Answer
$LaF_3$ $solubility$ $\approx 2 \times 10^{-13}M$
Work Step by Step
$Kps (LaF_3) = 2 \times 10^{-19}$
1. Find the inital concentration of $F^-$ in the solution:
$[F^-]_{(initial)} = [KF] = 0.01M $
2. Write the Kps equation:
$Kps = [La^{3+}][F^-]^3$
$LaF_3 solubility = x$
$La^{3+} solubility = x$
$F^- solubility = 3x$
3. Find x:
$2 \times 10^{-19} = [x][x + 0.01]^3$
*Since x has a very small value compared to 0.01:
$2 \times 10^{-19} = [x][0.01]^3$
$2 \times 10^{-19} = [x][10^{-6}]$
$\frac{2 \times 10^{-19}}{10^{-6}} = x$
$x \approx 2 \times 10^{-13}M$
