To have a pH = 1, the resultant concentration of $[H^+]$ must be $10^{-1} \text{ M} = 0.1 \text{ M}$. For a mixture of a strong acid and a strong base, the net $[H^+]$ concentration is given by: $[H^+] = \frac{M_1V_1 - M_2V_2}{V_1 + V_2}$

Let's calculate $[H^+]$ for all given options:

Option A: $60 \text{ mL of } \frac{M}{10} \text{ HCl} + 40 \text{ mL of } \frac{M}{10} \text{ NaOH}$ Milli-moles of $H^+ = 60 \times 0.1 = 6 \text{ mmol}$ Milli-moles of $OH^- = 40 \times 0.1 = 4 \text{ mmol}$ $[H^+] = \frac{6 - 4}{60 + 40} = \frac{2}{100} = 0.02 \text{ M} \implies \text{pH} = 1.7$

Option B: $55 \text{ mL of } \frac{M}{10} \text{ HCl} + 45 \text{ mL of } \frac{M}{10} \text{ NaOH}$ Milli-moles of $H^+ = 55 \times 0.1 = 5.5 \text{ mmol}$ Milli-moles of $OH^- = 45 \times 0.1 = 4.5 \text{ mmol}$ $[H^+] = \frac{5.5 - 4.5}{55 + 45} = \frac{1}{100} = 0.01 \text{ M} \implies \text{pH} = 2.0$

Option C: $75 \text{ mL of } \frac{M}{5} \text{ HCl} + 25 \text{ mL of } \frac{M}{5} \text{ NaOH}$ Milli-moles of $H^+ = 75 \times 0.2 = 15 \text{ mmol}$ Milli-moles of $OH^- = 25 \times 0.2 = 5 \text{ mmol}$ $[H^+] = \frac{15 - 5}{75 + 25} = \frac{10}{100} = 0.1 \text{ M} \implies \text{pH} = 1.0$

Option D: $100 \text{ mL of } \frac{M}{10} \text{ HCl} + 100 \text{ mL of } \frac{M}{10} \text{ NaOH}$ Both neutralize each other completely, so the solution is neutral. $\text{pH} = 7.0$

Hence, the correct mixture is the one in Option C.