Let the volume of each solution be $V$. Moles of $NaOH$ = $0.1 \times V = 0.1V$ Moles of $HCl$ = $0.01 \times V = 0.01V$ The neutralization reaction is $NaOH + HCl \rightarrow NaCl + H_2O$. Since $NaOH$ and $HCl$ react in a 1:1 molar ratio, $HCl$ is the limiting reagent. Moles of $NaOH$ left unreacted = $0.1V - 0.01V = 0.09V$ Total volume of the resulting mixture = $V + V = 2V$ Concentration of $OH^-$ in the final solution = $\frac{0.09V}{2V} = 0.045\text{ M}$ $pOH = -\log[OH^-] = -\log(0.045) = -\log(4.5 \times 10^{-2}) = 2 - \log(4.5) = 2 - 0.653 = 1.347$ $pH = 14 - pOH = 14 - 1.347 = 12.653 \approx 12.65$.