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Q. Which of the following solution(s) have $pH$ between $6$ and $7 ?$
1. $2 \times 10^{-6} M\, NaOH$
2. $2 \times 10^{-6} M \,HCl$
3. $10^{-8} M\, HCl$
4. $10^{-13} M \,NaOH$

Equilibrium

Solution:

$pH$ of $10^{-8} M HCl$ is not $8$ but it is less than $7$ because in this case contribution of $H ^{+}$from water is not neglected.
Total $H ^{+}=10^{-8}$ (from acid) $+10^{-7}$ (from water)
$=10^{-8}(1+10)$
$=11 \times 10^{-8} \,M$
$pH =-\log \left[ H ^{+}\right]=-\log \left[11 \times 10^{-8}\right]$
$=-\left[\log 11+\log 10^{-8}\right]$
$=-[1.0414-8]$
$=6.9586 \simeq 6.96$