Question

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

• A. 8
• B. -8
• C. between 7 and 8
• D. between 6 and 7

Answer 1

The correct option is D between 6 and 7

(i) pH of acid cannot be more than 7.
(ii) When the concentration of $\left[H^{+}\right]$ is low, calculation of pH should include the contribution $\left[H^{+}\right]$ from water also.
Considering the given molar concentration of $\left[H^{+}\right]$ from acid only, $pH=8$. But this is possible as pH of an acidic solution can not be more than 7.
Let us include the concentration of $\left[H^{+}\right]$ coming from $H_{2}O$ also.
Total $\left[H^{+}\right]=[H^{+}{]}_{HCl}+[H^{+}{]}_{H_{2}O}$
Water ionizes as :$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$
Let $x$ be the conc. of $\left[H^{+}\right]$ from $H_{2}O$
$[\because [H^{+}]=[OH{]}^{-}\text{in water}$
$\therefore \left[H^{+}\right]=x=\left[O{H}^{-}\right]$
But, $K_w$ =${10}^{-14}$
$\Rightarrow \left[H^{+}\right]\left[O{H}^{-}\right]=[{10}^{-8}+x]\left[x\right]$
Solving this quardratic equation for x,
x = $9.5\times {10}^{-8}$
$\text{Total}\left[H^{+}\right]={10}^{-8}+9.5\times {10}^{-8}=10.5\times {10}^{-8}pH=-\log (10.5\times {10}^{-8})=6.98$
It is between 6 and 7.