Answer to Question #155520 in General Chemistry for lilly

Question

Solubility product of lithium fluoride is K_sp = 0.02719 (standard state c_st = 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.

Answer 1

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

$K_{sp} = [Li^+] [F^-]\\ 0.02719 = s^2\\ s = 0.165M$

s (g/L) = s(M) × molar mass

s = 0.165 × 25.968

s = 4.28 g.L^-1.