The $\mathrm{pH}$ of a $10^{-8}$ molar solution of $\mathrm{HCl}$ in water is

The $\mathrm{pH}$ of a $10^{-8}$ molar solution of $\mathrm{HCl}$ in water is
  1. 8
  2. between $7 \& 8$
  3. between $6 \& 7$
  4. none of these

Solution

$\mathrm{pH}=-\log \left(\mathrm{H}^{+}ight)=-\log 10^{-8}=8$
As the solution is acidic, $\mathrm{pH} < 7$. This is because $\left[\mathrm{H}^{+}ight]$from $\mathrm{H} 20\left[10^{-7}ight]$ cannot be neglected in comparison to $10^{-8}$.
It is not possible for acid, so it is $\left[\mathrm{H}^{+}ight]$, the $\left[\mathrm{H}^{+}ight]$of water is also added
Total $\left[\mathrm{H}^{+}ight]$in solution
$=\left[\mathrm{H}^{+}ight]$of $\mathrm{HCl}+\left[\mathrm{H}^{+}ight]$of water
$=\left(1 \times 10^{-8}+1 \times 10^{-7}ight) \mathrm{M}$
$=(1+10) \times 10^{-8}=11 \times 10^{-8} \mathrm{M}$
$\therefore \mathrm{pH}=-\log \left[\mathrm{H}^{+}ight]=-\log 11 \times 10^{-8}$
$=-\log 11+8 \log 10=-1.0414+8=6.9586$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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