Answer to Question #155520 in General Chemistry for lilly

Question #155520

Solubility product of lithium fluoride is Ksp = 0.02719 (standard state cst = 1 mol/L).

What is the solubility of LiF in g·L-1?

M(LiF) = 25.968 g·mol-1

Present your numerical answer to 3 significant figures.


1
Expert's answer
2021-01-15T06:38:08-0500

"LiF \\rightleftarrows Li^+ + F^-"


"K_{sp} = [Li^+] [F^-]\\\\\n0.02719 = s^2\\\\\ns = 0.165M"


s (g/L) = s(M) × molar mass

s = 0.165 × 25.968

s = 4.28 g.L-1.


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