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Q. pH of $ 1\times 10^{-8}M\,HNO_{3} $ solution will be:

Uttarkhand PMTUttarkhand PMT 2006

Solution:

$\left[H^{+}\right]$in $H N O_{3}=1 \times 10^{-8} M$
$\left[H^{+}\right]$in water
$=1 \times 10^{-7} M$
Total $\left[H^{+}\right]$in solution $=\left(1 \times 10^{-8}+1 \times 10^{-7}\right) M$
$=11 \times 10^{-8}$
$= pH =-\log \left[H^{+}\right]$
$=-\log \left[11 \times 10^{-8}\right]$
$=-\log 11+8 \log 10$
$=-1.0414+8$
$=6.9586$