If the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$,the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:
- A$1.9 \times 10^{-18} \ M$
- B$1.9 \times 10^{-13} \ M$
- C$1.9 \times 10^{-15} \ M$
- D$1.9 \times 10^{-14} \ M$
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