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,

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

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

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

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

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