Answer
$ K_{sp} (M_2X_3) = (3.3 \times 10^{-93})$
Work Step by Step
1. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 3.6\times 10^{- 17}}{ 288}$
$n(moles) = 1.25\times 10^{- 19}$
2. Find the concentration in mol/L:
$1.25 \times 10^{-19}$ mol in 1L: $1.25 \times 10^{-19} M$
3. Write the $K_{sp}$ expression:
$ M_2X_3(s) \lt -- \gt 2{M}^{3+}(aq) + 3{X}^{2-}(aq)$
$ K_{sp} = [{M}^{3+}]^ 2[{X}^{2-}]^ 3$
4. Determine the ion concentrations:
$[{M}^{3+}] = [M_2X_3] * 2 = [1.25 \times 10^{-19}] * 2 = 2.5 \times 10^{-19}$
$[{X}^{2-}] = [M_2X_3] * 3 = 3.75 \times 10^{-19}$
5. Calculate the $K_{sp}$:
$ K_{sp} = (2.5 \times 10^{-19})^ 2 \times (3.75 \times 10^{-19})^ 3$
$ K_{sp} = (6.25 \times 10^{-38}) \times (5.273 \times 10^{-56})$
$ K_{sp} = (3.296 \times 10^{-93}) = 3.3 \times 10^{-93}$
