The concentration of $H^+$ ions can be calculated from the pH values: $[H^+] = 10^{-pH}$ For the first solution (pH = 3): $[H^+]_1 = 10^{-3}\text{ M}$ For the second solution (pH = 4): $[H^+]_2 = 10^{-4}\text{ M}$ For the third solution (pH = 5): $[H^+]_3 = 10^{-5}\text{ M}$ Let the volume of each solution be $V$. Total moles of $H^+$ in the mixture = $(10^{-3} + 10^{-4} + 10^{-5}) \times V = (100 \times 10^{-5} + 10 \times 10^{-5} + 1 \times 10^{-5}) \times V = 111 \times 10^{-5} \times V = 1.11 \times 10^{-3} \times V$. Total volume of the mixture = $V + V + V = 3V$. The final $H^+$ ion concentration = $\frac{\text{Total moles of } H^+}{\text{Total volume}} = \frac{1.11 \times 10^{-3} \times V}{3V} = 0.37 \times 10^{-3}\text{ M} = 3.7 \times 10^{-4}\text{ M}$.