Solve this following


$0.01$ moles of a weak acid $\mathrm{HA}\left(\mathrm{K}_{\mathrm{a}}=2.0 \times 10^{-6}\right)$ is dissolved in $1.0 \mathrm{~L}$ of $0.1 \mathrm{M} \mathrm{HCl}$ solution. The degree of dissociation of HA is $\times 10^{-5}$ (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation $<<1$ ]


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

Initial conc. $0.01 \mathrm{M} \quad 0.1 \mathrm{M} \quad 0$

Equ. conc. $(0.01-\mathrm{x})(0.1+\mathrm{x}) \mathrm{xM}$

$\approx 0.01 \mathrm{M} \approx 0.1 \mathrm{M}$

Now, $\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{x}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]} \Rightarrow 2 \times 10^{-6}=\frac{0.1 \times \mathrm{x}}{0.01}$

$\therefore \quad x=2 \times 10^{-7}$

Now, $\alpha=\frac{\mathrm{x}}{0.01}=\frac{2 \times 10^{-7}}{0.01}=2 \times 10^{-5}$


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