600 mL of 0.01 M HCl is mixed with 400 mL of 0.01 M H$$_2$$SO$$_4$$. The pH of the mixture is ______ $$\times 10^{-2}$$. (Nearest integer)
[Given log2 = 0.30, log3 = 0.48, log5 = 0.69, log7 = 0.84, log11 = 1.04]

600 mL of 0.01 M HCl and 400 mL of 0.01 M H$$_2$$SO$$_4$$ are mixed. The moles of H$$^+$$ from HCl = 0.6 × 0.01 = 0.006 mol, and the moles of H$$^+$$ from H$$_2$$SO$$_4$$ = 0.4 × 0.01 × 2 = 0.008 mol (diprotic acid), giving a total H$$^+$$ = 0.006 + 0.008 = 0.014 mol.

The total volume is 600 + 400 = 1000 mL = 1 L, so [H$$^+$$] = 0.014 M.

$$pH = -\log(0.014) = -\log(14 \times 10^{-3}) = 3 - \log 14$$

$$= 3 - \log 2 - \log 7 = 3 - 0.30 - 0.84 = 1.86$$

pH = 1.86 = $$186 \times 10^{-2}$$, so the answer is $$\boxed{186}$$.