Question

What is the pH of the following solutions?
(a) ${10}^{-3}M$ $HCl$
(b) $0.0001M$ $NaOH$
(c) $0.0001M$ $H_2{SO}_4$

Answer 1

$HCl$ is a strong electrolyte and is completely ionised
$HCl\rightleftharpoons H^{+}+{Cl}^{-}$
So, $\left[H^{+}\right]={10}^{-3}M$
$pH=-\log \left[H^{+}\right]=-\log {10}^{-3}=3$
(b) $NaOH$ is a strong electrolyte and is completely ionised
$NaOH\rightleftharpoons {Na}^{+}+{OH}^{-}$
so, $\left[{OH}^{+}\right]=0.0001M={10}^{-4}M$
$pOH=-\log {10}^{-4}=4$
(c) $H_2{SO}_4$ is a strong electrolyte and is completely ionised
$H_2{SO}_4\rightleftharpoons {2H}^{+}+{SO}_4^{2-}$
One molecule of $H_2{SO}_4$ furnishes $2H^{+}$ ions
So, $\left[H^{+}\right]=2\times {10}^{-4}M$
$pH=-\log \left[H^{+}\right]$
$=3.70$