At $$25^oC$$, the solubility product of $$Mg(OH)_2$$ is $$1.0\times 10^{-11}$$. At what $$pH$$, will $$Mg^{2+}$$ ions start precipitating in the form of $$Mg(OH)_2$$ from a solution of $$0.001\ M\ Mg^{2+}$$ ions?
Correct option is B. $$10$$
Given:
Concentration of $$\mathrm{{Mg}^{+2}}$$ = $$\mathrm{10^{-3} M}$$
Solubility product of $$\mathrm{Mg(OH)_2}$$ = $$1 \times 10^{-11}$$
$$\implies \mathrm{[{Mg}^{+2}] [OH^-]^2 = 1\times 10^{-11}}$$
$$\implies \mathrm{[OH^-] = 10^{-4} M}$$
$$\implies \mathrm{pOH = 4}$$
$$\because \ \mathrm{pH +pOH = 14}$$
$$\implies \mathrm{pH = 10}$$
So, when concentration of $$\mathrm{[OH^-] }$$ becomes $$\mathrm{10^{-4} \ M}$$, $$\mathrm{Mg(OH)_2}$$ starts to precipitate.
Hence, Option "B" is the correct answer.