To determine which solution has a pH of 1, we need to calculate the concentration of $$H$^+$ ions in each mixture.$\\$The pH is given by:$\\$pH$= -\log[$H$^+] \\$Let's analyze each solution:$\\$a. $60$ mL $\frac{M}{10}$ HCl $+ 40$ mL $\frac{M}{10}$ NaOH$\\$• Moles of HCl $= \frac{M}{10} \times 60 \times 10^{-3} = 0.006$ mol$\\$• Moles of NaOH $= \frac{M}{10} \times 40 \times 10^{-3} = 0.004$ mol$\\$• Excess moles of HCl $= 0.006 - 0.004 = 0.002$ mol$\\$• Total volume $= 60 + 40 = 100$ mL$= 0.1$ L$\\ [$H$^+] = \frac{0.002}{0.1} = 0.02$ M$\\$pH$= -\log(0.02) \approx 1.7 \\$b. $55$ mL $\frac{M}{10}$ HCl $+ 45$ mL $\frac{M}{10}$ NaOH$\\$• Moles of HCl $= \frac{M}{10} \times 55 \times 10^{-3} = 0.0055$ mol$\\$• Moles of NaOH $= \frac{M}{10} \times 45 \times 10^{-3} = 0.0045$ mol$\\$• Excess moles of HCl $= 0.0055 - 0.0045 = 0.001$ mol$\\$• Total volume $= 55 + 45 = 100$ mL$= 0.1$ L$\\ [$H$^+] = \frac{0.001}{0.1} = 0.01$ M$\\$pH$= -\log(0.01) = 2 \\$c. $75$ mL $\frac{M}{5}$ HCl $+ 25$ mL $\frac{M}{5}$ NaOH$\\$• Moles of HCl $= \frac{M}{5} \times 75 \times 10^{-3} = 0.015$ mol$\\$• Moles of NaOH $= \frac{M}{5} \times 25 \times 10^{-3} = 0.005$ mol$\\$• Excess moles of HCl $= 0.015 - 0.005 = 0.01$ mol$\\$• Total volume $= 75 + 25 = 100$ mL$= 0.1$ L$\\ [$H$^+] = \frac{0.01}{0.1} = 0.1$ M$\\$pH$= -\log(0.1) = 1 \\$d. $100$ mL $\frac{M}{10}$ HCl $+ 100$ mL $\frac{M}{10}$ NaOH$\\$• Moles of HCl $= \frac{M}{10} \times 100 \times 10^{-3} = 0.01$ mol$\\$• Moles of NaOH $= \frac{M}{10} \times 100 \times 10^{-3} = 0.01$ mol$\\$• No excess moles of HCl or NaOH, solution is neutral.$\\$pH$= 7 \\$Therefore, the solution in option c has a pH of 1.$\\$The correct option is 4.