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

Question

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

Aicharya · Accepted Answer

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

What is the $\mathrm{pH}$ of a $2.6 \times 10^{-8} \mathrm{M~} \mathrm{H}^{+}$ion solution? $(\log 2.6=0…

Solution