Question

Calculate the pH of the following solutions:
(i) $1.0\times {10}^{-8}M$ $HCl$
(ii) $1.0\times {10}^{-8}M$ $NaOH$

Answer 1

(i) The neutral water has $\left[H^{+}\right]=1\times {10}^{-7}M$
By adding $1.0\times {10}^{-8}M$ $HCl$ a concentration of $1.0\times {10}^{-8}M$ $H^{+}$ ions has increased in solution.
Thus, total $\left[H^{+}\right]=(1\times {10}^{-7}+1\times {10}^{-8})M=1.1\times {10}^{-7}M$
$pH=-\log (1.1\times {10}^{-7})=-[\log 1.1+\log {10}^{-7}]=6.9586$
(ii) The neutral water has $\left[{OH}^{-}\right]=(1\times {10}^{-7}M)$
By adding $1\times {10}^{-8}M$ $NAOH$, a concentration of $1\times {10}^{-8}M$ ${OH}^{-}$ ions has increased in solution.
Thus, total $\left[{OH}^{-}\right]=(1\times {10}^{-7}+1\times {10}^{-8})M=1.1\times {10}^{-7}M$
$pOH=-\log 1.1\times {10}^{-7}=6.9586$
$pH=(14-pOH)=14-6.9586)=7.0414$