The pH at which $$Mg(OH)_2 [K_{sp} = 1 \times 10^{-11}]$$ begins to precipitate from a solution containing $$0.10 \text{ M } Mg^{2+}$$ ions is ______.

We need to find the pH at which $$Mg(OH)_2$$ begins to precipitate from a solution containing $$0.10 \text{ M } Mg^{2+}$$ ions.

The solubility product expression for $$Mg(OH)_2$$ is:

$$K_{sp} = [Mg^{2+}][OH^-]^2$$

Precipitation begins when the ionic product just equals the solubility product.

Substituting the known values:

$$1 \times 10^{-11} = (0.10)[OH^-]^2$$

Solving for $$[OH^-]$$:

$$[OH^-]^2 = \frac{1 \times 10^{-11}}{0.10} = 1 \times 10^{-10}$$

$$[OH^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \text{ M}$$

Finding pOH:

$$pOH = -\log[OH^-] = -\log(10^{-5}) = 5$$

Finding pH:

$$pH = 14 - pOH = 14 - 5 = 9$$

The answer is pH = 9.