Assuming complete dissociation, calculate the pH of the following solutions:

Solution:

(i) 0.003MHCl:

$\mathrm{H}_{2} \mathrm{O}+\mathrm{HCl} \longleftrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}$

Since HCl is completely ionized,

$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=[\mathrm{HCl}] .$

 

$\Rightarrow\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=0.003$

Now,

$\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log (.003)$

$=2.52$

Hence, the $\mathrm{pH}$ of the solution is $2.52$.

(ii) $0.005 \mathrm{MNaOH}$ :

$\mathrm{NaOH}_{(a q)} \longleftrightarrow \mathrm{Na}_{(a q)}^{+}+\mathrm{HO}_{(a q)}^{-}$

$\left[\mathrm{HO}^{-}\right]=[\mathrm{NaOH}]$

 

$\Rightarrow\left[\mathrm{HO}^{-}\right]=.005$

$\mathrm{pOH}=-\log \left[\mathrm{HO}^{-}\right]=-\log (.005)$

$\mathrm{pOH}=2.30$

$\therefore \mathrm{pH}=14-2.30$

$=11.70$

Hence, the $\mathrm{pH}$ of the solution is $11.70$.

(iii) $0.002 \mathrm{HBr}$ :

$\mathrm{HBr}+\mathrm{H}_{2} \mathrm{O} \longleftrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Br}^{-}$

$\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=[\mathrm{HBr}]$

$\Rightarrow\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=.002$

$\therefore \mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]$

$=-\log (0.002)$

$=2.69$

Hence, the pH of the solution is 2.69.

(iv) $0.002 \mathrm{M} \mathrm{KOH}$ :

$\mathrm{KOH}_{(a q)} \longleftrightarrow \mathrm{K}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-}$

$\left[\mathrm{OH}^{-}\right]=[\mathrm{KOH}]$

$\Rightarrow\left[\mathrm{OH}^{-}\right]=.002$

Now, pOH $=-\log \left[\mathrm{OH}^{-}\right]$

$=2.69$

$\therefore \mathrm{pH}=14-2.69$

$=11.31$

Hence, the $\mathrm{pH}$ of the solution is $11.31$.