Chemistry - Ionic Equilibrium Question with Solution | TestHub

At pH$=3,\left[{\mathrm{H}}^{+}\right]=10^{-3}\mathrm{M}$

At pH$=4,\left[{\mathrm{H}}^{+}\right]=10^{-4}\mathrm{M}$

When in equal volume of the two solutions are mixed, the$\left[{\mathrm{H}}^{+}\right]$

$=\frac{10^{-3}+10^{-4}}{2}=\frac{10^{-3}\left[1+0.1\right]}{2}=\frac{1.1}{2}\times 10^{-3}$

$\left[{\mathrm{H}}^{+}\right]=5.5\times 10^{-4}$

$-\text{log}\left[{\mathrm{H}}^{+}\right]=-\text{log}\left(5.5\right)-\text{log1}{0}^{-4}$

pH$=-0.\text{74}04+4=3.\text{26}$