$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)
- A$0.056$
- B$0.12$
- C$0.24$
- D$0.025$
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