The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)$$_2$$ is (Assume complete neutralization)

Given: 10.0 mL of 0.01 M $$Ba(OH)_2$$ is to be neutralized by 0.02 M aqueous HBr.

The balanced neutralization reaction is:

$$Ba(OH)_2 + 2HBr \rightarrow BaBr_2 + 2H_2O$$

Moles of $$Ba(OH)_2$$:

$$n_{Ba(OH)_2} = 0.01 \times 10 = 0.1 \text{ mmol}$$

Since $$Ba(OH)_2$$ is dibasic, the moles of $$OH^-$$ ions:

$$n_{OH^-} = 2 \times 0.1 = 0.2 \text{ mmol}$$

For complete neutralization, moles of HBr required = moles of $$OH^-$$ = 0.2 mmol.

Volume of HBr solution needed:

$$V = \frac{n_{HBr}}{M_{HBr}} = \frac{0.2}{0.02} = 10.0 \text{ mL}$$

Therefore, the correct answer is Option C: $$\mathbf{10.0 \text{ mL}}$$.