Question

The $pH$ of $1.0\times {10}^{-8}M$ $HCl$ is:
(Given, $\log \left(1.1\right)=0.04$)

• A. $8$
• B. $6$
• C. $1$
• D. $6.96$

Answer 1

The correct option is D $6.96$

$HCl$ is a strong acid hence $pH$ of the solution can't be more than $7$.
${10}^{-8}\text{M}$ solution of $HCl$ means the solution is very very dilute. Hence, concentration of $H^{+}$ will be taken from both $H_{2}O$ and $HCl$.
$\left[H^{+}\right]={10}^{-7}\left(H_{2}O\right)+{10}^{-8}\left(HCl\right)$
$\left[H^{+}\right]={10}^{-7}[1+{10}^{-1}]$
$={10}^{-7}\left[1.1\right]$
$pH$ $=-\log \left[H^{+}\right]=-\log {10}^{-7}+(-\log [1.1\left]\right)=7-0.04$
$\Rightarrow pH=6.96$
$pH$ of ${10}^{-8}MHCl$ solution will be $6.96$.