-5

Expert Solution

Let the molar solubility of BaF2 be x

BaF2 ---------> Ba²⁺ + 2 F^-

Initial - 0 0.0870

Change - + x + 2x

Equilibrium - x ( 0.0870 + 2x)

[Ba²⁺] x [ F-]² = K_sp

(x) ( 0.0870 + 2x ) = 1.6 x 10^-5

2x² + 0.0870 x = 0.000016

2x² + 0.0870x - 0.000016 = 0

x = 0.000032 ( using quadratic formula)

x = 3.2 x 10^-5

The required molar solubility is 3.2 x 10^-5 M.

queen_honey_blossom answered 1 year ago

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