What is the \(\mathrm{pH}\) of a \(2.6 \times 10^{-8} \mathrm{M} \mathrm{H}^{+}\)ion solution? \((\log 2.6=0.4150)\)

\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\)
As \(\left(\mathrm{H}^{+}\right)\)ion conc. Is less, [ \(\left.\mathrm{H}^{+}\right]\)obtained from \(\mathrm{H}_2 \mathrm{O}\) will be considered \(\left[\mathrm{H}^{+}\right]\)be considered
\(\left[\mathrm{H}^{+}\right]\)be considered
\(\left[\mathrm{H}^{+}\right]\)be considered
\(\begin{aligned} & {\left[\mathrm{H}^{+}\right]=2.6 \times 10^{-8}+1 \times 10^{-7}} \\ & =1.26 \times 10^{-7}\end{aligned}\)
\(=1.26 \times 10^{-7}\)
\(\begin{aligned} & \therefore \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & =\log \left(1.26 \times 10^{-7}\right) \\ & =7-\log 1.26 \\ & =7-0.1 \\ & \mathrm{pH}=6.9\end{aligned}\)