Answer
$Ba(IO_3)_2$ $ Solubility: 5.313 \times 10^{-4}M$
Work Step by Step
1. Write the Kps formula:
$K_{ps} = [Ba^{+2}]*[{IO_3}^-]^2$
2. Calculate x:
$[Ba(IO_3)_2] = x$
$[Ba^{2+}] = x $
$[{IO_3}^-] = 2x$
$K_{ps} = x * (2x)^2$
$K_{ps} = x * 4x^2$
$6 \times 10^{-10} = 4x^3$
$x^3 = 1.5 \times 10^{-10}$
$x = 5.313 \times 10^{-4} M$
