Answer
$LaF_3$ $solubility$ $\approx 5.29 \times 10^{-7}$
Work Step by Step
$Kps (LaF_3) = 2 \times 10^{-19}$
1. Find the inital concentration of $La^{3+}$ in the solution:
$[La^{3+}]_{(initial)} = [LaCl_3] = 0.05M $
2. Write the Kps equation:
$Kps = [La^{3+}][F^-]^3$
$LaF_3 solubility = x$
$La^{3+} solubility = x$
$F^- solubility = 3x$
3. Find x:
$2 \times 10^{-19} = [x + 0.05][3x]^3$
*Since x has a very small value compared to 0.05:
$2 \times 10^{-19} = [0.05][3x]^3$
$2 \times 10^{-19} = [0.05][27x^3] = 1.35x^3$
$\frac{2 \times 10^{-19}}{1.35} = x^3$
$x^3 \approx 1.48 \times 10^{-19}$
$x \approx 5.29 \times 10^{-7}$
