Answer
$ K_{sp} (Ni(OH)_2)(20^{\circ}C) = (1.4 \times 10^{-8})$
Work Step by Step
1. Calculate the molar mass:
58.69* 1 + 2 * ( 16* 1 + 1.01* 1 ) = 92.71g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.14}{ 92.71}$
$n(moles) = 1.51\times 10^{- 3}$
3. Find the concentration in mol/L:
$1.51 \times 10^{-3}$ mol in 1L: $1.51 \times 10^{-3} M$
4. Write the $K_{sp}$ expression:
$ Ni(OH)_2(s) \lt -- \gt 1Ni^{2+}(aq) + 2{OH}^{-}(aq)$
$ K_{sp} = [Ni^{2+}]^ 1[{OH}^{-}]^ 2$
5. Determine the ions concentrations:
$[Ni^{2+}] = [Ni(OH)_2] * 1 = [1.51 \times 10^{-3}] * 1 = 1.51 \times 10^{-3}$
$[{OH}^{-}] = [Ni(OH)_2] * 2 = 3.02 \times 10^{-3}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (1.51 \times 10^{-3})^ 1 \times (3.02 \times 10^{-3})^ 2$
$ K_{sp} = (1.51 \times 10^{-3}) \times (9.121 \times 10^{-6})$
$ K_{sp} = (1.377 \times 10^{-8}) \approx 1.4 \times 10^{-8}$
