Question

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

(i) 0.003 M HCl

(ii) 0.005 M NaOH

(iii) 0.002 M HBr

(iv) 0.002 M KOH

Answer 1

(i) 0.003MHCl:
$H_{2}O+HCl\leftrightarrow H_3{O}^{+}+C{l}^{-}$
Since HCl is completely ionized,
$\left[H_3{O}^{+}\right]=\left[HCl\right].\Rightarrow \left[H_3{O}^{+}\right]=0.003$
Now,
$pH=-lgo\left[H_3{O}^{+}\right]=-log\left(.003\right)=2.52$

(ii) 0.005MNaOH:

$NaO{H}_{\left(aq\right)}\leftrightarrow N{a}_{aq}^{+}+H{O}_{\left(aq\right)}^{-}\left[H{O}^{-}\right]=\left[NaOH\right]\left[H{O}^{-}\right]=\left[NaOH\right]\Rightarrow \left[H{O}^{-}\right]=.005pOH=-log\left[H{O}^{-}\right]=-log\left(.005\right)pOH=2.30\therefore pH=14-2.30=11.70$
Hence, the pH of the solution is 11.70.

(iii)

0.002 HBr:

$HBr+H_{2}O\leftrightarrow H_3{O}^{+}+B{r}^{-}\left[H_3{O}^{+}\right]=\left[HBr\right]\Rightarrow \left[H_3{O}^{+}\right]=.002\therefore pH=-log\left[H_O^{+}\right]=-log\left(.002\right)=2.69$
Hence, the pH of the solution is 2.69.

(iv)

0.002 M KOH

$KO{H}_{\left(aq\right)}\leftrightarrow K_{\left(aq\right)}^{+}+O{H}_{\left(aq\right)}^{-}\left[O{H}^{-}\right]=\left[KOH\right]\Rightarrow \left[O{H}^{-}\right]=.002$
Now, $pOH=-log\left[O{H}^{-}\right]=2.69\therefore pH=14-2.69=11.31$
Hence, the pH of the solution is 11.31.