Question

$0.01$ moles of a weak acid $\mathrm{HA}(\mathrm{Ka}=2.0\mathrm{x}{10}^{-6})$ is dissolved in $1.0\mathrm{L}\mathrm{of}0.1\mathrm{M}\mathrm{HCl}$ solution. The degree of dissociation of $\mathrm{HA}$ is_______ $\mathrm{x}{10}^{-5}$ (Round off to the Nearest Integer). Assume the degree of dissociation $<<1$

Answer 1

Step $1$: Dissociation of weak and strong acid:

• The weak acid will dissociate as

$\mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-}$

• The strong acid will dissociate as

$\mathrm{HCl}\to {\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}$

The actual formula is $\mathrm{Ka}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

Step $2$: Finding the concentration terms:

$$\begin{gathered} \mathrm{C}=\frac{\mathrm{n}}{\mathrm{V}} \\ \mathrm{V}=1\mathrm{L}\mathrm{of}\mathrm{HCl} \\ \mathrm{so},\mathrm{C}=\mathrm{n} \end{gathered}$$

so, $$\begin{gathered} \mathrm{HCl}\to {\mathrm{H}}^{+}+{\mathrm{Cl}}^{-} \\ 0.010.010.01 \end{gathered}$$

So, the concentration of reactant and product at initial and at equilibrium concentration.

$$\begin{gathered} \mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{-} \\ \mathrm{C}0.010.10 \\ {\mathrm{C}}_{\mathrm{eq}}0.01(1-\mathrm{\alpha })0.1+0.01\mathrm{\alpha }0.01\mathrm{\alpha } \\ \approx 0.01\approx 0.1 \end{gathered}$$

Step $3$: Finding the dissociation constant value

On substituting the values and on rearranging the equation we get,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ =\frac{[0.1][0.01\alpha ]}{0.01} \\ 2\mathrm{x}{10}^{-6}=0.1\mathrm{x}\mathrm{\alpha } \\ 2\mathrm{x}{10}^{-6}={10}^{-1}\mathrm{x}\mathrm{\alpha } \\ \mathrm{\alpha }=\frac{2\mathrm{x}{10}^{-6}}{{10}^{-1}} \\ \mathrm{\alpha }=2\mathrm{x}{10}^{-5} \end{gathered}$$

Hence, the value of the degree of dissociation of $\mathrm{HA}$ is $2\times {10}^{-5}$.