Question

What is pH of $2\times {10}^{-8}$ M $HCl$ solution?

(Here, log $2=0.301$ and log $3=0.477$)

• A. $6.92$
• B. $9.5$
• C. $5.4$
• D. $7.7$

Answer 1

The correct option is A $6.92$

As solution is very dilute,
$\therefore \left[H^{+}\right]=[H^{+}{]}_{water}+[H^{+}{]}_{HCl}$

$={10}^{-7}+2\times {10}^{-8}=12\times {10}^{-8}$

$\therefore pH=-log[12\times {10}^{-18}]$

$=-log[2^2\times 3\times {10}^{-8}]$

$=-\left[2log\right(2)+log(3)+log({10}^{-8})=6.92$.

Option A is correct.