Chemistry - Ionic Equilibrium Question with Solution | TestHub

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

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

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

$=\frac{10^{-3}+10^{-4}}{2}$

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

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

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

$\mathrm{pH}=-0.7404+4=$3.26