x mg of $$Mg(OH)_2$$ (molar mass = 58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K. The value of x is ____ mg. (Nearest integer)
(Given : $$Mg(OH)_{2}$$  is assumed to dissociate completely in $$H_{2}O$$) 

The required pH is $$10.0$$, so

$$\text{pOH}=14-\text{pH}=14-10=4$$

Hence the hydroxide-ion concentration is

$$[OH^-]=10^{-4}\ \text{mol L}^{-1}$$

Magnesium hydroxide dissociates as

$$Mg(OH)_2 \rightarrow Mg^{2+}+2\,OH^-$$

From the stoichiometry, $$1$$ mole of $$Mg(OH)_2$$ releases $$2$$ moles of $$OH^-$$.
Therefore moles of $$Mg(OH)_2$$ that must dissolve are

$$n=\frac{[OH^-]}{2} =\frac{10^{-4}}{2} =5\times10^{-5}\ \text{mol}$$

Molar mass of $$Mg(OH)_2$$ is $$58\ \text{g mol}^{-1}$$, so the required mass is

$$m = n \times \text{molar mass} = 5\times10^{-5}\times58 = 2.9\times10^{-3}\ \text{g}$$

Converting to milligrams,

$$m = 2.9\ \text{mg}$$

The nearest integer value of $$x$$ is $$3\ \text{mg}$$.

Answer: $$\displaystyle 3\ \text{mg}$$