Question

Calculate the pH of $0.08M$ solution of hypochlorous acid,
$HOCl$. The ionization constant of the acid is $2.5\times {10}^{-5}$. Determine the percent dissociation of $HOCl$.

Answer 1

Given :
$\left[HOCl\right]=0.08m$
$K_a=2.5\times {10}^{-5}$

Step 1: Ionization reaction
Ionization of $\text{HOCl}$ can be written as-
$HOCl\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+Cl{O}^{-}\left(aq\right)$

Step :2 Equilibrium concentration
For the given reaction , equilibrium concentration can be calculated as:


$\begin{array}{cccccc}& HOCl\left(aq\right)& +H_{2}O\left(l\right)& \rightleftharpoons & H_3{O}^{+}\left(aq\right)& +Cl{O}^{-}\left(aq\right)\\ \text{Initial concentration}& 0.08& & & 0& 0\\ \text{Equilibrium concentration}& 0.08-x& & & x& x\end{array}$


Step :3 Ionization constant

$K_a=\frac{\left[H_3{O}^{+}\right]\left[Cl{o}^{-}\right]}{\left[HOCl\right]}$
$=\frac{x^2}{0.08-x)}$

Since $x<<0.08$, therefore $0.08-x\simeq 0.08$

So, $\frac{x^2}{0.08}=2.5\times {10}^{-5}$
$\Rightarrow x^2=2.0\times {10}^{-6}$
$x=1.41\times {10}^{-3}$

Therefore,
$\left[H^{+}\right]=1.41\times {10}^{-3}M$

Step $4$ pH calculation
We know,
$pH=-\log \left[H^{+}\right]=-\log (1.41\times {10}^{-3})=2.85$

Step $5$ Percent dissociation
$\%\text{dissociation}=\frac{\text{Amount dissociated}}{\text{Amount taken}}\times 100$

For given reaction,
$\%\text{dissociation}=\frac{[HOCl{]}_{\text{dissociated}}}{[HOCl{]}_{\text{initial}}}\times 100$
$=1.41\times {10}^3\times \frac{{10}^2}{0.08}=1.76\%$

Final answer:
$pH=2.85$
$\%\text{dissociation}=1.76\%$