Answer
$ K_{sp} (MX) = (1.791 \times 10^{-10})$
Work Step by Step
1. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 4.63\times 10^{- 3}}{ 346}$
$n(moles) = 1.338\times 10^{- 5}$
2. Find the concentration in mol/L:
$1.338 \times 10^{-5}$ mol in 1L: $1.338 \times 10^{-5} M$
3. Write the $K_{sp}$ expression:
$ MX(s) \lt -- \gt 1M^+(aq) + 1X^-(aq)$
$ K_{sp} = [M^+]^ 1[X^-]^ 1$
4. Determine the ion concentrations:
$[M^+] = [MX] * 1 = [1.338 \times 10^{-5}] * 1 = 1.338 \times 10^{-5}$
$[X^-] = [MX] * 1 = 1.338 \times 10^{-5}$
5. Calculate the $K_{sp}$:
$ K_{sp} = (1.338 \times 10^{-5}) \times (1.338 \times 10^{-5})$
$ K_{sp} = (1.791 \times 10^{-10})$
