**Answer: 3.3**

*If $\rho H = 5 H^+ = 10^{-\rho H} = 10^{-5}$*

*If $\rho H = 3 H^+ =10^{-3}$*

*Total $[H^+] = \frac{10^{-5} + 10^{-3}}{2}$ = $\frac{.01 \times 10^{-3} + 1 \times 10^{-3}}{2}$ **= $\frac{(1 + 0.01) \times 10^{-3}}{2}$*

*$\therefore [H^+] = \frac{1.01 \times 10^{-3}}{2} = 0.505 \times 10^{-3}$*

*$\rho H = - log (0.505 \times 10^{-3}) = 3.3$*