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  4. Calculate the pH of 10-8 M HCl .( log 11=1.0414)

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Q. Calculate the $pH$ of $10^{-8} M HCl .(\log 11=1.0414)$

Equilibrium

Solution:

image
$\left[ H ^{+}\right]_{\text {total }} =\left(10^{-8}+10^{-7}\right) M =10^{-8}(1+10) $
$=11 \times 10^{-8}$ (Neglecting common -ion effect )
$pH =\log \left( H ^{+}\right)=-\log \left(11 \times 10^{-8}\right)$
$=6.96=7$