Question

What is pH of $2\times 10^{-8}$ molar HCl solution? Here log2 $=0.301$ and log3 $=0.477$

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

Answer 1

Answer: (C)

As solution is very dilute,

$\therefore \left[H^{+}\right]={\left[H^{+}\right]}_{\text{water }}+{\left[H^{+}\right]}_{HCl}$

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

$\therefore pH=-\log \left[12\times 10^{-8}\right]$

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

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