Answer
$K_{ps} = 2.268 \times 10^{-9}$
Work Step by Step
1. Conver the mass to nº of moles:
$n(moles) CaC_2O_4= 40.08*1 + 12.01*2 + 16*4 $
$40.08 + 24.02 + 64 = 128.1 g/mol$
$n(moles) = \frac{mass(g)}{mm}$
$n(moles) = \frac{0.0061}{128.1}$
$n(moles) = 4.762 \times 10^{-5}$
2. Since there are $4.762 \times 10^{-5}$ moles in 1L, the concentration is: $4.762 \times 10^{-5}M$
3. Write the Kps equation:
$K_{ps} = [Ca^{2+}] *[{C_2O_4}^{2-}]$
$K_{ps} = 4.762 \times 10^{-5} * 4.762 \times 10^{-5}$
$K_{ps} = 2.268 \times 10^{-9}$
