Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\because$pH = 6

$\therefore \left[{\mathrm{H}}^{+}\right]={10}^{-6}\mathrm{ }\mathrm{M}$

After dilution$\left[{\mathrm{H}}^{+}\right]=\frac{{10}^{-6}}{1000}={10}^{-9}\mathrm{ }\mathrm{M}$

$\therefore \left[{\text{H}}^{+}\right]$from${\text{H}}_2\text{O}$cannot be neglected.

Total$\left[{\mathrm{H}}^{+}\right]={10}^{-9}+{10}^{-7}$(neglect common ion effect on${\text{H}}_2\text{O}$)

$={10}^{-7}\left({10}^{-2}+1\right)$

$={10}^{-7}\left(1.01\right)$

$\mathrm{p}\mathrm{H}=\mathrm{ }-\log \left(1.01\times {10}^{-7}\right)$

$=7-0.0043$

$=6.9957$