Answer
$ K_{sp} (NiCO_3) = (1.3 \times 10^{-7})$
Work Step by Step
1. Calculate the molar mass $(NiCO_3)$:
58.69* 1 + 12.01* 1 + 16* 3 = 118.7g/mol
2. Calculate the number of moles $(NiCO_3)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.042}{ 118.7}$
$n(moles) = 3.538\times 10^{- 4}$
3. Find the concentration in mol/L $(NiCO_3)$:
$3.538 \times 10^{-4}$ mol in 1L: $3.538 \times 10^{-4} M (NiCO_3)$
4. Write the $K_{sp}$ expression:
$ NiCO_3(s) \lt -- \gt 1Ni^{2+}(aq) + 1CO_3^{2-}(aq)$
$ K_{sp} = [Ni^{2+}]^ 1[CO_3^{2-}]^ 1$
5. Determine the ion concentrations:
$[Ni^{2+}] = [NiCO_3] * 1 = [3.538 \times 10^{-4}] * 1 = 3.538 \times 10^{-4}$
$[CO_3^{2-}] = [NiCO_3] * 1 = 3.538 \times 10^{-4}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (3.538 \times 10^{-4}) \times (3.538 \times 10^{-4})$
$ K_{sp} = (1.3 \times 10^{-7})$
